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Consider the sequence $n^2+3$ with $n \ge 1$.
If we write down the first terms of this sequence we get:
$4, 7, 12, 19, 28, 39, 52, 67, 84, 103, 124, 147, 172, 199, 228, 259, 292, 327, 364,$... .
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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