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Problem 445


Problem 445


Retractions A

For every integer n>1, the family of functions fn,a,b is defined by fn,a,b(x)≡ax+b mod n for a,b,x integer and 0<a<n, 0≤b<n, 0≤x<n.
We will call fn,a,b a retraction if fn,a,b(fn,a,b(x))≡fn,a,b(x) mod n for every 0≤x<n.
Let R(n) be the number of retractions for n.

You are given that
∑ R(c) for c=C(100 000,k), and 1 ≤ k ≤99 999 ≡628701600 (mod 1 000 000 007).
(C(n,k) is the binomial coefficient).

Find ∑ R(c) for c=C(10 000 000,k), and 1 ≤k≤ 9 999 999.
Give your answer modulo 1 000 000 007.


撤销函数 A

对于任意整数n>1,函数族fn,a,b按如下方式定义:fn,a,b(x)≡ax+b mod n,其中a,b,x都是整数,且0<a<n,0≤b<n,0≤x<n。
我们称fn,a,b为撤销函数,当对于0≤x<n均有在模n意义下fn,a,b(fn,a,b(x))≡fn,a,b(x)。
记R(n)是给定n下撤销函数的数目。

已知
对于c=C(100 000,k)且1 ≤ k ≤99 999,∑ R(c)≡628701600 (mod 1 000 000 007)。
(C(n,k)是二项式系数)。

对于c=C(100 000,k)且1 ≤ k ≤9 999 999,求∑ R(c)。
给出其对1 000 000 007取模的余数。