Contenu supprimé Contenu ajouté
|
|
|
: <math>P_n = \sum_{i=1}^{2^{2^{n}}
<math>P_n = \sum_{i=1}^{2^{2^{n}}}\left(\left\lfloor[\frac {1+ \sum_{m=1}^{i}{\left( 1-\left\lfloor[{\frac{\left\lfloor[{\frac{\left(m!\right)^2}{m^3}}\right\rfloor]}{\frac{\left(m!\right)^2}{m^3}}}\right\rfloor]\right)}}{n+1}\right\rfloor]\times{\left\lfloor[\frac{n+1}{1+ \sum_{m=1}^{i}{\left( 1-\left\lfloor[{\frac{\left\lfloor[{\frac{\left(m!\right)^2}{m^3}}\right\rfloor]}{\frac{\left(m!\right)^2}{m^3}}}\right\rfloor]\right)}}\right\rfloor]}\times{i}\times{\left({1-\left\lfloor[{\frac{\left\lfloor[(\frac{\left(i!\right)^2}{i^3}\right\rfloor]}{\frac{\left(i!\right)^2}{i^3}}}\right\rfloor]}\right)}\right) </math>
|
