Ant yon antye n ki pi gran ke 1 e doub antye sila a, toujou gen yon nonm premye.
se yon teyorèm ki te konjektire pa Bertrand e ki demontre an 1850 pou lapremyè fwa pa matematisyen ris Tchebycheff
Gen plizyè varyant pou demontre teyorèm oubyen postila a. Men de ladan yo ki privilejye apwòch elemantè.
Premyè demach la ap mennen nou redemontre tankou yon lèm ke
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{\displaystyle {\frac {2^{2n}}{2{\sqrt {n}}}}<N<{\frac {2^{2n}}{\sqrt {2n}}}}
N
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{\displaystyle N=C_{2n}^{n}={{2n} \choose n}}
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{\displaystyle N={\frac {\left(\left(2n\right)!\right)}{\left(n!\right)\times \left(2n-n\right)!}}}
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{\displaystyle N={\frac {\left(\left(2n\right)!\right)}{\left(n!\right)\times \left(n\right)!}}}
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{\displaystyle N={\frac {\left(\left(2n\right)!\right)}{\left(n!\right)^{2}}}}
Nou pale de redemontre paske se yon demonstrasyon ki te la deja ....
An nou konsidere pwodui tout antye enpè yo divize pa pwodui tout antye pè yo ki enferyè oubyen egal ak 2n
Konfòmeman ak Arnel Mercier ak Jean Marie de Koninck an desiye pwodui sa pa P
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{\displaystyle P={\frac {1\times \times 3\times ...\times \left(2n-3\right)\times \left(2n-1\right)}{2\times 4\times ...\times \left(2n-2\right)\times 2n}}}
An nou fè parèt anlè ya pwodui ki anba a
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{\displaystyle P={\frac {1\times \times 3\times ...\times \left(2n-3\right)\times \left(2n-1\right)}{2\times 4\times ...\times \left(2n-2\right)\times 2n}}\times {\frac {2\times 4\times 6\times ...\times \left(2n-2\right)\times 2n}{2\times 4\times 6\times ...\times \left(2n-2\right)\times 2n}}}
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{\displaystyle P={\frac {\left(1\times \times 3\times ...\times \left(2n-3\right)\times \left(2n-1\right)\right)\times \left(2\times 4\times 6\times ...\times \left(2n-2\right)\times 2n\right)}{2\times 4\times ...\times \left(2n-2\right)\times 2n\times \left(2\times 4\times 6\times ...\times \left(2n-2\right)\times 2n\right)}}}
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{\displaystyle P={\frac {\left(2n\right)!}{\left(2\times 4\times 6\times ...\times \left(2n-2\right)\times 2n\right)^{2}}}}
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{\displaystyle P={\frac {\left(2n\right)!}{{\left(2\times 1\times 2\times 2\times 2\times 3\times ...\times 2\times \left(n-1\right)\times 2\times n\right)}^{2}}}}
P
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{\displaystyle P={\frac {\left(2n\right)!}{{\left(2^{n}\times n!\right)}^{2}}}}
P
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{\displaystyle P={\frac {\left(2n\right)!}{{\left(2^{n}\right)}^{2}\times {\left(n!\right)}^{2}}}}
P
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{\displaystyle P={\frac {\left(2n\right)!}{2^{2n}\times {\left(n!\right)}^{2}}}}
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{\displaystyle P\times {2^{2n}}={\frac {\left(2n\right)!}{2^{2n}\times {\left(n!\right)}^{2}}}\times {2^{2n}}}
P
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{\displaystyle P\times {2^{2n}}={\frac {\left(2n\right)!}{{\left(n!\right)}^{2}}}}
P
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{\displaystyle P\times {2^{2n}}=N}
An nou konsidere
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{\displaystyle \prod _{i=1}^{n}{\left(1-{\frac {1}{{\left(2i\right)}^{2}}}\right)}{}}
si nou devlope pwodui nap genyen
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{\displaystyle \prod _{i=1}^{n}{\left(1-{\frac {1}{{\left(2i\right)}^{2}}}\right)}=\left(1-{\frac {1}{2^{2}}}\right)\times \left(1-{\frac {1}{4^{2}}}\right)\times \left(1-{\frac {1}{6^{2}}}\right)\times ...\times \left(1-{\frac {1}{{\left(2n-2\right)}^{2}}}\right)\times \left(1-{\frac {1}{{\left(2n\right)}^{2}}}\right)}
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{\displaystyle \prod _{i=1}^{n}{\left(1-{\frac {1}{{\left(2i\right)}^{2}}}\right)}{<1}}
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{\displaystyle \prod _{i=1}^{n}{\left({\frac {{\left(4i\right)}^{2}-1}{{\left(2i\right)}^{2}}}\right)}{<1}}
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{\displaystyle \prod _{i=1}^{n}{\left({\frac {\left(2i-1\right)\left(2i+1\right)}{{\left(2i\right)}^{2}}}\right)}{<1}}
sa ki bay
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{\displaystyle \left({\frac {1\times 3}{2^{2}}}\right)\times \left({\frac {3\times 5}{4^{2}}}\right)\times \left({\frac {5\times 7}{6^{2}}}\right)\times ...\times \left({\frac {\left(2n-3\right)\times \left(2n-1\right)}{{\left(2n-2\right)}^{2}}}\right)\times \left({\frac {\left(2n-1\right)\times \left(2n+1\right)}{{\left(2n\right)}^{2}}}\right){<1}}
lè nou diseke nimeratè yo ak denominatè yo :
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{\displaystyle \left({\frac {1}{2}}\times {\frac {3}{4}}\times {\frac {5}{6}}\times ...\times {\frac {2n-3}{2n-2}}\times {\frac {2n-1}{2n}}\right)\times \left({\frac {3}{2}}\times {\frac {5}{4}}\times {\frac {7}{6}}\times ...\times {\frac {2n-1}{2n-2}}\times {\frac {2n+1}{2n}}\right){<1}}
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{\displaystyle {\frac {1\times 3\times 5\times ...\times \left(2n-3\right)\times \left(2n-1\right)}{2\times 4\times 6\times ...\times \left(2n-2\right)\times 2n}}\times {\frac {3\times 5\times ...\times \left(2n-1\right)\times \left(2n+1\right)}{2\times 4\times 6\times ...\times \left(2n-2\right)\times 2n}}{<1}}
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{\displaystyle {\frac {1\times 3\times 5\times ...\times \left(2n-3\right)\times \left(2n-1\right)}{2\times 4\times 6\times ...\times \left(2n-2\right)\times 2n}}\times {\frac {1\times 3\times 5\times ...\times \left(2n-3\right)\times \left(2n-1\right)}{2\times 4\times 6\times ...\times \left(2n-2\right)\times 2n}}\times \left(2n+1\right){<1}}
P
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{\displaystyle P\times P\times \left(2n+1\right){<1}}
P
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{\displaystyle P^{2}\times \left(2n+1\right){<1}}
P
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{\displaystyle P^{2}\times 2n{<}P^{2}\times \left(2n+1\right){<1}}
kidonk
P
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{\displaystyle {P^{2}\times 2n}{<}1}
Nou te wè ke
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{\displaystyle 2^{2n}P=N}
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{\displaystyle {\left(2^{2n}\right)}^{2}P^{2}=N^{2}}
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{\displaystyle {\left(2^{2n}\right)}^{2}P^{2}\times 2n{<}{\left(2^{2n}\right)}^{2}}
N
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{\displaystyle N^{2}\times {2n}<{\left(2^{2n}\right)}^{2}}
N
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{\displaystyle N^{2}<{\frac {{\left(2^{2n}\right)}^{2}}{2n}}}
Fonksyon rasin kare kwasan, nou genyen :
N
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{\displaystyle N{<}{\sqrt {\frac {{\left(2^{2n}\right)}^{2}}{2n}}}}
N
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{\displaystyle N<{\frac {2^{2n}}{\sqrt {2n}}}}
Inegalite dwat la demontre
li pa fè answa apèl a oken nosyon nonm premye jiska prezan sinon ke faktoryèl la se nosyon de predileksyon nonm premye
Kounye an nou redemontre dezyèm pati inegalite, inegalite gòch la.
An nou konsidere pwodui sa :
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{\displaystyle \prod _{i=2}^{n}{\left(1-{\frac {1}{{\left(2i-1\right)}^{2}}}\right)}}
chak faktè yo pozitif e enferyè ak 1, kidonk pwodui ya globalman enferyè ak 1.
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<
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{\displaystyle \prod _{i=2}^{n}{\left(1-{\frac {1}{{\left(2i-1\right)}^{2}}}\right)}{<1}}
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{\displaystyle \prod _{i=2}^{n}{\left({\frac {{\left(2i-1\right)}^{2}-1}{{\left(2i-1\right)}^{2}}}\right)}{<1}}
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{\displaystyle \prod _{i=2}^{n}{\left({\frac {\left(2i-1-1\right)\left(2i-1+1\right)}{{\left(2i-1\right)}^{2}}}\right)}{<1}}
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{\displaystyle \prod _{i=2}^{n}{\left({\frac {\left(2i-2\right)\times 2i}{{\left(2i-1\right)}^{2}}}\right)}<1}
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{\displaystyle {\prod _{i=2}^{n}{\left({\frac {2i-2}{2i-1}}\right)}}\times {\prod _{i=2}^{n}{\left({\frac {2i}{2i-1}}\right)}}{<1}}
sa ki bay
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{\displaystyle \left({\frac {2}{3}}\times {\frac {4}{5}}\times ...\times {\frac {2n-4}{2n-3}}\times {\frac {2n-2}{2n-1}}\right)\times \left({\frac {4}{3}}\times {\frac {6}{5}}\times ...\times {\frac {2n-2}{2n-3}}\times {\frac {2n}{2n-1}}\right)<1}
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{\displaystyle {\frac {2\times 4\times ...\times \left(2n-4\right)\times \left(2n-2\right)}{3\times 5\times ...\times \left(2n-3\right)\times \left(2n-1\right)}}\times {\frac {4\times 6\times ...\times \left(2n-2\right)\times 2n}{3\times 5\times ...\times \left(2n-3\right)\times \left(2n-1\right)}}{<1}}
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<
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{\displaystyle {\frac {2\times 4\times ...\times \left(2n-4\right)\times \left(2n-2\right)}{3\times 5\times ...\times \left(2n-3\right)\times \left(2n-1\right)}}\times 2n\times 2\times {\frac {4\times 6\times ...\times \left(2n-2\right)\times 2n}{3\times 5\times ...\times \left(2n-3\right)\times \left(2n-1\right)}}{<2\times 2n}}
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{\displaystyle {\frac {2\times 4\times ...\times \left(2n-2\right)\times 2n}{3\times 5\times ...\times \left(2n-3\right)\times \left(2n-1\right)}}\times {\frac {2\times 4\times ...\times \left(2n-2\right)\times 2n}{3\times 5\times ...\times \left(2n-3\right)\times \left(2n-1\right)}}{<4n}}
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{\displaystyle \left({\frac {2\times 4\times ...\times \left(2n-2\right)\times 2n}{1\times 3\times 5\times ...\times \left(2n-3\right)\times \left(2n-1\right)}}\right)^{2}{<4n}}
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{\displaystyle {\left({\frac {1}{\frac {1\times 3\times 5\times ...\times \left(2n-3\right)\times \left(2n-1\right)}{2\times 4\times ...\times \left(2n-2\right)\times 2n}}}\right)}^{2}{<4n}}
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{\displaystyle {\left({\frac {1}{P}}\right)}^{2}{<4n}}
1
P
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{\displaystyle {\frac {1}{P^{2}}}{<4n}}
1
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P
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>
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{\displaystyle {\frac {1}{\frac {1}{P^{2}}}}{>}{\frac {1}{4n}}}
p
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n
{\displaystyle p^{2}>{\frac {1}{4n}}}
nou genyen
2
2
n
×
P
=
N
{\displaystyle {2^{2n}\times P}=N}
P
=
N
2
2
n
{\displaystyle P={\frac {N}{2^{2n}}}}
(
N
2
2
n
)
2
>
1
4
n
{\displaystyle {\left({\frac {N}{2^{2n}}}\right)}^{2}{>}{\frac {1}{4n}}}
(
N
2
2
n
)
>
1
4
n
{\displaystyle \left({\frac {N}{2^{2n}}}\right){>}{\sqrt {\frac {1}{4n}}}}
(
N
2
2
n
)
>
1
4
n
{\displaystyle \left({\frac {N}{2^{2n}}}\right){>}{\frac {1}{\sqrt {4n}}}}
(
N
2
2
n
)
>
1
2
n
{\displaystyle \left({\frac {N}{2^{2n}}}\right){>}{\frac {1}{2{\sqrt {n}}}}}
(
N
2
2
n
)
×
2
2
n
>
1
2
n
×
2
2
n
{\displaystyle {\left({\frac {N}{2^{2n}}}\right)}\times {2^{2n}}{>}{\frac {1}{2{\sqrt {n}}}}\times {2^{2n}}}
N
>
2
2
n
2
n
{\displaystyle N{>}{\frac {2^{2n}}{2{\sqrt {n}}}}}
2
2
n
2
n
<
N
{\displaystyle {\frac {2^{2n}}{2{\sqrt {n}}}}{<}N}
Inegalite goch la demontre
kidonk lè nou konbine inegalite goch la ak inegalite dwat la nou gen premye lèm
2
2
n
2
n
<
N
<
2
2
n
2
n
{\displaystyle {\frac {2^{2n}}{2{\sqrt {n}}}}<N<{\frac {2^{2n}}{\sqrt {2n}}}}
An nou demontre yon dezyèm pwopozisyon sa ki va sèvi kòm lèm
θ
(
n
)
<
2
n
L
n
2
{\displaystyle \theta \left(n\right){<}2nLn2}
An soulye ke fonksyon
θ
{\displaystyle \theta }
θ
(
x
)
=
∑
p
≤
x
L
n
p
{\displaystyle \theta \left(x\right)=\sum _{p\leq x}^{}{Lnp}}
x
≤
0
{\displaystyle x\leq 0}
Nan kad sa a nou sipoze ke n
∈
N
∗
{\displaystyle \in \mathbb {N} ^{*}}
θ
(
1
)
=
0
<
2
×
1
×
L
n
2
{\displaystyle \theta \left(1\right)=0<2\times 1\times Ln2}
θ
(
2
)
=
L
n
2
<
2
×
2
×
L
n
2
{\displaystyle \theta \left(2\right)=Ln2{<}2\times 2\times Ln2}
An nou sipoze li vrè pou yon sèten
n
>
2
{\displaystyle n>2}
N
2
{\displaystyle {\frac {N}{2}}}
N
2
=
1
2
N
=
1
2
(
2
n
n
)
=
(
2
n
)
!
2
×
(
n
!
)
×
n
!
=
(
(
2
n
−
1
)
!
)
×
2
n
2
×
(
n
!
)
×
n
!
=
(
(
2
n
−
1
)
!
)
×
n
(
n
!
)
×
(
(
n
−
1
)
!
)
×
n
=
(
2
n
−
1
)
!
(
(
n
−
1
)
!
)
×
n
!
=
(
2
n
−
1
)
(
(
n
−
1
)
!
)
×
(
(
2
n
−
1
)
−
(
n
−
1
)
)
!
