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Problem 659

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Largest prime

Consider the sequence $n^2+3$ with $n \ge 1$.
If we write down the first terms of this sequence we get:
$4, 7, 12, 19, 28, 39, 52, 67, 84, 103, 124, 147, 172, 199, 228, 259, 292, 327, 364,\ldots$
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 $\displaystyle \sum_{k=1}^{10,000,000} P(k)$.

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最大的素数

考虑序列$n^2+3$，其中$n \ge 1$，该序列的一开始几项分别是：
$4, 7, 12, 19, 28, 39, 52, 67, 84, 103, 124, 147, 172, 199, 228, 259, 292, 327, 364,\ldots$
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$\displaystyle \sum_{k=1}^{10,000,000} P(k)$的最后$18$位数字。

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