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[数论] 分拆数 $p(n)$ 模 13

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hbghlyj Posted at 2025-3-4 23:55:11 |Read mode
Last edited by hbghlyj at 2025-3-5 00:01:49分拆数 $p(n)$ 定义为将 $n$ 写成正整数之和的方法数,无关加数的顺序。$p(n)$ 的生成函数为
\begin{align*}
\frac{1}{(q ; q)_{\infty}} & =\left(1+q+q^2+q^3+\cdots\right)\left(1+q^2+q^4+q^6+\cdots\right)\left(1+q^3+q^6+q^9+\cdots\right) \cdots \\
& =\sum_{n=0}^{\infty} p(n) q^n
\end{align*}
如何证明以下性质?
  • $p(5 n+4) \equiv 0 \bmod 5$
  • $p(7 n+5) \equiv 0 \bmod 7$
  • $p(11 n+6) \equiv 0 \bmod 11$
  • $p(59^4\ 13 n+111247) \equiv 0 \bmod 13$

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2025-4-20 22:02 GMT+8

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