Project Euler Lab - Problem 560

#560 - Coprime Nim

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Coprime Nim is just like ordinary normal play Nim, but the players may only remove a number of stones from a pile that is coprime with the current size of the pile. Two players remove stones in turn. The player who removes the last stone wins.

Let \(L(n, k)\) be the number of losing starting positions for the first player, assuming perfect play, when the game is played with \(k\) piles, each having between \(1\) and \(n - 1\) stones inclusively.

For example, \(L(5, 2) = 6\) since the losing initial positions are \((1, 1)\), \((2, 2)\), \((2, 4)\), \((3, 3)\), \((4, 2)\) and \((4, 4)\).
You are also given \(L(10, 5) = 9964\), \(L(10, 10) = 472400303\), \(L(10^3, 10^3) \bmod 1\,000\,000\,007 = 954021836\).

Find \(L(10^7, 10^7)\bmod 1\,000\,000\,007\).

This problem is taken from Project Euler, Problem 560.
Problem text © Project Euler, licensed under CC BY-NC-SA 4.0. Original: projecteuler.net/problem=560. Published Saturday, 14th May 2016, 07:00 pm. Solved by 406 members at time of mirroring.

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