## Faulhaber's Formulas

### Problem 545

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`(if

_{i}`a`= 0, we shall consider

_{i}`q`= 1).

_{i}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 `q`_{1} for the sum of `k`^{th} powers (i.e. the denominator of the reduced fraction `a`_{1}).

Define F(`m`) as the `m`^{th} value of `k` ≥ 1 for which D(`k`) = 20010.

You are given D(4) = 30 (since `a`_{1} = -1/30), D(308) = 20010, F(1) = 308, F(10) = 96404.

Find F(10^{5}).