# 五邊形數定理

${\displaystyle \prod _{n=1}^{\infty }(1-x^{n})=\sum _{k=-\infty }^{\infty }(-1)^{k}x^{\frac {k(3k-1)}{2}}=\sum _{k=0}^{\infty }(-1)^{k}x^{\frac {k(3k\pm 1)}{2}}}$

${\displaystyle (1-x)(1-x^{2})(1-x^{3})\cdots =1-x-x^{2}+x^{5}+x^{7}-x^{12}-x^{15}+x^{22}+x^{26}+\cdots .}$

## 和分割函數的關係

${\displaystyle {\frac {1}{\phi (x)}}=\sum _{k=0}^{\infty }p(k)x^{k}}$

${\displaystyle (1-x-x^{2}+x^{5}+x^{7}-x^{12}-x^{15}+x^{22}+x^{26}+\cdots )(1+p(1)x+p(2)x^{2}+p(3)x^{3}+\cdots )=1}$

${\displaystyle p(n)-p(n-1)-p(n-2)+p(n-5)+p(n-7)+\cdots =0}$

${\displaystyle p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+\cdots }$

${\displaystyle p(10)=p(9)+p(8)-p(5)-p(3)=30+22-7-3=42}$

## 參考資料

1. ^ 原文為Euler, Leonhard. Evolutio producti infiniti ${\displaystyle (1-x)(1-xx)(1-x^{3})(1-x^{4})(1-x^{5})}$  etc. in seriem simplicem. Acta Academiae Scientarum Imperialis Petropolitinae. 1775, 1780: 47–55.
2. ^ 英文翻譯版為Bell, J在2004-12-4翻譯的《The Expansion of the Infinite Product ${\displaystyle (1-x)(1-xx)(1-x^{3})(1-x^{4})(1-x^{5})(1-x^{6})}$  etc. into a Single Series》，http://www.arxiv.org/abs/math.HO/0411454/.