=
C
2
n
−
1
n
−
1
=
(
2
n
−
1
n
−
1
)
{\displaystyle {\frac {N}{2}}={\frac {1}{2}}N={\frac {1}{2}}{{2n} \choose n}={\frac {\left(2n\right)!}{2\times {\left(n!\right)}\times {n!}}}={\frac {\left(\left(2n-1\right)!\right)\times 2n}{2\times \left(n!\right)\times n!}}={\frac {\left(\left(2n-1\right)!\right)\times n}{\left(n!\right)\times {\left(\left(n-1\right)!\right)}\times n}}={\frac {\left(2n-1\right)!}{\left(\left(n-1\right)!\right)\times n!}}={\frac {\left(2n-1\right)}{\left(\left(n-1\right)!\right)\times \left(\left(2n-1\right)-\left(n-1\right)\right)!}}=C_{2n-1}^{n-1}={{2n-1} \choose {n-1}}}
An nou gade byen nimeratè a ak denominatè a
(
2
n
−
1
)
!
(
n
!
)
×
(
(
n
−
1
)
!
)
=
1
×
2
×
3
×
.
.
.
×
(
2
n
−
2
)
×
(
2
n
−
1
)
(
1
×
2
×
3
×
.
.
.
×
(
n
−
1
)
×
n
)
×
(
1
×
2
×
3
×
.
.
.
×
(
n
−
2
)
×
(
n
−
1
)
)
{\displaystyle {\frac {\left(2n-1\right)!}{\left(n!\right)\times \left(\left(n-1\right)!\right)}}={\frac {1\times 2\times 3\times ...\times \left(2n-2\right)\times \left(2n-1\right)}{\left(1\times 2\times 3\times ...\times {\left(n-1\right)}\times n\right)\times {\left(1\times 2\times 3\times ...\times \left(n-2\right)\times \left(n-1\right)\right)}}}}
Menm konsiderasyon an ka fèt pou denominatè a jiska n .
An nou konsidere nonm premye p ki siperyè ak n e ki enferyè oubyen egal ak 2n-1.
n
<
p
≤
2
n
−
1
{\displaystyle n{<p}\leq {2n-1}}
p
∤
(
(
n
!
)
×
(
(
n
−
1
)
!
)
)
{\displaystyle p\nmid {\left(\left(n!\right)\times \left(\left(n-1\right)!\right)\right)}}
p
∣
(
2
n
−
1
)
!
{\displaystyle p\mid \left(2n-1\right)!}
kidonk yon faktè premye p ki tèl ke
n
<
p
≤
2
n
−
1
{\displaystyle n<p\leq {2n-1}}
N
2
{\displaystyle {\frac {N}{2}}}
n
<
p
≤
2
n
−
1
{\displaystyle n<p\leq {2n-1}}
N
2
{\displaystyle {\frac {N}{2}}}
N
2
≥
∏
n
<
p
≤
2
n
−
1
p
{\displaystyle {\frac {N}{2}}\geq {\prod _{n<p\leq {2n-1}}{p}}}
L
n
N
2
≥
L
n
∏
n
<
p
≤
2
n
−
1
p
{\displaystyle {Ln{\frac {N}{2}}}\geq {Ln{\prod _{n<p\leq {2n-1}}{p}}}}
L
n
N
2
≥
∑
n
<
p
≤
2
n
−
1
L
n
p
{\displaystyle {Ln{\frac {N}{2}}}\geq {\sum _{n<p\leq {2n-1}}{Lnp}}}
L
n
N
2
≥
∑
1
≤
p
≤
n
L
n
p
−
∑
1
≤
p
≤
n
L
n
p
+
∑
n
<
p
≤
2
n
−
1
L
n
p
{\displaystyle {Ln{\frac {N}{2}}}\geq {\sum _{1\leq {p}\leq {n}}{Lnp}}-{\sum _{1\leq {p}\leq {n}}{Lnp}}+{\sum _{n{<p}\leq {2n-1}}{Lnp}}}
L
n
N
2
≥
(
∑
1
≤
p
≤
n
L
n
p
+
∑
n
<
p
≤
2
n
−
1
L
n
p
)
−
∑
1
≤
p
≤
n
L
n
p
{\displaystyle {Ln{\frac {N}{2}}}\geq \left({\sum _{1\leq {p}\leq {n}}{Lnp}}+{\sum _{n{<p}\leq {2n-1}}{Lnp}}\right)-{\sum _{1\leq {p}\leq {n}}{Lnp}}}
L
n
N
2
≥
∑
1
≤
p
≤
2
n
−
1
L
n
p
−
∑
1
≤
p
≤
n
L
n
p
{\displaystyle {Ln{\frac {N}{2}}}\geq {\sum _{1\leq {p}\leq {2n-1}}{Lnp}}-{\sum _{1\leq {p}\leq {n}}{Lnp}}}
L
n
N
2
≥
θ
(
2
n
−
1
)
−
θ
(
n
)
{\displaystyle {Ln{\frac {N}{2}}}\geq \theta \left(2n-1\right)-\theta \left(n\right)}
θ
(
2
n
−
1
)
−
θ
(
n
)
≤
L
n
N
2
{\displaystyle \theta \left(2n-1\right)-\theta \left(n\right)\leq {Ln{\frac {N}{2}}}}
An nou konsidere inegalite dwat premye lèn lan oubyen dezyèm inegalite premye lèm lan
N
<
2
2
n
2
n
{\displaystyle {N}{<}{\frac {2^{2n}}{\sqrt {2n}}}}
nou kapab ekri li
N
<
2
2
n
(
2
n
)
1
2
{\displaystyle {N}<{\frac {2^{2n}}{{\left(2n\right)}^{\frac {1}{2}}}}}
Kòm se yon fonksyon kwasant, nap genyen
L
n
N
<
2
2
n
(
2
n
)
1
2
{\displaystyle LnN<{\frac {2^{2n}}{{\left(2n\right)}^{\frac {1}{2}}}}}
L
n
N
<
L
n
2
2
n
−
L
n
(
2
n
)
1
2
{\displaystyle LnN<{Ln2^{2n}}-Ln{\left(2n\right)}^{\frac {1}{2}}}
L
n
N
<
2
n
L
n
2
−
1
2
L
n
2
n
{\displaystyle LnN{<}2nLn2-{\frac {1}{2}}Ln2n}
An nou konbine
{
θ
(
2
n
−
1
)
−
θ
(
n
)
≤
L
n
N
2
L
n
N
<
2
n
L
n
2
−
1
2
L
n
2
n
{\displaystyle \{_{\theta \left(2n-1\right)-\theta \left(n\right)\leq Ln{\frac {N}{2}}}^{LnN<2nLn2-{\frac {1}{2}}Ln{2n}}}
θ
(
2
n
−
1
)
−
θ
(
n
)
+
L
n
N
<
L
n
N
2
+
2
n
L
n
N
−
1
2
L
n
2
n
{\displaystyle \theta \left(2n-1\right)-\theta \left(n\right)+LnN{<}{Ln{\frac {N}{2}}+2nLnN-{\frac {1}{2}}Ln{2n}}}
θ
(
2
n
−
1
)
−
θ
(
n
)
<
L
n
N
2
−
L
n
N
+
2
n
L
n
2
−
1
2
L
n
2
n
{\displaystyle \theta \left(2n-1\right)-\theta \left(n\right)<Ln{\frac {N}{2}}-LnN+2nLn2-{\frac {1}{2}}Ln{2n}}
θ
(
2
n
−
1
)
−
θ
(
n
)
<
L
n
N
2
N
+
2
n
L
o
g
2
−
1
2
L
n
2
n
{\displaystyle \theta \left(2n-1\right)-\theta \left(n\right){<}Ln{\frac {\frac {N}{2}}{N}}+2nLog2-{\frac {1}{2}}Ln{2n}}
θ
(
2
n
−
1
)
−
θ
(
n
)
<
L
n
1
2
+
2
n
L
n
2
−
1
2
L
n
2
n
{\displaystyle \theta \left(2n-1\right)-\theta \left(n\right)<Ln{\frac {1}{2}}+2nLn2-{\frac {1}{2}}Ln{2n}}
θ
(
2
n
−
1
)
−
θ
(
n
)
<
−
L
n
2
+
2
n
L
n
2
−
1
2
L
n
2
n
{\displaystyle \theta \left(2n-1\right)-\theta \left(n\right)<-Ln2+2nLn2-{\frac {1}{2}}Ln{2n}}
θ
(
2
n
−
1
)
−
θ
(
n
)
<
(
2
n
−
1
)
L
n
2
−
1
2
L
n
2
n
{\displaystyle \theta \left(2n-1\right)-\theta \left(n\right)<\left(2n-1\right)Ln2-{\frac {1}{2}}Ln{2n}}
θ
(
n
)
<
2
n
L
n
2
{\displaystyle \theta \left(n\right){<}2nLn2}
θ
(
2
n
−
1
)
−
θ
(
n
)
+
θ
(
n
)
<
(
2
n
−
1
)
L
n
2
−
1
2
L
n
2
n
+
2
n
L
n
2
{\displaystyle \theta \left(2n-1\right)-\theta \left(n\right)+\theta \left(n\right){<}\left(2n-1\right)Ln2-{\frac {1}{2}}Ln2n+2nLn2}
θ
(
2
n
−
1
)
<
(
4
n
−
1
)
L
n
2
−
1
2
L
n
2
n
{\displaystyle \theta \left(2n-1\right){<}\left(4n-1\right)Ln2-{\frac {1}{2}}Ln2n}
....
kòm
n
>
2
{\displaystyle n>2}
2
n
>
2
×
2
{\displaystyle 2n{>}{2\times 2}}
2
n
>
2
2
{\displaystyle 2n{>}{2^{2}}}
L
n
2
n
>
L
n
2
2
{\displaystyle Ln2n{>}{Ln{2^{2}}}}
L
n
2
n
>
2
L
n
2
{\displaystyle Ln2n{>}2Ln2}
1
2
L
n
2
n
>
L
n
2
{\displaystyle {\frac {1}{2}}Ln2n{>}Ln2}
−
1
2
L
n
2
n
<
−
L
n
2
{\displaystyle -{\frac {1}{2}}Ln2n{<}-Ln2}
(
4
n
−
1
)
L
n
2
−
1
2
L
n
2
n
<
(
4
n
−
1
)
L
n
2
−
L
n
2
{\displaystyle \left(4n-1\right)Ln2-{\frac {1}{2}}Ln2n{<}\left(4n-1\right)Ln2-Ln2}
(
4
n
−
1
)
L
n
2
−
1
2
L
n
2
n
<
(
4
n
−
2
)
L
n
2
{\displaystyle \left(4n-1\right)Ln2-{\frac {1}{2}}Ln2n{<}\left(4n-2\right)Ln2}
(
4
n
−
1
)
L
n
2
−
1
2
L
n
2
n
<
2
(
2
n
−
1
)
L
n
2
{\displaystyle \left(4n-1\right)Ln2-{\frac {1}{2}}Ln2n{<}2\left(2n-1\right)Ln2}
θ
(
2
n
−
1
)
<
(
4
n
−
1
)
L
n
2
−
1
2
L
n
2
n
<
2
(
2
n
−
1
)
L
n
2
{\displaystyle \theta \left(2n-1\right){<}\left(4n-1\right)Ln2-{\frac {1}{2}}Ln2n<2\left(2n-1\right)Ln2}
θ
(
2
n
−
1
)
<
2
(
2
n
−
1
)
L
n
2
{\displaystyle {\theta \left(2n-1\right)}<{2\left(2n-1\right)Ln2}}
si nou gen
n
≥
2
{\displaystyle n\geq 2}
2
n
≥
4
{\displaystyle 2n\geq 4}
2
n
≠
2
{\displaystyle 2n\neq 2}
2
n
∉
P
{\displaystyle 2n\notin \mathbb {P} }
{
n
∈
N
n
≥
2
⇒
2
n
∉
P
⇔
∄
p
∈
P
/
2
n
−
1
<
p
≤
2
n
{\displaystyle {\{_{n\in \mathbb {N} }^{n\geq 2}}\Rightarrow {2n\notin \mathbb {P} }\Leftrightarrow \not \exists p\in \mathbb {P} {/}2n-1{<}p\leq 2n}
θ
(
2
n
)
=
∑
1
≤
p
≤
2
n
L
n
p
=
∑
1
≤
p
≤
2
n
−
1
L
n
p
+
∑
2
n
−
1
≤
p
≤
2
n
L
n
p
=
∑
1
≤
p
≤
2
n
−
1
L
n
p
+
0
=
∑
1
≤
p
≤
2
n
−
1
L
n
p
=
θ
(
2
n
)
{\displaystyle \theta \left(2n\right)=\sum _{1\leq p\leq 2n}Lnp=\sum _{1\leq p\leq 2n-1}Lnp+\sum _{2n-1\leq p\leq 2n}Lnp=\sum _{1\leq p\leq 2n-1}{Lnp}+0=\sum _{1\leq p\leq 2n-1}{Lnp}=\theta \left(2n\right)}
n
≥
2
⇒
θ
(
2
n
−
1
)
=
θ
(
2
n
)
{\displaystyle n\geq 2\Rightarrow \theta \left(2n-1\right)=\theta \left(2n\right)}
pou tout
n
≥
2
{\displaystyle n\geq 2}
si lèm lan vre pou n li vre pou 2n - 1
an nou pwouve rapidman ke inegalite vre pou 2n tou.
