---

Problem 545

---

Faulhaber’s Formulas

The sum of the k^th powers of the first n positive integers can be expressed as a polynomial of degree k+1 with rational coefficients, the Faulhaber’s Formulas:
$$1^k+2^k+\dots +n^k=\sum _{i=1}^n{i}^k=\sum _{i=1}^{k+1}{a}_i{n}^i=a_{1}n+a_2{n}^2+\dots +a_k{n}^k+a_{k+1}{n}^{k+1},$$
where a_i‘s are rational coefficients that can be written as reduced fractions p_i/q_i (if a_i = 0, we shall consider q_i = 1).

For example, $1^4+2^4+\dots +n^4=-\frac{1}{30}n+\frac{1}{3}{n}^3+\frac{1}{2}{n}^4+\frac{1}{5}{n}^5$.

Define D(k) as the value of q₁ for the sum of k^th powers (i.e. the denominator of the reduced fraction a₁).
Define F(m) as the m^th value of k ≥ 1 for which D(k) = 20010.
You are given D(4) = 30 (since a₁ = -1/30), D(308) = 20010, F(1) = 308, F(10) = 96404.

Find F(10⁵).

---

福尔哈贝尔公式

前n个正整数的k次幂的和可以用一个k+1次有理系数多项式来表示，称为福尔哈贝尔公式：
$$1^k+2^k+\dots +n^k=\sum _{i=1}^n{i}^k=\sum _{i=1}^{k+1}{a}_i{n}^i=a_{1}n+a_2{n}^2+\dots +a_k{n}^k+a_{k+1}{n}^{k+1},$$
每个系数a_i都是有理数，因而可以写成最简形式p_i/q_i（如果a_i = 0，我们约定q_i = 1）。

例如，$1^4+2^4+\dots +n^4=-\frac{1}{30}n+\frac{1}{3}{n}^3+\frac{1}{2}{n}^4+\frac{1}{5}{n}^5$。

记D(k)为k次幂的福尔哈贝尔公式中q₁的值（也即a₁化简后的分母）。
记F(m)为使得D(k) = 20010的第m个k值。
已知D(4) = 30（因为a₁ = -1/30），D(308) = 20010，F(1) = 308，F(10) = 96404。

求F(10⁵)。

---