{"app":{"locale":"en-GB"},"problem":{"content":"

The number 209 can be expressed as $a^2 + 3ab + b^2$ in two distinct ways:

$ \\qquad 209 = 8^2 + 3\\cdot 8\\cdot 5 + 5^2$
$ \\qquad 209 = 13^2 + 3\\cdot13\\cdot 1 + 1^2$

Let $f(n,r)$ be the number of integers $k$ not exceeding $n$ that can be expressed as $k=a^2 + 3ab + b^2$, with $a\\gt b>0$ integers, in exactly $r$ different ways.

You are given that $f(10^5, 4) = 237$ and $f(10^8, 6) = 59517$.

Find $f(10^{15}, 40)$.

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