θ
(
2
n
−
1
)
<
2
(
2
n
−
1
)
L
n
2
{\displaystyle \theta \left(2n-1\right){<}2\left(2n-1\right)Ln2}
θ
(
2
n
)
<
2
(
2
n
−
1
)
L
n
2
{\displaystyle \theta \left(2n\right){<}2\left(2n-1\right)Ln2}
θ
(
2
n
)
<
4
n
L
n
2
−
2
L
n
2
{\displaystyle \theta \left(2n\right){<}4nLn2-2Ln2}
θ
(
2
n
)
<
4
n
L
n
2
−
2
L
n
2
<
4
n
L
n
2
{\displaystyle \theta \left(2n\right){<}4nLn2-2Ln2{<}4nLn2}
θ
(
2
n
)
<
4
n
L
n
2
{\displaystyle \theta \left(2n\right){<}4nLn2}
θ
(
2
n
)
<
2
×
2
n
L
n
2
{\displaystyle \theta \left(2n\right)<2\times 2nLn2}
si inegalite a vre pou
n
≥
2
{\displaystyle n\geq 2}
An nou montre kounye a si fòmil la vrè pou entèval
]
2
r
−
1
,
2
r
]
{\displaystyle ]{2^{r-1}},2^{r}]}
]
2
r
,
2
r
+
1
]
{\displaystyle ]2^{r},2^{r+1}]}
Avan nou montre sa an nou verifye ke ingalite a vrè pou entèval
]
2
,
4
]
{\displaystyle ]2,4]}
Inegalite a vrè pou n = 2 selon enplikasyon avan an li vrè pou
2
×
2
−
1
e
t
p
o
u
2
×
2
{\displaystyle 2\times 2-1etpou2\times 2}
[
2
,
4
]
{\displaystyle [2,4]}
]
2
,
4
]
{\displaystyle ]2,4]}
]
2
1
,
2
2
]
{\displaystyle ]2^{1},2^{2}]}
An nou konsidere ke inegalite sou tout entèval
]
2
r
−
1
,
2
r
]
{\displaystyle ]2^{r-1},2^{r}]}
]
2
r
,
2
r
+
1
]
{\displaystyle ]2^{r},2^{r+1}]}
An nou pran yon antye m kèlkon sou entèval
]
2
r
,
2
r
+
1
]
{\displaystyle ]2^{r},2^{r+1}]}
sa vle di
m
∈
]
2
r
,
2
r
+
1
]
{\displaystyle m\in ]2^{r},2^{r+1}]}
m
∈
]
2
r
,
2
r
+
1
]
⇔
2
r
<
m
≤
2
r
+
1
{\displaystyle m\in ]2^{r},2^{r+1}]\Leftrightarrow {2^{r}<m\leq {2^{r+1}}}}
m se yon nonm pè ou m se yon nonm enpè
1)
m pè
⇔
∃
n
∈
N
/
m
=
2
n
{\displaystyle \Leftrightarrow \exists n\in \mathbb {N} {/}m=2n}
{
m
=
2
n
2
r
<
m
≤
2
r
+
1
⇒
2
r
<
2
n
≤
2
r
+
1
⇒
2
×
2
r
−
1
<
2
n
≤
2
×
2
r
⇒
2
r
−
1
<
n
≤
2
r
⇒
n
∈
]
2
r
−
1
,
2
r
]
{\displaystyle \{_{m={2n}}^{2^{r}{<}m\leq {2^{r+1}}}\Rightarrow {2^{r}{<}{2n}\leq {2^{r+1}}}\Rightarrow {2\times {2^{r-1}}{<}{2n}\leq {2\times 2^{r}}}\Rightarrow {2^{r-1}{<}n\leq {2^{r}}}\Rightarrow n\in ]2^{r-1},2^{r}]}
ò pa ipotèz rekirans, inegalite a vrè pou tout n ki nan entèval presedan an. Inegalite a vrè pou
θ
(
n
)
{\displaystyle \theta \left(n\right)}
θ
(
2
n
)
{\displaystyle \theta \left(2n\right)}
]
2
r
,
2
r
+
1
]
{\displaystyle ]{2^{r}},{2^{r+1}}]}
2)
m enpè
⇔
∃
n
∈
N
/
m
=
2
n
−
1
{\displaystyle \Leftrightarrow \exists n\in \mathbb {N} {/}m={2n-1}}
{
m
=
2
n
−
1
2
r
<
m
≤
2
r
+
1
⇒
2
r
<
2
n
−
1
≤
2
r
+
1
⇒
2
r
+
1
<
2
n
≤
2
r
+
1
+
1
⇒
{
2
r
<
2
n
<
2
r
+
1
+
1
2
n
≠
2
r
+
1
+
1
⇒
2
r
<
2
n
≤
2
r
+
1
⇒
2
r
2
<
n
≤
2
r
+
1
2
⇒
2
r
−
1
<
n
≤
2
r
⇒
n
∈
]
2
r
−
1
,
2
r
]
{\displaystyle \{_{m={2n-1}}^{2^{r}{<}m\leq {2^{r+1}}}\Rightarrow {2^{r}{<}{2n-1}\leq {2^{r+1}}}\Rightarrow {{2^{r}+1}<{2n}\leq {2^{r+1}+1}}\Rightarrow {\{_{2^{r}<{2n}<2^{r+1}+1}^{{2n}\neq 2^{r+1}+1}}\Rightarrow {2^{r}{<}{2n}\leq {2^{r+1}}}\Rightarrow {\frac {2^{r}}{2}}{<}n\leq {\frac {2^{r+1}}{2}}\Rightarrow {2^{r-1}{<}n\leq {2^{r}}}\Rightarrow n\in ]2^{r-1},2^{r}]}
θ
(
n
)
<
2
n
L
n
2
{\displaystyle \theta \left(n\right){<}{2n}Ln2}
inegalite a vrè pou tout n ki nan entèval presedan an. Inegalite a vrè pou
θ
(
n
)
{\displaystyle \theta \left(n\right)}
Kidonk selon premyè dediksyon an li vrè pou
θ
(
2
n
−
1
)
{\displaystyle \theta \left(2n-1\right)}
]
2
r
,
2
r
+
1
]
{\displaystyle ]{2^{r}},{2^{r+1}}]}
n
≥
2
{\displaystyle {n}\geq {2}}
∀
n
≥
2
,
θ
(
n
)
<
2
n
L
n
2
{\displaystyle \forall n\geq 2,\theta \left(n\right){<}{2n}Ln2}
Si nou gen yon nonm premye fiks, pi gwo ekspozan k tèl ke
p
k
∣
n
!
{\displaystyle {p^{k}}\mid n!}
k
=
∑
i
=
1
∞
[
n
p
i
]
{\displaystyle k=\sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}
p
k
‖
n
!
{\displaystyle p^{k}\|{n!}}
An nou demontre sa pa rekirans
fòmil la vrè pou
n
=
1
{\displaystyle n=1}
∑
i
=
1
∞
[
1
p
i
]
=
0
{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {1}{p^{i}}}\right]}}=0}
p
i
>
1
{\displaystyle p^{i}>1}
1
p
i
<
1
{\displaystyle {\frac {1}{p^{i}}}<1}
an sipoze fòmil la bon pou
n
{\displaystyle n}
n
+
1
{\displaystyle n+1}
p
α
‖
n
+
1
{\displaystyle p^{\alpha }\|{n+1}}
∑
i
=
1
∞
[
n
+
1
p
i
]
=
∑
i
=
1
α
[
n
+
1
p
i
]
+
∑
i
=
α
+
1
∞
[
n
+
1
p
i
]
{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\alpha }{\left[{\frac {n+1}{p^{i}}}\right]}+\sum _{i=\alpha +1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}}
∑
i
=
1
∞
[
n
+
1
p
i
]
=
∑
i
=
1
α
(
[
n
p
i
]
+
1
)
+
∑
i
=
α
+
1
∞
[
n
p
i
]
{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\alpha }{\left(\left[{\frac {n}{p^{i}}}\right]+1\right)}+\sum _{i=\alpha +1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}}
∑
i
=
1
∞
[
n
+
1
p
i
]
=
∑
i
=
1
α
[
n
p
i
]
+
∑
i
=
1
α
1
+
∑
i
=
α
+
1
∞
[
n
p
i
]
{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\alpha }{\left[{\frac {n}{p^{i}}}\right]}+\sum _{i=1}^{\alpha }{1}+\sum _{i=\alpha +1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}}
∑
i
=
1
∞
[
n
+
1
p
i
]
=
∑
i
=
1
α
[
n
p
i
]
+
α
+
∑
i
=
α
+
1
∞
[
n
p
i
]
{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\alpha }{\left[{\frac {n}{p^{i}}}\right]}+\alpha +\sum _{i=\alpha +1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}}
∑
i
=
1
∞
[
n
+
1
p
i
]
=
∑
i
=
1
α
[
n
p
i
]
+
∑
i
=
α
+
1
∞
[
n
p
i
]
+
α
{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\alpha }{\left[{\frac {n}{p^{i}}}\right]}+\sum _{i=\alpha +1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}+\alpha }}
∑
i
=
1
∞
[
n
+
1
p
i
]
=
∑
i
=
1
∞
[
n
p
i
]
+
α
{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}+{\alpha }}
∑
i
=
1
∞
[
n
+
1
p
i
]
=
α
+
k
{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\alpha +k}}
ò
(
n
+
1
)
!
=
(
n
+
1
)
×
n
!
{\displaystyle {\left(n+1\right)!}={\left(n+1\right)\times {n!}}}
p
k
‖
n
!
{\displaystyle {p^{k}}\|{n!}}
ak
p
α
‖
n
+
1
{\displaystyle {p^{\alpha }}\|{n+1}}
enplike
p
α
×
p
k
‖
(
n
+
1
)
×
n
!
{\displaystyle {{p^{\alpha }}\times {p^{k}}}\|{{\left(n+1\right)}\times {n!}}}
Sa ki enplike
p
α
+
k
‖
(
n
+
1
)
!
{\displaystyle {p^{\alpha +k}}\|{{\left(n+1\right)}!}}
n
{\displaystyle n}
n
+
1
{\displaystyle n+1}
modifye
An nou montre an premye lye ke
θ
(
2
n
)
−
θ
(
n
)
>
0
{\displaystyle \theta \left(2n\right)-\theta \left(n\right)>0}
pou chak
n
>
2
6
{\displaystyle n>2^{6}}
]
n
,
2
n
[
{\displaystyle ]{n},{2n}[}
n
>
2
6
{\displaystyle n>{2^{6}}}
An nou konsidere yon lòt fwa nonm
N
=
(
2
n
n
)
{\displaystyle N={\dbinom {2n}{n}}}
N
=
(
2
n
n
)
=
(
2
n
)
!
n
!
×
(
2
n
−
n
)
!
=
(
2
n
)
!
(
n
!
)
2
{\displaystyle N={\dbinom {2n}{n}}={\frac {{\left(2n\right)}!}{{n!}\times \left(2n-n\right)!}}={\frac {{\left(2n\right)}!}{{\left(n!\right)}^{2}}}}
Oken nonm premye ki siperyè ak 2n pa yon divizè de N .
Tandiske tout nonm premye ki konpri ant n ekstrikteman ak 2n lajman divize N yon sèl fwa.
An nou konsidere n ak 2n pou nou wè kisa ki rive.
N
=
(
2
n
)
!
(
n
!
)
×
(
n
!
)
{\displaystyle N={\frac {\left(2n\right)!}{\left(n!\right)\times \left(n!\right)}}}
si n premye, li parèt de fwa anlè a sou fòm n ak
2
×
n
{\displaystyle 2\times n}
menm jan an li parèt 2 fwa anba a sèlman. Kidonk, n pa yon faktè de N.
Pou 2n , si
n
>
1
{\displaystyle n>1}
Nan entèval
]
n
,
2
n
[
{\displaystyle ]n,2n[}
]
2
,
2
n
]
{\displaystyle ]2,2n]}
De tout fason nou ap gen pou nou retounen sou konsiderasyon sa a.
Si nou konsidere teyorèm Legendre lan nou kapab ekri :
N
=
2
n
!
n
!
×
(
n
!
)
=
∏
p
≤
2
n
(
p
∑
i
=
1
∞
[
2
n
p
i
]
)
∏
p
≤
n
(
p
2
×
∑
i
=
1
∞
[
n
p
i
]
)
{\displaystyle N={\frac {{2n}!}{{n!}\times \left(n!\right)}}={\frac {\prod _{p\leq {2n}}{\left(p^{\sum _{i=1}^{\infty }{\left[{\frac {2n}{p^{i}}}\right]}}\right)}}{\prod _{p\leq n}{\left(p^{2\times \sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}\right)}}}}
An nou konsidere
∏
n
<
p
≤
2
n
p
2
×
∑
i
=
1
∞
(
[
n
p
i
]
)
{\displaystyle \prod _{n<p\leq {2n}}{p^{2\times \sum _{i=1}^{\infty }{\left(\left[{\frac {n}{p^{i}}}\right]\right)}}}}
Pwodui sa vo 1 paske ekspozan toujou vo 0.
pou tout antye n > 1 , 2n pa premye kidon
∀
n
>
1
,
n
<
p
≤
2
n
⇒
n
<
p
<
2
n
⇒
1
2
n
<
1
p
<
1
n
⇒
n
2
n
<
n
p
<
n
n
⇒
1
2
<
n
p
<
1
⇒
[
n
p
]
=
0
{\displaystyle \forall n>1,{n<p\leq {2n}}\Rightarrow {n<p<{2n}}\Rightarrow {{\frac {1}{2n}}<{\frac {1}{p}}<{\frac {1}{n}}}\Rightarrow {{\frac {n}{2n}}<{\frac {n}{p}}<{\frac {n}{n}}}\Rightarrow {{\frac {1}{2}}<{\frac {n}{p}}<1}\Rightarrow {\left[{\frac {n}{p}}\right]=0}}
pou tout
i
≥
1
{\displaystyle i\geq 1}
n
p
i
<
n
p
{\displaystyle {\frac {n}{p^{i}}}<{\frac {n}{p}}}
[
n
p
i
]
=
0
{\displaystyle \left[{\frac {n}{p^{i}}}\right]=0}
i
≥
1
{\displaystyle i\geq 1}
∏
n
<
p
≤
2
n
p
2
×
∑
i
=
1
∞
(
[
n
p
i
]
)
=
∏
n
<
p
≤
2
n
p
2
×
0
=
1
{\displaystyle \prod _{n<p\leq {2n}}{p^{2\times \sum _{i=1}^{\infty }{\left(\left[{\frac {n}{p^{i}}}\right]\right)}}}=\prod _{n<p\leq {2n}}{p^{2\times 0}}=1}
N
=
2
n
!
n
!
×
(
n
!
)
=
∏
p
≤
2
n
(
p
∑
i
=
1
∞
[
2
n
p
i
]
)
∏
p
≤
n
(
p
2
×
∑
i
=
1
∞
[
n
p
i
]
)
{\displaystyle N={\frac {{2n}!}{{n!}\times \left(n!\right)}}={\frac {\prod _{p\leq {2n}}{\left(p^{\sum _{i=1}^{\infty }{\left[{\frac {2n}{p^{i}}}\right]}}\right)}}{\prod _{p\leq n}{\left(p^{2\times \sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}\right)}}}}
N
=
2
n
!
n
!
×
(
n
!
)
=
∏
p
≤
2
n
(
p
∑
i
=
1
∞
[
2
n
p
i
]
)
∏
p
≤
n
(
p
2
×
∑
i
=
1
∞
[
n
p
i
]
)
×
∏
n
<
p
≤
2
n
(
p
2
×
∑
i
=
1
∞
[
n
p
i
]
)
{\displaystyle N={\frac {{2n}!}{{n!}\times \left(n!\right)}}={\frac {\prod _{p\leq {2n}}{\left(p^{\sum _{i=1}^{\infty }{\left[{\frac {2n}{p^{i}}}\right]}}\right)}}{{\prod _{p\leq n}{\left(p^{2\times \sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}\right)}}\times {\prod _{n<p\leq {2n}}{\left(p^{2\times \sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}\right)}}}}}
N
=
2
n
!
n
!
×
(
n
!
)
=
∏
p
≤
2
n
(
p
∑
i
=
1
∞
[
2
n
p
i
]
)
∏
p
≤
2
n
(
p
2
×
∑
i
=
1
∞
[
n
p
i
]
)
{\displaystyle N={\frac {{2n}!}{{n!}\times \left(n!\right)}}={\frac {\prod _{p\leq {2n}}{\left(p^{\sum _{i=1}^{\infty }{\left[{\frac {2n}{p^{i}}}\right]}}\right)}}{\prod _{p\leq {2n}}{\left(p^{2\times \sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}\right)}}}}
N
=
2
n
!
n
!
×
(
n
!
)
=
∏
p
≤
2
n
(
p
(
∑
i
=
1
∞
[
2
n
p
i
]
−
2
∑
i
=
1
∞
[
n
p
i
]
)
)
{\displaystyle N={\frac {{2n}!}{{n!}\times \left(n!\right)}}=\prod _{p\leq {2n}}{\left(p^{\left(\sum _{i=1}^{\infty }{\left[{\frac {2n}{p^{i}}}\right]}-2\sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}\right)}\right)}}
N
=
∏
p
≤
2
n
(
p
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
)
{\displaystyle N=\prod _{p\leq {2n}}{\left(p^{\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\right)}}
An nou pran logarithm Neperyen de manm egalite ya.
L
n
N
=
L
n
(
∏
p
≤
2
n
(
p
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
)
)
{\displaystyle LnN=Ln\left(\prod _{p\leq {2n}}{\left(p^{\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\right)}\right)}
L
n
N
=
∑
p
≤
2
n
(
L
n
(
(
p
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
)
)
)
{\displaystyle LnN=\sum _{p\leq {2n}}{\left(Ln\left({\left(p^{\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\right)}\right)\right)}}
L
n
N
=
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
{\displaystyle LnN=\sum _{p\leq {2n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}
An nou diseke sòm sa an 4 tèrm .
