Problem 326

Modulo Summations

Let a_n be a sequence recursively defined by: $a_{1} = 1, a_{n} = (\sum_{k = 1}^{n - 1} k \cdot a_{k}) mod n$ .

So the first 10 elements of a_n are: 1,1,0,3,0,3,5,4,1,9.

Let f(N,M) represent the number of pairs (p,q) such that:

$1 \leq p \leq q \leq N and (\sum_{t = p}^{q} a_{i}) mod M = 0$

It can be seen that f(10,10)=4 with the pairs (3,3), (5,5), (7,9) and (9,10).

You are also given that f(10⁴,10³)=97158.

Find f(10¹²,10⁶).

同余加和

数列a_n按如下方式递归定义： $a_{1} = 1, a_{n} = (\sum_{k = 1}^{n - 1} k \cdot a_{k}) mod n$ 。

因此a_n的前10项分别是：1,1,0,3,0,3,5,4,1,9。

记f(N,M)为满足如下条件的数对(p,q)的数目：

$1 \leq p \leq q \leq N 且 (\sum_{t = p}^{q} a_{i}) mod M = 0$

可以看出f(10,10)=4，这4个数对分别是(3,3)、(5,5)、(7,9)和(9,10)。

此外还已知f(10⁴,10³)=97158。

求f(10¹²,10⁶)。