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Problem 717

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Summation of a Modular Formula

For an odd prime $p$, define $f(p) = \left\lfloor\frac{2^{(2^p)}}{p}\right\rfloor\bmod{2^p}$.
For example, when $p=3$, $\lfloor 2^8/3\rfloor = 85 \equiv 5 \pmod 8$ and so $f(3) = 5$.

Further define $g(p) = f(p)\bmod p$. You are given $g(31) = 17$.

Now define $G(N)$ to be the summation of $g(p)$ for all odd primes less than $N$.
You are given $G(100) = 474$ and $G(10^4) = 2819236$.

Find $G(10^7)$.

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余数求和

对于奇素数$p$，记$f(p) = \left\lfloor\frac{2^{(2^p)}}{p}\right\rfloor\bmod{2^p}$。
例如，若$p=3$，$\lfloor 2^8/3\rfloor = 85 \equiv 5 \pmod 8$，因此$f(3) = 5$。

记$g(p) = f(p)\bmod p$。已知$g(31) = 17$。

记$G(N)$为所有小于$N$的奇素数$p$所对应$g(p)$之和。
已知$G(100) = 474$，$G(10^4) = 2819236$。

求$G(10^7)$。

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