L
n
N
=
∑
n
<
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
+
∑
2
n
3
<
p
≤
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
+
∑
2
n
<
p
≤
2
n
3
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
+
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
{\displaystyle LnN=\sum _{n<p\leq {2n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+\sum _{{\frac {2n}{3}}<p\leq {n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}
pou premye tèm lan nou genyen :
∑
n
<
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
=
∑
n
<
p
≤
2
n
(
L
n
p
)
{\displaystyle \sum _{n<p\leq {2n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{n<p\leq {2n}}{\left(Lnp\right)}}
....
kidonk
∑
n
<
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
=
∑
1
≤
p
≤
2
n
(
L
n
p
)
−
∑
1
≤
p
≤
n
(
L
n
p
)
=
θ
(
2
n
)
−
θ
(
n
)
{\displaystyle \sum _{n<p\leq {2n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{1\leq {p}\leq {2n}}{\left(Lnp\right)}-\sum _{1\leq {p}\leq {n}}{\left(Lnp\right)}=\theta \left(2n\right)-\theta \left(n\right)}
n
<
p
≤
2
n
⇒
1
2
n
≤
1
p
<
1
n
⇒
n
2
n
≤
n
p
<
n
n
⇒
1
2
≤
n
p
<
1
⇒
[
n
p
]
=
0
{\displaystyle n<p\leq {2n}\Rightarrow {{\frac {1}{2n}}\leq {\frac {1}{p}}<{\frac {1}{n}}}\Rightarrow {{\frac {n}{2n}}\leq {\frac {n}{p}}<{\frac {n}{n}}}\Rightarrow {{\frac {1}{2}}\leq {\frac {n}{p}}<1}\Rightarrow \left[{\frac {n}{p}}\right]=0}
∀
i
>
0
,
n
p
i
{\displaystyle \forall i>0,{\frac {n}{p^{i}}}}
[
n
p
i
]
=
0
{\displaystyle \left[{\frac {n}{p^{i}}}\right]=0}
n
<
p
≤
2
n
⇒
1
2
n
≤
1
p
<
1
n
⇒
2
n
2
n
≤
2
n
p
<
2
n
n
⇒
1
≤
2
n
p
<
2
⇒
[
2
n
p
]
=
1
{\displaystyle n<p\leq {2n}\Rightarrow {{\frac {1}{2n}}\leq {\frac {1}{p}}<{\frac {1}{n}}}\Rightarrow {{\frac {2n}{2n}}\leq {\frac {2n}{p}}<{\frac {2n}{n}}}\Rightarrow {1\leq {\frac {2n}{p}}<2}\Rightarrow \left[{\frac {2n}{p}}\right]=1}
∀
p
,
p
≤
2
{\displaystyle \forall p,p\leq 2}
p
≤
2
⇒
1
p
≤
1
2
{\displaystyle p\leq 2\Rightarrow {\frac {1}{p}}\leq {\frac {1}{2}}}
nou genyen sou baz ipotèz la
1
p
≤
2
n
p
2
<
2
p
≤
1
{\displaystyle {\frac {1}{p}}\leq {\frac {2n}{p^{2}}}<{\frac {2}{p}}\leq 1}
2
n
p
2
<
1
{\displaystyle {\frac {2n}{p^{2}}}<1}
pou tout
i
>
2
,
[
2
n
p
i
]
=
0
{\displaystyle i>2,\left[{\frac {2n}{p^{i}}}\right]=0}
∑
2
n
3
<
p
≤
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
{\displaystyle \sum _{{\frac {2n}{3}}<p\leq {n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}
2
n
3
<
p
≤
n
⇒
1
n
≤
1
p
<
3
2
n
⇒
n
n
≤
n
p
<
3
n
2
n
⇒
1
≤
n
p
<
3
2
⇒
[
n
p
]
=
1
{\displaystyle {\frac {2n}{3}}<p\leq n\Rightarrow {{\frac {1}{n}}\leq {\frac {1}{p}}<{\frac {3}{2n}}}\Rightarrow {{\frac {n}{n}}\leq {\frac {n}{p}}<{\frac {3n}{2n}}}\Rightarrow 1\leq {\frac {n}{p}}<{\frac {3}{2}}\Rightarrow \left[{\frac {n}{p}}\right]=1}
2
n
3
<
p
≤
n
⇒
1
n
≤
1
p
<
3
2
n
⇒
2
n
n
≤
2
n
p
<
3
⇒
2
≤
2
n
p
<
3
⇒
[
2
n
p
]
=
2
{\displaystyle {\frac {2n}{3}}<p\leq n\Rightarrow {{\frac {1}{n}}\leq {\frac {1}{p}}<{\frac {3}{2n}}}\Rightarrow {{\frac {2n}{n}}\leq {\frac {2n}{p}}<3}\Rightarrow {2\leq {\frac {2n}{p}}<3}\Rightarrow \left[{\frac {2n}{p}}\right]=2}
[
2
n
p
]
−
2
[
n
p
]
=
2
−
2
×
1
=
0
{\displaystyle \left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]=2-2\times 1=0}
sou yon lòt ang ,
1
p
≤
n
p
2
<
3
2
p
<
3
2
×
2
{\displaystyle {\frac {1}{p}}\leq {\frac {n}{p^{2}}}<{\frac {3}{2p}}<{\frac {3}{2\times 2}}}
[
n
p
2
]
=
0
{\displaystyle \left[{\frac {n}{p^{2}}}\right]=0}
2
p
≤
2
n
p
2
<
3
×
2
2
×
p
{\displaystyle {\frac {2}{p}}\leq {\frac {2n}{p^{2}}}<{\frac {3\times 2}{2\times p}}}
2
p
≤
2
n
p
2
<
3
p
{\displaystyle {\frac {2}{p}}\leq {\frac {2n}{p^{2}}}<{\frac {3}{p}}}
1
≤
2
n
2
2
<
3
2
{\displaystyle 1\leq {\frac {2n}{2^{2}}}<{\frac {3}{2}}}
nap genyen lè sa a :
[
2
n
p
2
]
=
1
{\displaystyle \left[{\frac {2n}{p^{2}}}\right]=1}
1
≤
n
2
<
3
2
{\displaystyle 1\leq {\frac {n}{2}}<{\frac {3}{2}}}
kidonk tou nou tap genyen
n
<
3
{\displaystyle n<3}
n
≤
3
{\displaystyle n\leq 3}
An nou gade ki sa nou ap genyen
2
n
≥
3
×
2
{\displaystyle 2n\geq 3\times 2}
2
n
≥
6
{\displaystyle 2n\geq 6}
2
n
3
≥
2
{\displaystyle {\frac {2n}{3}}\geq 2}
2
≤
2
n
3
<
p
≤
n
{\displaystyle 2\leq {\frac {2n}{3}}<p\leq n}
kidonk
p
>
2
{\displaystyle p>2}
p
>
3
{\displaystyle p>3}
2
n
p
2
<
3
p
{\displaystyle {\frac {2n}{p^{2}}}<{\frac {3}{p}}}
1
p
<
1
3
{\displaystyle {\frac {1}{p}}<{\frac {1}{3}}}
3
p
<
1
{\displaystyle {\frac {3}{p}}<1}
2
n
p
2
<
3
p
<
1
{\displaystyle {\frac {2n}{p^{2}}}<{\frac {3}{p}}<1}
2
n
p
2
<
1
{\displaystyle {\frac {2n}{p^{2}}}<1}
[
2
n
p
2
]
=
0
{\displaystyle \left[{\frac {2n}{p^{2}}}\right]=0}
pou
n
≤
3
{\displaystyle n\leq 3}
i
≥
2
{\displaystyle i\geq 2}
[
2
n
p
i
]
=
0
{\displaystyle \left[{\frac {2n}{p^{i}}}\right]=0}
[
n
p
i
]
=
0
{\displaystyle \left[{\frac {n}{p^{i}}}\right]=0}
[
2
n
p
i
]
−
2
[
n
p
i
]
=
0
{\displaystyle \left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]=0}
n
≥
3
{\displaystyle n\geq 3}
∑
2
n
3
<
p
≤
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
=
0
{\displaystyle \sum _{{\frac {2n}{3}}<p\leq {n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=0}
An nou konsidere twazyèm sòm lan
∑
2
n
<
p
≤
2
n
3
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
{\displaystyle \sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}
An konsidere nan ki kondisyon inegalite ki anba somasyon ak gen sans :
2
n
≤
2
n
3
{\displaystyle {\sqrt {2n}}\leq {\frac {2n}{3}}}
2
n
≤
4
n
2
9
{\displaystyle 2n\leq {\frac {4n^{2}}{9}}}
18
n
≤
4
n
2
{\displaystyle 18n\leq 4n^{2}}
18
n
−
4
n
2
≤
0
{\displaystyle 18n-4n^{2}\leq 0}
2
n
(
9
−
2
n
)
≤
0
{\displaystyle 2n\left(9-2n\right)\leq 0}
9
−
2
n
≤
0
{\displaystyle 9-2n\leq 0}
9
≤
2
n
{\displaystyle 9\leq 2n}
9
2
≤
n
{\displaystyle {\frac {9}{2}}\leq n}
n
≥
9
2
{\displaystyle n\geq {\frac {9}{2}}}
n
≥
5
{\displaystyle n\geq 5}
[
2
×
n
p
r
]
−
2
[
n
p
r
]
{\displaystyle \left[2\times {\frac {n}{p^{r}}}\right]-2\left[{\frac {n}{p^{r}}}\right]}
[
2
x
]
−
2
[
x
]
{\displaystyle \left[2x\right]-2\left[x\right]}
x
=
[
x
]
+
{
x
}
{\displaystyle x=\left[x\right]+\{x\}}
2
x
=
2
[
x
]
+
2
{
x
}
{\displaystyle 2x=2\left[x\right]+2\{x\}}
[
2
x
]
=
[
2
[
x
]
+
2
{
x
}
]
{\displaystyle \left[2x\right]=\left[2\left[x\right]+2\{x\}\right]}
[
2
x
]
=
2
[
x
]
+
[
2
{
x
}
]
{\displaystyle \left[2x\right]=2\left[x\right]+\left[2\{x\}\right]}
[
2
x
]
−
2
[
x
]
=
[
2
{
x
}
]
{\displaystyle \left[2x\right]-2\left[x\right]=\left[2\{x\}\right]}
0
≤
{
x
}
<
1
{\displaystyle 0\leq \{x\}<1}
0
≤
2
{
x
}
<
2
{\displaystyle 0\leq 2\{x\}<2}
[
2
x
]
−
2
[
x
]
=
[
2
{
x
}
]
{\displaystyle \left[2x\right]-2\left[x\right]=\left[2\{x\}\right]}
[
2
n
p
i
]
−
2
[
n
p
i
]
{\displaystyle \left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]}
i
=
1
{\displaystyle i=1}
An nou montre ke li vo 0 nan tout lòt ka .
2
n
<
p
≤
2
n
3
⇒
2
n
<
p
2
≤
4
n
2
9
⇒
9
4
n
2
≤
1
p
2
<
1
2
n
⇒
9
n
4
n
2
≤
n
p
2
<
n
2
n
⇒
9
n
4
n
2
≤
n
p
2
<
1
2
⇒
[
n
p
2
]
=
0
{\displaystyle {\sqrt {2n}}<p\leq {\frac {2n}{3}}\Rightarrow 2n<p^{2}\leq {\frac {4n^{2}}{9}}\Rightarrow {\frac {9}{4n^{2}}}\leq {\frac {1}{p^{2}}}<{\frac {1}{2n}}\Rightarrow {\frac {9n}{4n^{2}}}\leq {\frac {n}{p^{2}}}<{\frac {n}{2n}}\Rightarrow {\frac {9n}{4n^{2}}}\leq {\frac {n}{p^{2}}}<{\frac {1}{2}}\Rightarrow \left[{\frac {n}{p^{2}}}\right]=0}
2
n
<
p
≤
2
n
3
⇒
2
n
<
p
2
≤
4
n
2
9
⇒
9
4
n
2
≤
1
p
2
<
1
2
n
⇒
9
n
4
n
2
≤
n
p
2
<
n
2
n
⇒
9
n
2
n
2
≤
2
n
p
2
<
1
⇒
2
n
p
2
+
j
<
1
,
∀
j
∈
N
{\displaystyle {\sqrt {2n}}<p\leq {\frac {2n}{3}}\Rightarrow 2n<p^{2}\leq {\frac {4n^{2}}{9}}\Rightarrow {\frac {9}{4n^{2}}}\leq {\frac {1}{p^{2}}}<{\frac {1}{2n}}\Rightarrow {\frac {9n}{4n^{2}}}\leq {\frac {n}{p^{2}}}<{\frac {n}{2n}}\Rightarrow {\frac {9n}{2n^{2}}}\leq {\frac {2n}{p^{2}}}<1\Rightarrow {\frac {2n}{p^{2+j}}}<1,\forall j\in \mathbb {N} }
∀
i
≥
2
,
2
n
p
i
<
1
{\displaystyle \forall i\geq 2,{\frac {2n}{p^{i}}}<1}
∀
i
≥
2
,
[
2
n
p
i
]
=
0
{\displaystyle \forall i\geq 2,\left[{\frac {2n}{p^{i}}}\right]=0}
∀
n
≥
5
{\displaystyle \forall n\geq 5}
∑
2
n
<
p
≤
2
n
3
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
=
∑
2
n
<
p
≤
2
n
3
(
(
[
2
n
p
]
−
2
[
n
p
]
)
×
L
n
P
)
+
∑
2
n
<
p
≤
2
n
3
(
(
∑
i
=
2
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
=
∑
2
n
<
p
≤
2
n
3
(
(
[
2
n
p
]
−
2
[
n
p
]
)
×
L
n
P
)
+
0
=
∑
2
n
<
p
≤
2
n
3
(
(
[
2
n
p
]
−
2
[
n
p
]
)
×
L
n
P
)
{\displaystyle \sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}+\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=2}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}+0=\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}}
ò
[
2
n
p
]
−
2
[
n
p
]
{\displaystyle \left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]}
Kidonk
[
2
n
p
]
−
2
[
n
p
]
≤
1
{\displaystyle \left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\leq 1}
(
(
[
2
n
p
]
−
2
[
n
p
]
)
×
L
n
P
)
≤
L
n
p
{\displaystyle {\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}\leq Lnp}
∑
2
n
<
p
≤
2
n
3
(
(
[
2
n
p
]
−
2
[
n
p
]
)
×
L
n
P
)
≤
∑
2
n
<
p
≤
2
n
3
L
n
p
{\displaystyle {\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}}\leq {\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{Lnp}}}
∑
2
n
<
p
≤
2
n
3
L
n
p
=
θ
(
2
n
3
)
−
θ
(
2
n
)
{\displaystyle {\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{Lnp}}=\theta \left({\frac {2n}{3}}\right)-\theta \left({\sqrt {2n}}\right)}
∑
2
n
<
p
≤
2
n
3
(
(
[
2
n
p
]
−
2
[
n
p
]
)
×
L
n
P
)
≤
θ
(
2
n
3
)
−
θ
(
2
n
)
{\displaystyle {\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}}\leq {\theta \left({\frac {2n}{3}}\right)-\theta \left({\sqrt {2n}}\right)}}
∑
2
n
<
p
≤
2
n
3
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
≤
θ
(
2
n
3
)
−
θ
(
2
n
)
{\displaystyle \sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq {\theta \left({\frac {2n}{3}}\right)-\theta \left({\sqrt {2n}}\right)}}
θ
(
2
n
)
=
L
n
2
+
L
n
3
+
.
.
.
