---

Problem 586

---

Binary Quadratic Form

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

$$\begin{aligned}
209 &= 8^2 + 3\cdot 8\cdot 5 + 5^2 \\
209 &= 13^2 + 3\cdot13\cdot 1 + 1^2
\end{aligned}
$$

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

---

二元二次型

数$209$可以用两种方式表达为$a^2 + 3ab + b^2$：

$$\begin{aligned}
209 &= 8^2 + 3\cdot 8\cdot 5 + 5^2 \\
209 &= 13^2 + 3\cdot13\cdot 1 + 1^2
\end{aligned}
$$

记$f(n,r)$为不超过$n$且恰好可以用$r$种方式表达为$k=a^2 + 3ab + b^2$的整数$k$的数目，其中整数$a\gt b>0$。

已知$f(10^5, 4) = 237$以及$f(10^8, 6) = 59517$。

求$f(10^{15}, 40)$。

---