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Problem 545

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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:
$$1^k + 2^k + … + n^k = \sum_{i=1}^n i^k = \sum_{i=1}^{k+1} a_{i} n^i = a_{1} n + a_{2} n^2 + … + a_{k} n^k + a_{k+1} n^{k + 1},$$
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, $1^4 + 2^4 + … + n^4 = -\frac{1}{30}n + \frac{1}{3}n^3 + \frac{1}{2}n^4 + \frac{1}{5}n^5$.

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).

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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：
$$1^k + 2^k + … + n^k = \sum_{i=1}^n i^k = \sum_{i=1}^{k+1} a_{i} n^i = a_{1} n + a_{2} n^2 + … + a_{k} n^k + a_{k+1} n^{k + 1},$$
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，$1^4 + 2^4 + … + n^4 = -\frac{1}{30}n + \frac{1}{3}n^3 + \frac{1}{2}n^4 + \frac{1}{5}n^5$。

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)。

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