+
L
n
p
(
π
(
2
n
)
)
≥
π
(
2
n
)
×
L
n
2
{\displaystyle \theta \left({\sqrt {2n}}\right)=Ln2+Ln3+...+Lnp_{\left(\pi \left({\sqrt {2n}}\right)\right)}\geq {\pi \left({\sqrt {2n}}\right)\times {Ln2}}}
−
θ
(
2
n
)
≤
−
π
(
2
n
)
×
L
n
2
{\displaystyle -\theta \left({\sqrt {2n}}\right)\leq {-\pi \left({\sqrt {2n}}\right)\times {Ln2}}}
θ
(
2
n
3
)
−
θ
(
2
n
)
≤
θ
(
2
n
3
)
−
π
(
2
n
)
×
L
n
2
{\displaystyle \theta \left({\frac {2n}{3}}\right)-\theta \left({\sqrt {2n}}\right)\leq {\theta \left({\frac {2n}{3}}\right)-\pi \left({\sqrt {2n}}\right)\times {Ln2}}}
Pou tout
n
≥
5
{\displaystyle n\geq 5}
∑
2
n
<
p
≤
2
n
3
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
≤
θ
(
2
n
3
)
−
π
(
2
n
)
×
L
n
2
{\displaystyle \sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq {\theta \left({\frac {2n}{3}}\right)-\pi \left({\sqrt {2n}}\right)\times {Ln2}}}
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}
[
2
n
p
i
]
−
2
[
n
p
i
]
{\displaystyle \left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]}
yon lòt kote
[
2
n
p
i
]
≥
2
[
n
p
i
]
{\displaystyle \left[{\frac {2n}{p^{i}}}\right]\geq {2\left[{\frac {n}{p^{i}}}\right]}}
[
2
n
p
i
]
{\displaystyle \left[{\frac {2n}{p^{i}}}\right]}
[
2
n
p
i
]
=
0
⇔
2
n
p
i
<
1
⇔
2
n
<
p
i
⇔
p
i
>
2
n
⇔
L
n
p
i
>
L
n
2
n
⇔
i
L
n
p
>
L
n
2
n
⇔
i
>
L
n
2
n
L
n
p
⇔
i
>
[
L
n
2
n
L
n
p
]
⇔
i
≥
[
L
n
2
n
L
n
p
]
+
1
{\displaystyle \left[{\frac {2n}{p^{i}}}\right]=0\Leftrightarrow {\frac {2n}{p^{i}}}<1\Leftrightarrow {2n}<{p^{i}}\Leftrightarrow {p^{i}}>{2n}\Leftrightarrow Ln{p^{i}}>Ln{2n}\Leftrightarrow iLn{p}>Ln{2n}\Leftrightarrow i>{\frac {Ln2n}{Lnp}}\Leftrightarrow i>\left[{\frac {Ln2n}{Lnp}}\right]\Leftrightarrow i\geq \left[{\frac {Ln2n}{Lnp}}\right]+1}
i
≥
[
L
n
2
n
L
n
p
]
+
1
⇒
[
2
n
p
i
]
−
2
[
n
p
i
]
=
0
{\displaystyle i\geq \left[{\frac {Ln2n}{Lnp}}\right]+1\Rightarrow \left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]=0}
n
≥
5
{\displaystyle n\geq 5}
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
=
∑
p
≤
2
n
(
(
∑
i
=
1
[
L
n
2
n
L
n
p
]
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
+
∑
p
≤
2
n
(
(
∑
[
L
n
2
n
L
n
p
]
+
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\left[{\frac {Ln2n}{Lnp}}\right]}{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{\left[{\frac {Ln2n}{Lnp}}\right]+1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
=
∑
p
≤
2
n
(
(
∑
i
=
1
[
L
n
2
n
L
n
p
]
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
+
0
{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\left[{\frac {Ln2n}{Lnp}}\right]}{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+0}
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
=
∑
p
≤
2
n
(
(
∑
i
=
1
[
L
n
2
n
L
n
p
]
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
≤
∑
p
≤
2
n
(
(
∑
i
=
1
[
L
n
2
n
L
n
p
]
1
)
×
L
n
p
)
{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\left[{\frac {Ln2n}{Lnp}}\right]}{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq {\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\left[{\frac {Ln2n}{Lnp}}\right]}{1}\right)}\times Lnp\right)}}}
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
≤
∑
p
≤
2
n
(
[
L
n
2
n
L
n
p
]
×
L
n
P
)
{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{\left(\left[{\frac {Ln2n}{Lnp}}\right]\times LnP\right)}}
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
≤
∑
p
≤
2
n
(
[
L
n
2
n
L
n
p
]
×
L
n
p
)
≤
∑
p
≤
2
n
(
L
n
2
n
L
n
p
×
L
n
p
)
{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{\left(\left[{\frac {Ln2n}{Lnp}}\right]\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{\left({\frac {Ln2n}{Lnp}}\times Lnp\right)}}
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
≤
∑
p
≤
2
n
(
[
L
n
2
n
L
n
p
]
×
L
n
p
)
≤
∑
p
≤
2
n
L
n
2
n
{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{\left(\left[{\frac {Ln2n}{Lnp}}\right]\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{Ln2n}}
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
≤
∑
p
≤
2
n
(
[
L
n
2
n
L
n
p
]
×
L
n
p
)
≤
L
n
2
n
∑
p
≤
2
n
1
{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{\left(\left[{\frac {Ln2n}{Lnp}}\right]\times Lnp\right)}\leq Ln2n\sum _{p\leq {\sqrt {2n}}}{1}}
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
≤
π
(
2
n
)
L
n
2
n
{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq \pi \left({\sqrt {2n}}\right)Ln2n}
L
n
N
=
∑
n
<
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
+
∑
2
n
3
<
p
≤
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
+
∑
2
n
<
p
≤
2
n
3
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
+
∑
p
≤
2
n
(
(
∑
i
=
1
∞
(
[
2
n
p
i
]
−
2
[
n
p
i
]
)
)
×
L
n
p
)
{\displaystyle LnN=\sum _{n<p\leq {2n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+\sum _{{\frac {2n}{3}}<p\leq {n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}
lè nou anvizaje tout estimasyon yo oswa tout estime yo nou kapab ekri :
pou tout
n
≤
5
{\displaystyle n\leq 5}
L
n
N
≤
(
θ
(
2
n
)
−
θ
(
n
)
)
+
0
+
(
θ
(
2
n
3
)
−
π
(
2
n
)
×
L
n
2
)
+
(
π
(
2
n
)
×
L
n
2
n
)
{\displaystyle LnN\leq \left(\theta \left(2n\right)-\theta \left(n\right)\right)+0+\left(\theta \left({\frac {2n}{3}}\right)-\pi \left({\sqrt {2n}}\right)\times Ln2\right)+\left(\pi \left({\sqrt {2n}}\right)\times Ln2n\right)}
θ
(
2
n
)
−
θ
(
n
)
≥
L
n
N
−
θ
(
2
n
3
)
+
π
(
2
n
)
(
L
n
2
−
L
n
2
n
)
{\displaystyle \theta \left(2n\right)-\theta \left(n\right)\geq LnN-\theta \left({\frac {2n}{3}}\right)+\pi \left({\sqrt {2n}}\right)\left(Ln2-Ln2n\right)}
θ
(
2
n
)
−
θ
(
n
)
≥
L
n
N
−
θ
(
2
n
3
)
−
π
(
2
n
)
(
L
n
(
2
n
2
)
)
{\displaystyle \theta \left(2n\right)-\theta \left(n\right)\geq LnN-\theta \left({\frac {2n}{3}}\right)-\pi \left({\sqrt {2n}}\right)\left(Ln\left({\frac {2n}{2}}\right)\right)}
θ
(
2
n
)
−
θ
(
n
)
≥
L
n
N
−
θ
(
2
n
3
)
−
π
(
2
n
)
L
n
2
{\displaystyle \theta \left(2n\right)-\theta \left(n\right)\geq LnN-\theta \left({\frac {2n}{3}}\right)-\pi \left({\sqrt {2n}}\right)Ln2}
Li rete pou nou pwouve ke
L
n
N
−
θ
(
2
n
3
)
−
π
(
2
n
)
L
n
{\displaystyle LnN-\theta \left({\frac {2n}{3}}\right)-\pi \left({\sqrt {2n}}\right)Ln}
Nou te genyen
N
>
2
n
2
n
{\displaystyle N>{\frac {2^{n}}{2{\sqrt {n}}}}}
L
n
N
>
L
n
(
2
n
2
n
)
{\displaystyle LnN>Ln\left({\frac {2^{n}}{2{\sqrt {n}}}}\right)}
L
n
N
>
L
n
2
2
n
−
l
n
2
n
{\displaystyle LnN>Ln2^{2n}-ln{2{\sqrt {n}}}}
L
n
N
>
2
n
l
n
2
−
l
n
(
2
n
)
{\displaystyle LnN>2nln2-ln\left(2{\sqrt {n}}\right)}
L
n
N
>
2
n
L
n
2
−
L
n
2
−
L
n
n
{\displaystyle LnN>2nLn2-Ln2-Ln{\sqrt {n}}}
L
n
N
>
2
n
L
n
2
−
L
n
2
−
L
n
n
1
2
{\displaystyle LnN>2nLn2-Ln2-Ln{n^{\frac {1}{2}}}}
L
n
N
>
2
n
L
n
2
−
L
n
2
−
1
2
L
n
n
{\displaystyle LnN>2nLn2-Ln2-{\frac {1}{2}}Lnn}
Anplis lè
n
≤
2
{\displaystyle n\leq 2}
θ
(
2
n
3
)
=
θ
(
[
2
n
3
]
)
<
2
×
[
2
n
3
]
L
n
2
<
4
n
3
L
n
2
{\displaystyle \theta \left({\frac {2n}{3}}\right)=\theta \left(\left[{\frac {2n}{3}}\right]\right)<2\times \left[{\frac {2n}{3}}\right]Ln2<{\frac {4n}{3}}Ln2}
−
θ
(
2
n
3
)
>
−
4
n
3
L
n
2
{\displaystyle -\theta \left({\frac {2n}{3}}\right)>-{\frac {4n}{3}}Ln2}
yon lòt kote,
π
(
n
)
≤
n
2
,
∀
n
≥
8
{\displaystyle \pi \left(n\right)\leq {\frac {n}{2}},\forall n\geq 8}
Nou vin genyen :
θ
(
2
n
)
−
θ
(
n
)
>
2
n
L
n
2
−
L
n
(
2
n
)
−
4
n
3
L
n
2
−
2
n
2
L
n
n
{\displaystyle \theta \left(2n\right)-\theta \left(n\right)>2nLn2-Ln\left(2{\sqrt {n}}\right)-{\frac {4n}{3}}Ln2-{\frac {\sqrt {2n}}{2}}Lnn}
θ
(
2
n
)
−
θ
(
n
)
>
2
n
L
n
2
−
L
n
2
−
L
n
n
−
4
n
3
L
n
2
−
2
n
2
L
n
n
{\displaystyle \theta \left(2n\right)-\theta \left(n\right)>2nLn2-Ln2-Ln{\sqrt {n}}-{\frac {4n}{3}}Ln2-{\frac {\sqrt {2n}}{2}}Lnn}
θ
(
2
n
)
−
θ
(
n
)
>
2
n
L
n
2
−
L
n
2
−
1
2
L
n
n
−
4
n
3
L
n
2
−
2
n
2
L
n
n
{\displaystyle \theta \left(2n\right)-\theta \left(n\right)>2nLn2-Ln2-{\frac {1}{2}}Lnn-{\frac {4n}{3}}Ln2-{\frac {\sqrt {2n}}{2}}Lnn}
θ
(
2
n
)
−
θ
(
n
)
>
(
2
n
−
4
n
3
−
1
)
L
n
2
−
(
1
2
+
2
n
2
)
L
n
n
{\displaystyle \theta \left(2n\right)-\theta \left(n\right)>\left(2n-{\frac {4n}{3}}-1\right)Ln2-\left({\frac {1}{2}}+{\frac {\sqrt {2n}}{2}}\right)Lnn}
θ
(
2
n
)
−
θ
(
n
)
>
(
2
n
3
−
1
)
L
n
2
−
(
1
+
2
n
2
)
L
n
n
{\displaystyle \theta \left(2n\right)-\theta \left(n\right)>\left({\frac {2n}{3}}-1\right)Ln2-\left({\frac {1+{\sqrt {2n}}}{2}}\right)Lnn}
Li rete pou nou pwouve ke manm dwat la pi gran ke zero e konsa ant n ak 2n nap va asire ke gen o mwen yon nonm premye .
