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No.1263 Is the Order Math?

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タグ : (AC するまで非表示) / 解いたユーザー数 9
作問者 : PCTprobability / テスター : kozy
お気に入りにしたユーザー ProblemId : 5323 / 自分の提出
問題文最終更新日: 2020-10-17 00:28:44
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yukicoder contest 270 (mathematics contest) (順位表) の他の問題:

Problem Statement

For multiple zeta values, please refer to the following page.

https://ja.wikipedia.org/wiki/%E5%A4%9A%E9%87%8D%E3%82%BC%E3%83%BC%E3%82%BF%E5%80%A4

However, in the definition of multiple zeta values, the elements of the index are real numbers greater than or equal to $1$.

When $a$ is a real number greater than $1$ and $n$ is an integer greater than or equal to $2$, we define $S_{n,a}$ as
the sequence consisting of the products of zeta values, chosen from the zeta values of depth $1$ whose indices are all multiples of $a$ so that the weight is $na$,
and we denote the $i$-th element of $S_{n,a}$ by $(S_{n,a})_i$.
Here, it is guaranteed that the length of $S_{n,a}$ is $p(n)$ (the partition number of $n$).

  • ex: $S_{4,2}=$ { $ζ(2)^4,ζ(4)ζ(2)^2,ζ(6)ζ(2),ζ(4)^2,ζ(8)$ }
You are given a positive integer $x$ greater than or equal to $2$. For each sequence of rational numbers $f$ satisfying the following conditions, multiply the product of all its elements by $x!^{p(x)}$ (where $p(x)$ is the partition number of $x$), and output the sum of these values modulo $1000000007$. It is guaranteed that this value is an integer.
  • For every real number $y$ greater than $1$, \( \displaystyle \sum_{i=1}^{p(x)} (S_{x,y})_i×f_i \) $=ζ(y,y,...,y)(y$ appearing $x$ times) (21:54: corrected $S_{y,x}$ to $S_{x,y}$.)
  • \( \displaystyle \sum_{i=1}^{p(x)} f_i=0 \)
Here, it is guaranteed that the answer is uniquely determined regardless of the order of $S_{y,x}$. It can also be proved that sequences that can be $f$ exist and that they are finite in number.

Input

$x$

  • All inputs are integers.
  • $2 \le x \le 8$

Output

For each possible $f$, multiply the product of all its elements by $x!^{p(x)}$, and output the sum of these values modulo $1000000007$.

Sample

Sample 1
Input
2
Output
1000000006

$S_{2,a}={ζ(a)^2,ζ(2a)}$. In this case, $ζ(a,a)= \frac{ζ(a)^2}{2}- \frac{ζ(2a)}{2}$ holds, and it can be proved that no other sequence of rational numbers $f$ satisfies the conditions. Here $f=$ { $ \frac{1}{2} , -\frac{1}{2} $ }, so the product of the elements is $ -\frac{1}{4}$. Multiplying this by $2!^{p(2)}=4$ gives $-1$, and you should output its remainder modulo $1000000007$, which is $1000000006$.

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