Eccoci.

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 $C_{in}$ be the circle having the diameter $XY$.
Let $C_{out}$ be the circle having the diameter $WZ$.

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

• $C_i$ has no common interior points with any $C_j$ for $1 \leq i, j \leq k$ and $i \neq j$,
• $C_i$ is tangent to both $C_{in}$ and $C_{out}$ for $1 \leq i \leq k$,
• $C_i$ is tangent to $C_{i+1}$ for $1 \leq i \lt k$, 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 $b \leq n$. For example, $T(1) = 9$, $T(20) = 732$ and $T(3000) = 438106$.

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