(
2
n
3
−
1
)
L
n
2
−
(
1
+
2
n
2
)
L
n
n
>
0
⇔
2
n
−
3
2
L
n
n
L
n
2
−
3
2
L
n
2
L
n
4
n
4
n
>
0
{\displaystyle \left({\frac {2n}{3}}-1\right)Ln2-\left({\frac {1+{\sqrt {2n}}}{2}}\right)Lnn>0\Leftrightarrow {\sqrt {2n}}-{\frac {3}{2}}{\frac {Lnn}{Ln2}}-{\frac {3{\sqrt {2}}}{Ln2}}{\frac {Ln{\sqrt {4n}}}{\sqrt {4n}}}>0}
2
n
−
3
2
L
n
n
L
n
2
−
3
2
L
n
2
L
n
4
n
4
n
>
0
⇔
2
4
n
L
n
2
×
2
n
−
3
4
n
L
n
n
−
6
2
L
n
4
n
2
4
n
L
n
2
>
0
⇔
4
n
L
n
2
×
2
n
−
3
4
n
L
n
n
−
6
2
L
n
4
n
>
0
{\displaystyle {\sqrt {2n}}-{\frac {3}{2}}{\frac {Lnn}{Ln2}}-{\frac {3{\sqrt {2}}}{Ln2}}{\frac {Ln{\sqrt {4n}}}{\sqrt {4n}}}>0\Leftrightarrow {\frac {2{\sqrt {4n}}Ln2\times {\sqrt {2n}}-3{\sqrt {4n}}Lnn-6{\sqrt {2}}Ln{\sqrt {4n}}}{2{\sqrt {4n}}Ln2}}>0\Leftrightarrow {\sqrt {4n}}Ln2\times {\sqrt {2n}}-3{\sqrt {4n}}Lnn-6{\sqrt {2}}Ln{\sqrt {4n}}>0}
4
n
L
n
2
×
2
n
−
3
4
n
L
n
n
−
6
2
L
n
4
n
>
0
⇔
2
8
n
2
L
n
2
−
6
n
L
n
n
−
6
2
L
n
4
n
⇔
4
n
2
L
n
2
−
6
n
L
n
n
−
6
2
L
n
4
n
>
0
⇔
4
n
2
L
n
2
−
6
n
L
n
n
−
6
2
(
L
n
4
+
L
n
n
)
⇔
4
n
2
L
n
2
−
6
n
L
n
n
−
6
2
(
L
n
2
+
1
2
L
n
2
)
>
0
{\displaystyle {\sqrt {4n}}Ln2\times {\sqrt {2n}}-3{\sqrt {4n}}Lnn-6{\sqrt {2}}Ln{\sqrt {4n}}>0\Leftrightarrow 2{\sqrt {8n^{2}}}Ln2-6{\sqrt {n}}Lnn-6{\sqrt {2}}Ln{\sqrt {4n}}\Leftrightarrow 4n{\sqrt {2}}Ln2-6{\sqrt {n}}Lnn-6{\sqrt {2}}Ln{\sqrt {4n}}>0\Leftrightarrow 4n{\sqrt {2}}Ln2-6{\sqrt {n}}Lnn-6{\sqrt {2}}\left(Ln{\sqrt {4}}+Ln{\sqrt {n}}\right)\Leftrightarrow 4n{\sqrt {2}}Ln2-6{\sqrt {n}}Lnn-6{\sqrt {2}}\left(Ln2+{\frac {1}{2}}Ln2\right)>0}
⇔
4
n
2
L
n
2
−
6
n
L
n
n
−
6
2
(
L
n
2
+
1
2
L
n
2
)
>
0
⇔
4
n
6
2
L
n
2
−
n
L
n
n
−
2
L
n
2
−
2
2
L
n
n
>
0
⇔
2
n
3
2
L
n
2
−
n
L
n
n
−
2
L
n
2
−
2
2
L
n
n
>
0
⇔
2
×
(
2
n
3
L
n
2
−
2
2
n
L
n
n
−
L
n
2
−
L
n
n
)
>
0
{\displaystyle \Leftrightarrow 4n{\sqrt {2}}Ln2-6{\sqrt {n}}Lnn-6{\sqrt {2}}\left(Ln2+{\frac {1}{2}}Ln2\right)>0\Leftrightarrow {\frac {4n}{6}}{\sqrt {2}}Ln2-{\sqrt {n}}Lnn-{\sqrt {2}}Ln2-{\frac {\sqrt {2}}{2}}Lnn>0\Leftrightarrow {\frac {2n}{3}}{\sqrt {2}}Ln2-{\sqrt {n}}Lnn-{\sqrt {2}}Ln2-{\frac {\sqrt {2}}{2}}Lnn>0\Leftrightarrow {\sqrt {2}}\times \left({\frac {2n}{3}}Ln2-{\frac {\sqrt {2}}{2}}{\sqrt {n}}Lnn-Ln2-Ln{\sqrt {n}}\right)>0}
⇔
2
×
(
2
n
3
L
n
2
−
2
2
n
L
n
n
−
L
n
2
−
L
n
n
)
>
0
⇔
(
2
n
3
−
1
)
L
n
2
−
2
2
n
L
n
n
−
L
n
n
>
0
⇔
(
2
n
3
−
1
)
L
n
2
−
2
2
n
L
n
n
−
1
2
L
n
n
>
0
⇔
(
2
n
3
−
1
)
L
n
2
−
(
1
+
2
n
2
)
L
n
n
>
0
{\displaystyle \Leftrightarrow {\sqrt {2}}\times \left({\frac {2n}{3}}Ln2-{\frac {\sqrt {2}}{2}}{\sqrt {n}}Lnn-Ln2-Ln{\sqrt {n}}\right)>0\Leftrightarrow \left({\frac {2n}{3}}-1\right)Ln2-{\frac {\sqrt {2}}{2}}{\sqrt {n}}Lnn-Ln{\sqrt {n}}>0\Leftrightarrow \left({\frac {2n}{3}}-1\right)Ln2-{\frac {\sqrt {2}}{2}}{\sqrt {n}}Lnn-{\frac {1}{2}}Lnn>0\Leftrightarrow \left({\frac {2n}{3}}-1\right)Ln2-\left({\frac {1+{\sqrt {2n}}}{2}}\right)Lnn>0}
An nou deziye pa
f
(
n
)
{\displaystyle f\left(n\right)}
2
n
−
3
2
L
n
n
L
n
2
−
3
2
L
n
2
L
n
4
n
4
n
{\displaystyle {\sqrt {2n}}-{\frac {3}{2}}{\frac {Lnn}{Ln2}}-{\frac {3{\sqrt {2}}}{Ln2}}{\frac {Ln{\sqrt {4n}}}{\sqrt {4n}}}}
An nou montre ke li pozitif pou tout
n
≥
2
6
{\displaystyle n\geq 2^{6}}
R
{\displaystyle \mathbb {R} }
f
(
x
)
=
2
x
−
3
2
L
n
x
L
n
2
−
3
2
L
n
2
L
n
4
x
4
x
{\displaystyle f\left(x\right)={\sqrt {2x}}-{\frac {3}{2}}{\frac {Lnx}{Ln2}}-{\frac {3{\sqrt {2}}}{Ln2}}{\frac {Ln{\sqrt {4x}}}{\sqrt {4x}}}}
f
(
2
6
)
>
0
{\displaystyle f\left(2^{6}\right)>0}
f
(
2
6
)
=
2
3
×
2
−
3
2
×
6
L
n
2
L
n
2
−
3
2
L
n
2
×
L
n
4
×
2
6
4
×
2
6
=
8
2
−
9
−
3
2
L
n
2
×
4
L
n
2
2
4
=
8
2
−
9
−
12
2
2
4
=
8
2
−
9
−
3
2
4
=
32
2
−
36
−
3
2
4
=
29
2
−
36
4
{\displaystyle f\left(2^{6}\right)=2^{3}\times {\sqrt {2}}-{\frac {3}{2}}\times {\frac {6Ln2}{Ln2}}-{\frac {3{\sqrt {2}}}{Ln2}}\times {\frac {Ln{\sqrt {4\times 2^{6}}}}{\sqrt {4\times 2^{6}}}}=8{\sqrt {2}}-9-{\frac {3{\sqrt {2}}}{Ln2}}\times {\frac {4Ln2}{2^{4}}}=8{\sqrt {2}}-9-{\frac {12{\sqrt {2}}}{2^{4}}}=8{\sqrt {2}}-9-{\frac {3{\sqrt {2}}}{4}}={\frac {32{\sqrt {2}}-36-3{\sqrt {2}}}{4}}={\frac {29{\sqrt {2}}-36}{4}}}
f
(
2
6
)
=
29
2
−
36
4
=
(
29
2
−
36
)
×
(
29
2
+
36
)
4
×
1
29
2
+
36
=
(
29
×
29
×
2
−
36
×
36
)
4
×
1
29
2
+
36
=
386
4
×
1
29
2
+
36
>
0
{\displaystyle f\left(2^{6}\right)={\frac {29{\sqrt {2}}-36}{4}}={\frac {\left({29{\sqrt {2}}-36}\right)\times \left({29{\sqrt {2}}+36}\right)}{4}}\times {\frac {1}{29{\sqrt {2}}+36}}={\frac {\left(29\times 29\times 2-36\times 36\right)}{4}}\times {\frac {1}{29{\sqrt {2}}+36}}={\frac {386}{4}}\times {\frac {1}{29{\sqrt {2}}+36}}>0}
an nou pwouve tou ke
f
′
(
x
)
>
0
{\displaystyle f'\left(x\right)>0}
n
≥
2
6
{\displaystyle n\geq 2^{6}}
kidonk fonksyon f la estriteman kwasant
(
2
x
)
′
=
(
(
2
x
)
1
2
)
=
1
2
(
2
x
)
−
1
2
×
2
=
(
2
x
)
−
1
2
=
1
2
x
=
2
x
2
x
×
2
x
=
2
x
2
x
{\displaystyle \left({\sqrt {2x}}\right)'=\left(\left(2x\right)^{\frac {1}{2}}\right)={\frac {1}{2}}\left(2x\right)^{-{\frac {1}{2}}}\times 2=\left(2x\right)^{-{\frac {1}{2}}}={\frac {1}{\sqrt {2x}}}={\frac {\sqrt {2x}}{{\sqrt {2x}}\times {\sqrt {2x}}}}={\frac {\sqrt {2x}}{2x}}}
(
−
3
2
L
n
x
L
n
2
)
′
=
−
3
2
L
n
2
×
1
x
=
−
3
2
x
L
n
2
{\displaystyle \left(-{\frac {3}{2}}{\frac {Lnx}{Ln2}}\right)'=-{\frac {3}{2Ln2}}\times {\frac {1}{x}}={\frac {-3}{2xLn2}}}
(
4
x
)
′
=
(
(
4
x
)
1
2
)
′
=
1
2
×
(
4
x
)
1
2
×
4
=
2
×
(
4
x
)
1
2
=
2
4
x
=
2
4
x
4
x
=
4
x
2
x
=
x
x
{\displaystyle \left({\sqrt {4x}}\right)'=\left(\left(4x\right)^{\frac {1}{2}}\right)'={\frac {1}{2}}\times \left(4x\right)^{\frac {1}{2}}\times 4=2\times \left(4x\right)^{\frac {1}{2}}={\frac {2}{\sqrt {4x}}}={\frac {2{\sqrt {4x}}}{4x}}={\frac {\sqrt {4x}}{2x}}={\frac {\sqrt {x}}{x}}}
(
L
n
4
x
)
′
=
1
4
x
×
4
x
2
x
=
1
2
x
{\displaystyle \left(Ln{\sqrt {4x}}\right)'={\frac {1}{\sqrt {4x}}}\times {\frac {\sqrt {4x}}{2x}}={\frac {1}{2x}}}
oubyen
(
L
n
4
x
)
′
=
(
L
n
(
4
x
)
1
2
)
′
=
(
1
2
L
n
4
x
)
′
=
1
2
×
1
4
x
×
4
=
4
8
x
=
1
2
x
{\displaystyle \left(Ln{\sqrt {4x}}\right)'=\left(Ln\left(4x\right)^{\frac {1}{2}}\right)'=\left({\frac {1}{2}}Ln{4x}\right)'={\frac {1}{2}}\times {\frac {1}{4x}}\times 4={\frac {4}{8x}}={\frac {1}{2x}}}
(
L
n
4
x
4
x
)
′
=
1
2
x
×
4
x
−
4
x
2
x
×
L
n
4
x
(
4
x
)
2
=
4
x
−
4
x
×
L
n
4
x
2
x
4
x
=
4
x
(
1
−
L
n
4
x
)
8
x
2
{\displaystyle \left({\frac {Ln{\sqrt {4x}}}{\sqrt {4x}}}\right)'={\frac {{\frac {1}{2x}}\times {\sqrt {4x}}-{\frac {\sqrt {4x}}{2x}}\times Ln{\sqrt {4x}}}{\left({\sqrt {4x}}\right)^{2}}}={\frac {\frac {{\sqrt {4x}}-{\sqrt {4x}}\times Ln{\sqrt {4x}}}{2x}}{4x}}={\frac {{\sqrt {4x}}\left(1-Ln{\sqrt {4x}}\right)}{8x^{2}}}}
(
−
3
2
L
n
2
×
L
n
4
x
4
x
)
′
=
4
x
−
4
x
×
L
n
4
x
2
x
4
x
=
−
3
2
L
n
2
×
4
x
(
1
−
L
n
4
x
)
8
x
2
{\displaystyle \left({\frac {-3{\sqrt {2}}}{Ln2}}\times {\frac {Ln{\sqrt {4x}}}{\sqrt {4x}}}\right)'={\frac {\frac {{\sqrt {4x}}-{\sqrt {4x}}\times Ln{\sqrt {4x}}}{2x}}{4x}}={\frac {-3{\sqrt {2}}}{Ln2}}\times {\frac {{\sqrt {4x}}\left(1-Ln{\sqrt {4x}}\right)}{8x^{2}}}}
f
′
(
x
)
=
2
x
2
x
−
3
2
x
l
n
2
−
3
2
L
n
2
×
4
x
×
(
1
−
L
n
4
x
)
8
x
2
{\displaystyle f'\left(x\right)={\frac {\sqrt {2x}}{2x}}-{\frac {3}{2xln2}}-{\frac {3{\sqrt {2}}}{Ln2}}\times {\frac {{\sqrt {4x}}\times \left(1-Ln{\sqrt {4x}}\right)}{8x^{2}}}}
f
′
(
x
)
=
4
x
×
L
n
2
×
2
x
−
12
x
−
3
8
x
(
1
−
L
n
4
x
)
8
x
2
L
n
2
{\displaystyle f'\left(x\right)={\frac {4x\times Ln2\times {\sqrt {2x}}-12x-3{\sqrt {8x}}\left(1-Ln{\sqrt {4x}}\right)}{8x^{2}Ln2}}}
f
′
(
x
)
=
4
x
×
L
n
2
×
2
x
−
12
x
−
3
8
x
+
3
8
x
l
n
4
x
8
x
2
L
n
2
{\displaystyle f'\left(x\right)={\frac {4x\times Ln2\times {\sqrt {2x}}-12x-3{\sqrt {8x}}+3{\sqrt {8x}}ln{\sqrt {4x}}}{8x^{2}Ln2}}}
f
′
(
x
)
>
0
⇔
4
x
×
L
n
2
×
2
x
−
12
x
−
3
8
x
+
3
8
x
L
n
4
x
8
x
2
L
n
2
>
0
⇔
4
x
×
L
n
2
×
2
x
+
3
8
x
×
L
n
4
x
>
12
x
+
3
8
x
{\displaystyle f'\left(x\right)>0\Leftrightarrow {\frac {4x\times Ln2\times {\sqrt {2x}}-12x-3{\sqrt {8x}}+3{\sqrt {8x}}Ln{\sqrt {4x}}}{8x^{2}Ln2}}>0\Leftrightarrow 4x\times Ln2\times {\sqrt {2x}}+3{\sqrt {8x}}\times Ln{\sqrt {4x}}>12x+3{\sqrt {8x}}}
2
x
≠
0
{\displaystyle {\sqrt {2x}}\neq 0}
f
′
(
x
)
>
0
⇔
4
x
×
L
n
2
×
2
x
−
12
x
−
3
8
x
+
3
8
x
L
n
4
x
8
x
2
L
n
2
>
0
⇔
4
x
×
L
n
2
×
2
x
+
3
8
x
×
L
n
4
x
>
12
x
+
3
8
x
⇔
4
x
L
n
2
×
2
x
+
3
×
4
x
×
L
n
4
x
>
12
x
×
2
x
+
3
×
4
x
⇔
8
x
L
n
2
+
12
L
n
4
x
>
12
2
x
+
12
{\displaystyle f'\left(x\right)>0\Leftrightarrow {\frac {4x\times Ln2\times {\sqrt {2x}}-12x-3{\sqrt {8x}}+3{\sqrt {8x}}Ln{\sqrt {4x}}}{8x^{2}Ln2}}>0\Leftrightarrow 4x\times Ln2\times {\sqrt {2x}}+3{\sqrt {8x}}\times Ln{\sqrt {4x}}>12x+3{\sqrt {8x}}\Leftrightarrow 4xLn2\times 2x+3\times 4x\times Ln{\sqrt {4x}}>12x\times {\sqrt {2x}}+3\times 4x\Leftrightarrow 8xLn2+12Ln{\sqrt {4x}}>12{\sqrt {2x}}+12}
x
>
2
6
{\displaystyle x>2^{6}}
8
x
L
n
2
+
12
L
n
4
x
>
12
2
x
+
12
{\displaystyle 8xLn2+12Ln{\sqrt {4x}}>12{\sqrt {2x}}+12}
x
>
2
6
⇒
x
>
18
⇒
16
x
>
288
⇒
16
x
×
L
n
4
×
L
n
4
>
288
×
L
n
e
×
L
n
e
⇒
2
×
2
×
L
n
2
×
L
n
2
×
16
x
⇒
64
x
×
(
l
n
2
)
2
>
288
⇒
64
x
2
×
(
l
n
2
)
2
>
288
x
⇒
64
x
2
×
(
l
n
2
)
2
>
144
×
2
x
⇒
8
x
l
n
2
>
12
2
x
{\displaystyle x>2^{6}\Rightarrow x>18\Rightarrow 16x>288\Rightarrow 16x\times Ln4\times Ln4>288\times Lne\times Lne\Rightarrow 2\times 2\times Ln2\times Ln2\times 16x\Rightarrow 64x\times \left(ln2\right)^{2}>288\Rightarrow 64x^{2}\times \left(ln2\right)^{2}>288x\Rightarrow 64x^{2}\times \left(ln2\right)^{2}>144\times 2x\Rightarrow 8xln2>12{\sqrt {2x}}}
yon lòt kote nou genyen
12
L
n
4
x
>
12
⇔
L
n
4
x
>
1
⇔
s
q
r
t
4
x
>
e
⇔
4
x
>
e
2
⇔
x
>
e
2
4
{\displaystyle 12Ln{\sqrt {4x}}>12\Leftrightarrow Ln{\sqrt {4x}}>1\Leftrightarrow sqrt{4x}>e\Leftrightarrow 4x>e^{2}\Leftrightarrow x>{\frac {e^{2}}{4}}}
non genyen tou :
e
<
3
{\displaystyle e<3}
0
<
e
<
3
⇒
e
2
<
9
⇒
e
2
4
<
9
4
{\displaystyle 0<e<3\Rightarrow e^{2}<9\Rightarrow {\frac {e^{2}}{4}}<{\frac {9}{4}}}
2
6
>
9
4
>
e
2
4
{\displaystyle 2^{6}>{\frac {9}{4}}>{\frac {e^{2}}{4}}}
x
>
2
6
⇒
x
>
e
2
4
⇒
4
x
>
e
2
⇒
4
x
>
e
⇒
L
n
4
x
>
1
⇒
12
L
n
4
x
>
12
{\displaystyle x>2^{6}\Rightarrow x>{\frac {e^{2}}{4}}\Rightarrow 4x>e^{2}\Rightarrow {\sqrt {4x}}>e\Rightarrow Ln{\sqrt {4x}}>1\Rightarrow 12Ln{\sqrt {4x}}>12}
donk
x
>
2
6
⇒
8
x
L
n
2
+
12
L
n
4
x
>
12
2
x
+
12
{\displaystyle x>2^{6}\Rightarrow 8xLn2+12Ln{\sqrt {4x}}>12{\sqrt {2x}}+12}
modifye
3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61
modifye
P
n
≤
2
n
{\displaystyle P_{n}\leq {2^{n}}}
P
1
=
2
=
2
1
{\displaystyle P_{1}=2=2^{1}}
2
2
=
4
{\displaystyle 2^{2}=4}
P
2
=
3
<
2
2
{\displaystyle P_{2}=3<2^{2}}
P
3
=
5
{\displaystyle P_{3}=5}
3
2
{\displaystyle 3^{2}}
P
3
<
3
2
{\displaystyle P_{3}<3^{2}}
n
>
3
{\displaystyle n>3}
P
n
≤
2
n
⇒
π
(
P
n
)
≤
π
(
2
n
)
⇒
n
≤
π
(
2
n
)
≥
n
{\displaystyle P_{n}\leq 2^{n}\Rightarrow \pi \left(P_{n}\right)\leq \pi \left(2^{n}\right)\Rightarrow n\leq \pi \left(2^{n}\right)\geq n}
kidonk nan entèval \left[1, 2^n\right] gen o mwen n nonm premye selon ipotèz rekirans lan .
