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Problem 890

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Binary Partitions

Let $p(n)$ be the number of ways to write $n$ as the sum of powers of two, ignoring order.

For example, $p(7)=6$, the partitions being
$$\begin{array}{rl}7& =1+1+1+1+1+1+1\\ & =1+1+1+1+1+2\\ & =1+1+1+2+2\\ & =1+1+1+4\\ & =1+2+2+2\\ & =1+2+4\end{array}$$
You are also given $p(7^7)\equiv 144548435(\mathrm{mod}{10}^9+7)$.

Find $p(7^{777})$. Give your answer modulo ${10}^9+7$.

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二进制分拆

记$p(n)$为将$n$写成二的幂之和的方法数，不计顺序。

例如，$p(7)=6$，对应的分拆包括：
$$\begin{array}{rl}7& =1+1+1+1+1+1+1\\ & =1+1+1+1+1+2\\ & =1+1+1+2+2\\ & =1+1+1+4\\ & =1+2+2+2\\ & =1+2+4\end{array}$$
已知$p(7^7)\equiv 144548435(\mathrm{mod}{10}^9+7)$。

求$p(7^{777})$，并对${10}^9+7$取余作为你的答案。

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