--description--

Let $a$, $b$ and $c$ be positive numbers.

Let $W$, $X$, $Y$, $Z$ be four collinear points where $|WX| = a$, $|XY| = b$, $|YZ| = c$ and $|WZ| = a + b + c$.

Let $C_{\text{in}}$ be the circle having the diameter $XY$.

Let $C_{\text{out}}$ be the circle having the diameter $WZ$.

The triplet ($a$, $b$, $c$) is called a necklace triplet if you can place $k ≥ 3$ distinct circles $C_1, C_2, \ldots, C_k$ such that:

• $C_i$ has no common interior points with any $C_j$ for $1 ≤ i$, $j ≤ k$ and $i ≠ j$,
• $C_i$ is tangent to both $C_{\text{in}}$ and $C_{\text{out}}$ for $1 ≤ i ≤ k$,
• $C_i$ is tangent to $C_{i + 1}$ for $1 ≤ i < k$, and
• $C_k$ is tangent to $C_1$.

For example, (5, 5, 5) and (4, 3, 21) are necklace triplets, while it can be shown that (2, 2, 5) is not.

Let $T(n)$ be the number of necklace triplets $(a, b, c)$ such that $a$, $b$ and $c$ are positive integers, and $b ≤ n$. For example, $T(1) = 9$, $T(20) = 732$ and $T(3\,000) = 438\,106$.

Find $T(1\,000\,000\,000)$.

--hints--

necklace(1000000000) should return 747215561862.

assert.strictEqual(necklace(1000000000), 747215561862);

--seed--

--seed-contents--

function necklace(n) {

  return true;
}

necklace(1000000000)

--solutions--

// solution required