Proportionate Nim

 Published on Sunday, 14th April 2019, 10:00 am; Solved by 221;
Difficulty rating: 55%

Problem 665

Two players play a game with two piles of stones, alternating turns.

On each turn, the corresponding player chooses a positive integer $n$ and does one of the following:

  • removes $n$ stones from one pile;
  • removes $n$ stones from both piles; or
  • removes $n$ stones from one pile and $2n$ stones from the other pile.

The player who removes the last stone wins.

We denote by $(n,m)$ the position in which the piles have $n$ and $m$ stones remaining. Note that $(n,m)$ is considered to be the same position as $(m,n)$.

Then, for example, if the position is $(2,6)$, the next player may reach the following positions:
$(0,2)$, $(0,4)$, $(0,5)$, $(0,6)$, $(1,2)$, $(1,4)$, $(1,5)$, $(1,6)$, $(2,2)$, $(2,3)$, $(2,4)$, $(2,5)$

A position is a losing position if the player to move next cannot force a win. For example, $(1,3)$, $(2,6)$, $(4,5)$ are the first few losing positions.

Let $f(M)$ be the sum of $n+m$ for all losing positions $(n,m)$ with $n\le m$ and $n+m \le M$. For example, $f(10) = 21$, by considering the losing positions $(1,3)$, $(2,6)$, $(4,5)$.

You are given that $f(100) = 1164$ and $f(1000) = 117002$.

Find $f(10^7)$.