---

Problem 421

---

Prime factors of n¹⁵+1

Numbers of the form n¹⁵+1 are composite for every integer n > 1.
For positive integers n and m let s(n,m) be defined as the sum of the distinct prime factors of n¹⁵+1 not exceeding m.

E.g. 2¹⁵+1 = 3×3×11×331.
So s(2,10) = 3 and s(2,1000) = 3+11+331 = 345.

Also 10¹⁵+1 = 7×11×13×211×241×2161×9091.
So s(10,100) = 31 and s(10,1000) = 483.

Find ∑ s(n,10⁸) for 1 ≤ n ≤ 10¹¹.

---

n¹⁵+1的质因数

可以表示为n¹⁵+1的数，对于任意n > 1都是合数。
对于正整数n和m，记s(n,m)为n¹⁵+1所有不超过m的不同质因数之和。

例如，2¹⁵+1 = 3×3×11×331。
因此s(2,10) = 3，而s(2,1000) = 3+11+331 = 345。

同样地，10¹⁵+1 = 7×11×13×211×241×2161×9091。
因此s(10,100) = 31而s(10,1000) = 483。

对于1 ≤ n ≤ 10¹¹，求∑ s(n,10⁸)。

---