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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
$$
\begin{aligned}
7 & = 1+1+1+1+1+1+1 \\
& =1+1+1+1+1+2 \\
& =1+1+1+2+2 \\
& =1+1+1+4 \\
& =1+2+2+2 \\
& =1+2+4
\end{aligned}
$$
You are also given $p(7^7) \equiv 144548435 \pmod {10^9+7}$.

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：
$$
\begin{aligned}
7 & = 1+1+1+1+1+1+1 \\
& =1+1+1+1+1+2 \\
& =1+1+1+2+2 \\
& =1+1+1+4 \\
& =1+2+2+2 \\
& =1+2+4
\end{aligned}
$$
已知$p(7^7) \equiv 144548435 \pmod {10^9+7}$。

求$p(7^{777})$，并对$10^9 + 7$取余作为你的答案。

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