Problem 659

Largest prime

Consider the sequence $n^{2} + 3$ with $n \geq 1$ .
If we write down the first terms of this sequence we get:
$4, 7, 12, 19, 28, 39, 52, 67, 84, 103, 124, 147, 172, 199, 228, 259, 292, 327, 364, \dots$
We see that the terms for $n = 6$ and $n = 7$ ( $39$ and $52$ ) are both divisible by $13$ .
In fact $13$ is the largest prime dividing any two successive terms of this sequence.

Let $P (k)$ be the largest prime that divides any two successive terms of the sequence $n^{2} + k^{2}$ .

Find the last $18$ digits of $\sum_{k = 1}^{10, 000, 000} P (k)$ .

最大的素数

考虑序列 $n^{2} + 3$ ，其中 $n \geq 1$ ，该序列的一开始几项分别是：
$4, 7, 12, 19, 28, 39, 52, 67, 84, 103, 124, 147, 172, 199, 228, 259, 292, 327, 364, \dots$
可以看出， $n = 6$ 和 $n = 7$ 所对应的项（ $39$ 和 $52$ ）均能被 $13$ 整除。
实际上， $13$ 是能够整除该序列相邻两项的最大的素数。

对于序列 $n^{2} + k^{2}$ ，记 $P (k)$ 为能够整除该序列相邻两项的最大的素数。

求 $\sum_{k = 1}^{10, 000, 000} P (k)$ 的最后 $18$ 位数字。