]
2
n
,
2
n
+
1
[
{\displaystyle \left]2^{n},2^{n+1}\right[}
]
2
n
,
2
n
+
1
]
{\displaystyle \left]2^{n},2^{n+1}\right]}
π
(
2
n
+
1
)
−
π
(
2
n
)
≥
1
{\displaystyle \pi \left(2^{n+1}\right)-\pi \left(2^{n}\right)\geq 1}
π
(
2
n
+
1
)
≥
1
+
π
(
2
n
)
{\displaystyle \pi \left(2^{n+1}\right)\geq 1+\pi \left(2^{n}\right)}
π
(
2
n
+
1
)
≥
1
+
n
{\displaystyle \pi \left(2^{n+1}\right)\geq 1+n}
[
1
,
2
n
+
1
]
{\displaystyle \left[1,2^{n+1}\right]}
P
n
+
1
<
2
n
+
1
{\displaystyle P_{n+1}<2^{n+1}}
Anfendkont si majorasyon an vrè pou n, li vrè pou n + 1 .
∀
n
∈
N
,
P
n
≤
2
n
{\displaystyle \forall n\in \mathbb {N} ,P_{n}\leq 2^{n}}
.....
P
k
(
n
)
≤
n
×
2
k
{\displaystyle P_{k}\left(n\right)\leq {n\times {2^{k}}}}
P
k
(
n
)
{\displaystyle P_{k}\left(n\right)}
k
+
1
{\displaystyle k+1}
P
k
(
n
)
≤
n
×
2
k
{\displaystyle P_{k}\left(n\right)\leq n\times 2^{k}}
Anvan nou fè sipozisyon an , an nou verifye ke majorasyon vre pou k = 1
P
1
(
n
)
≤
2
n
{\displaystyle P_{1}\left(n\right)\leq 2n}
]
n
,
2
n
[
{\displaystyle \left]n,2n\right[}
Anfèt si n premye,
P
1
(
n
)
=
n
{\displaystyle P_{1}\left(n\right)=n}
si n pa premye
P
1
(
n
)
∈
]
n
,
2
n
]
{\displaystyle P_{1}\left(n\right)\in \left]n,2n\right]}
P
1
(
n
)
∈
[
n
,
2
n
]
{\displaystyle P_{1}{\left(n\right)}\in \left[n,2n\right]}
P
1
(
n
)
≤
2
n
{\displaystyle P_{1}\left(n\right)\leq 2n}
modifye
modifye
P
n
=
∑
m
=
1
2
n
(
[
1
+
∑
m
=
1
i
(
−
[
(
m
!
)
2
m
3
]
−
[
−
(
m
!
)
2
m
3
]
)
n
+
1
]
×
[
n
+
1
1
+
∑
m
=
1
i
(
−
[
(
m
!
)
2
m
3
]
−
[
−
(
m
!
)
2
m
3
]
)
]
×
i
×
(
−
[
(
i
!
)
2
i
3
]
−
[
−
(
i
!
)
2
i
3
]
)
)
{\displaystyle P_{n}=\sum _{m=1}^{2^{n}}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(-\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]-\left[-{\frac {\left(m!\right)^{2}}{m^{3}}}\right]\right)}}{n+1}}\right]\times \left[{\frac {n+1}{1+\sum _{m=1}^{i}{\left(-\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]-\left[-{\frac {\left(m!\right)^{2}}{m^{3}}}\right]\right)}}}\right]\times i\times \left(-\left[{\frac {\left(i!\right)^{2}}{i^{3}}}\right]-\left[-{\frac {\left(i!\right)^{2}}{i^{3}}}\right]\right)\right)}}
modifye
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{\displaystyle P_{K}\left(n\right)=\left(\sum _{i=1}^{2^{k}\times n}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{k+1+\sum _{m=1}^{n}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times \left[{\frac {k+1+\sum _{m=1}^{n}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times i\times \left(1-\left[{\frac {\left[{\frac {{\left(i!\right)}^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)\right)}\right)\times \left[{\frac {\left[{\frac {{\left(n!\right)}^{2}}{n^{3}}}\right]}{\frac {{\left(n!\right)}^{2}}{n^{3}}}}\right]+{\left(\sum _{i=1}^{2^{k}\times n}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{k+\sum _{m=1}^{n}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times \left[{\frac {k+\sum _{m=1}^{n}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times i\times \left(1-\left[{\frac {\left[{\frac {{\left(i!\right)}^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)\right)}\right)\times \left(1-\left[{\frac {\left[{\frac {{\left(n!\right)}^{2}}{n^{3}}}\right]}{\frac {{\left(n!\right)}^{2}}{n^{3}}}}\right]\right)}}
nan de evenman kontrè nou gen fòseman youn. Nou kapab di ke oubyen :
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nou gen asirans ke
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{\displaystyle P_{K}\left(n\right)}
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{\displaystyle i=P_{K}\left(n\right)}
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{\displaystyle P_{K}\left(n\right)=\left(\sum _{i=1}^{2^{k}\times n}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{k+1+\sum _{m=1}^{n-1}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times \left[{\frac {k+1+\sum _{m=1}^{n-1}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times i\times \left(1-\left[{\frac {\left[{\frac {{\left(i!\right)}^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)\right)}\right)}
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{\displaystyle P_{k}\left(n\right)=\sum _{i=n}^{2^{k}\times n}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}{k+1}}\right]\times \left[{\frac {k+1}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}}\right]\times i\times \left(1-\left[{\frac {\left[{\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)\right)}}
![{\displaystyle ]{2^{r-1}},2^{r}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8f591acbd3b711650b9a12d57fe28b72862c12c)
![{\displaystyle ]2^{r},2^{r+1}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f29a600eab9f8fca4ac82d19d84363d62b62e75b)
![{\displaystyle ]2,4]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15f612a16960e64caa44ba64284e443be479eec4)
![{\displaystyle [2,4]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e34ad1554b1763ca516be3e0343c4e5b7503a6a0)
![{\displaystyle ]2^{1},2^{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/133799dd2f395db0ec84cad65425515fe27156ed)
![{\displaystyle ]2^{r-1},2^{r}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9dd625d8dbf34f34325a41b1ce4a82b36023bc1)
![{\displaystyle m\in ]2^{r},2^{r+1}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6cc4d25532a61577e7ca9ce3148ba95495f409a9)
![{\displaystyle m\in ]2^{r},2^{r+1}]\Leftrightarrow {2^{r}<m\leq {2^{r+1}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f245c224637879c83c2e9ca41628d9781376edb4)
![{\displaystyle \{_{m={2n}}^{2^{r}{<}m\leq {2^{r+1}}}\Rightarrow {2^{r}{<}{2n}\leq {2^{r+1}}}\Rightarrow {2\times {2^{r-1}}{<}{2n}\leq {2\times 2^{r}}}\Rightarrow {2^{r-1}{<}n\leq {2^{r}}}\Rightarrow n\in ]2^{r-1},2^{r}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b304b5d3e9de27acc15800459afbe734f5aca41)
![{\displaystyle ]{2^{r}},{2^{r+1}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c4165e2688b7fa402d132de270fd663a463701b)
![{\displaystyle \{_{m={2n-1}}^{2^{r}{<}m\leq {2^{r+1}}}\Rightarrow {2^{r}{<}{2n-1}\leq {2^{r+1}}}\Rightarrow {{2^{r}+1}<{2n}\leq {2^{r+1}+1}}\Rightarrow {\{_{2^{r}<{2n}<2^{r+1}+1}^{{2n}\neq 2^{r+1}+1}}\Rightarrow {2^{r}{<}{2n}\leq {2^{r+1}}}\Rightarrow {\frac {2^{r}}{2}}{<}n\leq {\frac {2^{r+1}}{2}}\Rightarrow {2^{r-1}{<}n\leq {2^{r}}}\Rightarrow n\in ]2^{r-1},2^{r}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/803ad68ecf136c95cf8c0a7f3bc76014ad391915)
![{\displaystyle k=\sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fedb01263db24849146522ac2a86fc4bd0d2eb89)
![{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {1}{p^{i}}}\right]}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38198393e4c46c55e91ba299323abb66a603c1a7)
![{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\alpha }{\left[{\frac {n+1}{p^{i}}}\right]}+\sum _{i=\alpha +1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28a52a319fe12dcd11d5d9e9f19bf1e4c95c8e21)
![{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\alpha }{\left(\left[{\frac {n}{p^{i}}}\right]+1\right)}+\sum _{i=\alpha +1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/249cf00a9072f12dec96b7692f553c49482c1999)
![{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\alpha }{\left[{\frac {n}{p^{i}}}\right]}+\sum _{i=1}^{\alpha }{1}+\sum _{i=\alpha +1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfba51de17f45d9d6cdb267cdcb7691b562af130)
![{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\alpha }{\left[{\frac {n}{p^{i}}}\right]}+\alpha +\sum _{i=\alpha +1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/172407fc7046a400f58dff01c8fc9bb5942c4060)
![{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\alpha }{\left[{\frac {n}{p^{i}}}\right]}+\sum _{i=\alpha +1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}+\alpha }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6759ff1c870489fb97f94687cf2266f600977f1f)
![{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}+{\alpha }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c588a3a777916c363813a6d0085ae51be803b2f0)
![{\displaystyle {\sum _{i=1}^{\infty }{\left[{\frac {n+1}{p^{i}}}\right]}}={\alpha +k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9796c5f910f67f6c5c14a21f610b8bb551c5e6b)
![{\displaystyle ]{n},{2n}[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c333ef236e8d7fa7e323ec810d9021050c8c9bfe)
![{\displaystyle ]n,2n[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4437a9c0421c814c185fbf65dca24eb64eb0c75)
![{\displaystyle ]2,2n]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0435965d71f35e358a40cb670817a93f25aba8d7)
![{\displaystyle N={\frac {{2n}!}{{n!}\times \left(n!\right)}}={\frac {\prod _{p\leq {2n}}{\left(p^{\sum _{i=1}^{\infty }{\left[{\frac {2n}{p^{i}}}\right]}}\right)}}{\prod _{p\leq n}{\left(p^{2\times \sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}\right)}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ffafde4e7581d74be4d3086c7d0bf1861adca41)
![{\displaystyle \prod _{n<p\leq {2n}}{p^{2\times \sum _{i=1}^{\infty }{\left(\left[{\frac {n}{p^{i}}}\right]\right)}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86218acaf124bd0d12fb0aa389ee84fb30e65115)
![{\displaystyle \forall n>1,{n<p\leq {2n}}\Rightarrow {n<p<{2n}}\Rightarrow {{\frac {1}{2n}}<{\frac {1}{p}}<{\frac {1}{n}}}\Rightarrow {{\frac {n}{2n}}<{\frac {n}{p}}<{\frac {n}{n}}}\Rightarrow {{\frac {1}{2}}<{\frac {n}{p}}<1}\Rightarrow {\left[{\frac {n}{p}}\right]=0}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d81d8a45383ff1e42d94c4ebdfbc81d9f793fc5)
![{\displaystyle \left[{\frac {n}{p^{i}}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/940ba3f69b301f7a091e63cc8223bc09d450e43c)
![{\displaystyle \prod _{n<p\leq {2n}}{p^{2\times \sum _{i=1}^{\infty }{\left(\left[{\frac {n}{p^{i}}}\right]\right)}}}=\prod _{n<p\leq {2n}}{p^{2\times 0}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1765ae939a77710fba55fd5ef075a0750ab18ae)
![{\displaystyle N={\frac {{2n}!}{{n!}\times \left(n!\right)}}={\frac {\prod _{p\leq {2n}}{\left(p^{\sum _{i=1}^{\infty }{\left[{\frac {2n}{p^{i}}}\right]}}\right)}}{{\prod _{p\leq n}{\left(p^{2\times \sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}\right)}}\times {\prod _{n<p\leq {2n}}{\left(p^{2\times \sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}\right)}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/baf38f1ced8d3069c4988e7ed754dd0463a22bff)
![{\displaystyle N={\frac {{2n}!}{{n!}\times \left(n!\right)}}={\frac {\prod _{p\leq {2n}}{\left(p^{\sum _{i=1}^{\infty }{\left[{\frac {2n}{p^{i}}}\right]}}\right)}}{\prod _{p\leq {2n}}{\left(p^{2\times \sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}}\right)}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84ae7968f8b1dbc9cb3ca20a6460f78e14e06153)
![{\displaystyle N={\frac {{2n}!}{{n!}\times \left(n!\right)}}=\prod _{p\leq {2n}}{\left(p^{\left(\sum _{i=1}^{\infty }{\left[{\frac {2n}{p^{i}}}\right]}-2\sum _{i=1}^{\infty }{\left[{\frac {n}{p^{i}}}\right]}\right)}\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bbfc566628dcddafd96e8f154bc29e396cc5e72)
![{\displaystyle N=\prod _{p\leq {2n}}{\left(p^{\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/007116753caec853d344b0b2fb3ef2cc6c5d05e5)
![{\displaystyle LnN=Ln\left(\prod _{p\leq {2n}}{\left(p^{\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\right)}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16feb78703226cf7900cc70b5ffa5debc56f5264)
![{\displaystyle LnN=\sum _{p\leq {2n}}{\left(Ln\left({\left(p^{\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\right)}\right)\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1355dd2d2eb8dbba0881291b55fceb247efb7aa2)
![{\displaystyle LnN=\sum _{p\leq {2n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f0f0ae3801d6ac77471f9bc33b55c84106e245b)
![{\displaystyle LnN=\sum _{n<p\leq {2n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+\sum _{{\frac {2n}{3}}<p\leq {n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1860fb686f4c9116bb13417cdb8074042e3143aa)
![{\displaystyle \sum _{n<p\leq {2n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{n<p\leq {2n}}{\left(Lnp\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/25e7e4589951f730965d42c58213c3f335b3f5ad)
![{\displaystyle \sum _{n<p\leq {2n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{1\leq {p}\leq {2n}}{\left(Lnp\right)}-\sum _{1\leq {p}\leq {n}}{\left(Lnp\right)}=\theta \left(2n\right)-\theta \left(n\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/17dd467929fe570318939058bdce0efdcd53005a)
![{\displaystyle n<p\leq {2n}\Rightarrow {{\frac {1}{2n}}\leq {\frac {1}{p}}<{\frac {1}{n}}}\Rightarrow {{\frac {n}{2n}}\leq {\frac {n}{p}}<{\frac {n}{n}}}\Rightarrow {{\frac {1}{2}}\leq {\frac {n}{p}}<1}\Rightarrow \left[{\frac {n}{p}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90cc936a27d2072a6e3c34a212812d8ece5b06bc)
![{\displaystyle n<p\leq {2n}\Rightarrow {{\frac {1}{2n}}\leq {\frac {1}{p}}<{\frac {1}{n}}}\Rightarrow {{\frac {2n}{2n}}\leq {\frac {2n}{p}}<{\frac {2n}{n}}}\Rightarrow {1\leq {\frac {2n}{p}}<2}\Rightarrow \left[{\frac {2n}{p}}\right]=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/635e7b69ce89b70822a4b244dabfc5b4f0318ab7)
![{\displaystyle i>2,\left[{\frac {2n}{p^{i}}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edc232d51fc3c7eb69724eeeb2c7208378f73e9e)
![{\displaystyle \sum _{{\frac {2n}{3}}<p\leq {n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7dfd7f680a4d78e0ff063e4b11b22236e6694c82)
![{\displaystyle {\frac {2n}{3}}<p\leq n\Rightarrow {{\frac {1}{n}}\leq {\frac {1}{p}}<{\frac {3}{2n}}}\Rightarrow {{\frac {n}{n}}\leq {\frac {n}{p}}<{\frac {3n}{2n}}}\Rightarrow 1\leq {\frac {n}{p}}<{\frac {3}{2}}\Rightarrow \left[{\frac {n}{p}}\right]=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d16903c530b30c1a77a2dcc0251190db2aee7b0f)
![{\displaystyle {\frac {2n}{3}}<p\leq n\Rightarrow {{\frac {1}{n}}\leq {\frac {1}{p}}<{\frac {3}{2n}}}\Rightarrow {{\frac {2n}{n}}\leq {\frac {2n}{p}}<3}\Rightarrow {2\leq {\frac {2n}{p}}<3}\Rightarrow \left[{\frac {2n}{p}}\right]=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bb95c67628e331d0fe91584a6065e61f2dc7e24)
![{\displaystyle \left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]=2-2\times 1=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d1b84785bdcff13bbf5ccbd337c42e2b7b76497)
![{\displaystyle \left[{\frac {n}{p^{2}}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/055b36f5120d0dbce73c2bfa439afa92eed56b19)
![{\displaystyle \left[{\frac {2n}{p^{2}}}\right]=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fabaeff5598a96f919deebaac1553f1b8899c0e)
![{\displaystyle \left[{\frac {2n}{p^{2}}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41d8c3c326eee5a4f1db002640e98c6726a01b68)
![{\displaystyle \left[{\frac {2n}{p^{i}}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d169edbada5ebbe0782ae0af8a57ec849ab3c4b)
![{\displaystyle \left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c43a5d441fa009cf4043757e1791351e33e05103)
![{\displaystyle \sum _{{\frac {2n}{3}}<p\leq {n}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22bbfcff637a58f12e1549f2f97b5340866de1f8)
![{\displaystyle \sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2a091d2bfca3c601f5b71c143f410c195d2608a)
![{\displaystyle \left[2\times {\frac {n}{p^{r}}}\right]-2\left[{\frac {n}{p^{r}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd2c2e39c7a39588ee0e52ba1aef2a3c7faa3f50)
![{\displaystyle \left[2x\right]-2\left[x\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08b8e2e2b175827b732d43c881bad9722ec192be)
![{\displaystyle x=\left[x\right]+\{x\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29fd94f2c0e9df08ab3da5227b618e655ec2e8c3)
![{\displaystyle 2x=2\left[x\right]+2\{x\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1337b3a5bf14f00fc7970c1d6210b5f3b0586329)
![{\displaystyle \left[2x\right]=\left[2\left[x\right]+2\{x\}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/244be4adf487b695d0e2cb0d73d1712e55cbffa4)
![{\displaystyle \left[2x\right]=2\left[x\right]+\left[2\{x\}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c41730aadee35e75841835fcbbc236157d51fb05)
![{\displaystyle \left[2x\right]-2\left[x\right]=\left[2\{x\}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e00ec8b9042486a5e2de8f1b8034a6cabff85d05)
![{\displaystyle \left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2f7cca621891f60c519c163533346951711acbb)
![{\displaystyle {\sqrt {2n}}<p\leq {\frac {2n}{3}}\Rightarrow 2n<p^{2}\leq {\frac {4n^{2}}{9}}\Rightarrow {\frac {9}{4n^{2}}}\leq {\frac {1}{p^{2}}}<{\frac {1}{2n}}\Rightarrow {\frac {9n}{4n^{2}}}\leq {\frac {n}{p^{2}}}<{\frac {n}{2n}}\Rightarrow {\frac {9n}{4n^{2}}}\leq {\frac {n}{p^{2}}}<{\frac {1}{2}}\Rightarrow \left[{\frac {n}{p^{2}}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7a164eba37c999e4ce0c617a442a9f1000ed20b)
![{\displaystyle \forall i\geq 2,\left[{\frac {2n}{p^{i}}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cedfd824ae3bd9b9699da20f2c5d10be72bf780)
![{\displaystyle \sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}+\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=2}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}+0=\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c42293ebda088c8194e3d289faffded05e7d53a)
![{\displaystyle \left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b14b1a224cd38cd9665c950cd8e9f49bc96e687)
![{\displaystyle \left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\leq 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cde3fa1f039843122fbe4bee3dd79db87b153e2f)
![{\displaystyle {\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}\leq Lnp}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac1001db7ca2a15419992a50aedf1ce24906aa7b)
![{\displaystyle {\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}}\leq {\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{Lnp}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e6aef5af41501460587983f7af15d4762fcc0d1)
![{\displaystyle {\sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left(\left(\left[{\frac {2n}{p}}\right]-2\left[{\frac {n}{p}}\right]\right)\times LnP\right)}}\leq {\theta \left({\frac {2n}{3}}\right)-\theta \left({\sqrt {2n}}\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68092ccaeabe6166cb71e9dc1040d92a88fe773c)
![{\displaystyle \sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq {\theta \left({\frac {2n}{3}}\right)-\theta \left({\sqrt {2n}}\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c65e2737272864b5d9acd04a00d272ac31ad608f)
![{\displaystyle \sum _{{\sqrt {2n}}<p\leq {\frac {2n}{3}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq {\theta \left({\frac {2n}{3}}\right)-\pi \left({\sqrt {2n}}\right)\times {Ln2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2147348ae36edfbbf341731ac48f3216b5c9a2ff)
![{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b352b2cac65d9d25417429b0f685d014a3dab3ba)
![{\displaystyle \left[{\frac {2n}{p^{i}}}\right]\geq {2\left[{\frac {n}{p^{i}}}\right]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/296f995c0b064d8ba21f1a02d2037d50e2e4323e)
![{\displaystyle \left[{\frac {2n}{p^{i}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83478bd2bc7f5ec97f3438538f1cfb3c21e93b57)
![{\displaystyle \left[{\frac {2n}{p^{i}}}\right]=0\Leftrightarrow {\frac {2n}{p^{i}}}<1\Leftrightarrow {2n}<{p^{i}}\Leftrightarrow {p^{i}}>{2n}\Leftrightarrow Ln{p^{i}}>Ln{2n}\Leftrightarrow iLn{p}>Ln{2n}\Leftrightarrow i>{\frac {Ln2n}{Lnp}}\Leftrightarrow i>\left[{\frac {Ln2n}{Lnp}}\right]\Leftrightarrow i\geq \left[{\frac {Ln2n}{Lnp}}\right]+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/19211dbf341e6d76f91458b054b85b360dcbd37e)
![{\displaystyle i\geq \left[{\frac {Ln2n}{Lnp}}\right]+1\Rightarrow \left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0bde8d2961842aa298dbb6ec94c501854742349a)
![{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\left[{\frac {Ln2n}{Lnp}}\right]}{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{\left[{\frac {Ln2n}{Lnp}}\right]+1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ad31ec7315742c60459b14cb9283de9a23fc1c7)
![{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\left[{\frac {Ln2n}{Lnp}}\right]}{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}+0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62b1e9a2ad6aa3b13606f7429e47bb04c6977923)
![{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}=\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\left[{\frac {Ln2n}{Lnp}}\right]}{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq {\sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\left[{\frac {Ln2n}{Lnp}}\right]}{1}\right)}\times Lnp\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8162600b1d19e1bff4b31543d5cb5efddef8462)
![{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{\left(\left[{\frac {Ln2n}{Lnp}}\right]\times LnP\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4bca397289e0740b3d5dfe7cec18d5785be4adb1)
![{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{\left(\left[{\frac {Ln2n}{Lnp}}\right]\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{\left({\frac {Ln2n}{Lnp}}\times Lnp\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1864943a0778a5334875f369cc0bb4f2bdc3e4ab)
![{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{\left(\left[{\frac {Ln2n}{Lnp}}\right]\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{Ln2n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfbe2605b2f26146a19bea5118b3de0f63836e03)
![{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq \sum _{p\leq {\sqrt {2n}}}{\left(\left[{\frac {Ln2n}{Lnp}}\right]\times Lnp\right)}\leq Ln2n\sum _{p\leq {\sqrt {2n}}}{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70d6889bcb76c6377a35d8a8d5749c2749929cf2)
![{\displaystyle \sum _{p\leq {\sqrt {2n}}}{\left({\left(\sum _{i=1}^{\infty }{\left(\left[{\frac {2n}{p^{i}}}\right]-2\left[{\frac {n}{p^{i}}}\right]\right)}\right)}\times Lnp\right)}\leq \pi \left({\sqrt {2n}}\right)Ln2n}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b496053aa30c0462c917dcb93728db2dbfacd0b6)
![{\displaystyle \theta \left({\frac {2n}{3}}\right)=\theta \left(\left[{\frac {2n}{3}}\right]\right)<2\times \left[{\frac {2n}{3}}\right]Ln2<{\frac {4n}{3}}Ln2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfb5f1e337999dd5aba0392d6f42d2ff7dcbd1bd)
![{\displaystyle \left]2^{n},2^{n+1}\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab9afe5408138bee32d771491a4651944ab68a13)
![{\displaystyle \left]2^{n},2^{n+1}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/709d483da89287160ce5654c98170abd66457354)
![{\displaystyle \left[1,2^{n+1}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/370a3667b5e738949a5fc2c21104a972d872cee2)
![{\displaystyle \left]n,2n\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68936d7aa492b16979322c367610a36907414b23)
![{\displaystyle P_{1}\left(n\right)\in \left]n,2n\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9fc9697d8efa73a0a83669c49cfd3f8dbb48d40d)
![{\displaystyle P_{1}{\left(n\right)}\in \left[n,2n\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab65b0c47d432109266bf3f20eecdafbdac8ac39)
![{\displaystyle P_{n}=\sum _{m=1}^{2^{n}}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(-\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]-\left[-{\frac {\left(m!\right)^{2}}{m^{3}}}\right]\right)}}{n+1}}\right]\times \left[{\frac {n+1}{1+\sum _{m=1}^{i}{\left(-\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]-\left[-{\frac {\left(m!\right)^{2}}{m^{3}}}\right]\right)}}}\right]\times i\times \left(-\left[{\frac {\left(i!\right)^{2}}{i^{3}}}\right]-\left[-{\frac {\left(i!\right)^{2}}{i^{3}}}\right]\right)\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a51546556b07ed4545007cadcef0116535d96af6)
![{\displaystyle P_{K}\left(n\right)=\left(\sum _{i=1}^{2^{k}\times n}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{k+1+\sum _{m=1}^{n}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times \left[{\frac {k+1+\sum _{m=1}^{n}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times i\times \left(1-\left[{\frac {\left[{\frac {{\left(i!\right)}^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)\right)}\right)\times \left[{\frac {\left[{\frac {{\left(n!\right)}^{2}}{n^{3}}}\right]}{\frac {{\left(n!\right)}^{2}}{n^{3}}}}\right]+{\left(\sum _{i=1}^{2^{k}\times n}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{k+\sum _{m=1}^{n}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times \left[{\frac {k+\sum _{m=1}^{n}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times i\times \left(1-\left[{\frac {\left[{\frac {{\left(i!\right)}^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)\right)}\right)\times \left(1-\left[{\frac {\left[{\frac {{\left(n!\right)}^{2}}{n^{3}}}\right]}{\frac {{\left(n!\right)}^{2}}{n^{3}}}}\right]\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79936ac4419ced0b7d092e0b2ffcc361bb9ab28c)
![{\displaystyle P_{K}\left(n\right)=\left(\sum _{i=1}^{2^{k}\times n}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{k+1+\sum _{m=1}^{n-1}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times \left[{\frac {k+1+\sum _{m=1}^{n-1}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {{\left(m!\right)}^{2}}{m^{3}}}\right]}{\frac {{\left(m!\right)}^{2}}{m^{3}}}}\right]\right)}}}\right]\times i\times \left(1-\left[{\frac {\left[{\frac {{\left(i!\right)}^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)\right)}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0796be600d6f8556aa317c94abb27ff95f723cf7)
![{\displaystyle P_{k}\left(n\right)=\sum _{i=n}^{2^{k}\times n}{\left(\left[{\frac {1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}{k+1}}\right]\times \left[{\frac {k+1}{1+\sum _{m=1}^{i}{\left(1-\left[{\frac {\left[{\frac {\left(m!\right)^{2}}{m^{3}}}\right]}{\frac {\left(m!\right)^{2}}{m^{3}}}}\right]\right)}}}\right]\times i\times \left(1-\left[{\frac {\left[{\frac {\left(i!\right)^{2}}{i^{3}}}\right]}{\frac {\left(i!\right)^{2}}{i^{3}}}}\right]\right)\right)}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1571e3d9d84f9d3e6c61cdfcafa9cb69e704ff78)