The $n$th triangular number is

$$T(n) = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}.$$

Triangular numbers satisfy many interesting properties, including a *product rule *for $m, n \ge 1$:

$$T(mn) = T(m) T(n) + T(m-1) T(n-1).$$

This product rule can be interpreted geometrically by subdividing a large triangle into smaller triangles. This picture illustrates the case $m = 5$ and $n = 4$.

An interesting question (in my opinion) is whether there are other sequences that satisfy this identity. It turns out that there are exactly five such sequences:

- $T(n) = 0$ for all $n \ge 0$.
- $T(n) = \frac12$ for all $n \ge 0$.
- $T(0) = 0$ and $T(2n) = T(2n-1) = n$ for all $n \ge 1$.
- $T(3n) = T(3n+2) = 0$ and $T(3n+1) = 1$ for all $n \ge 0$.
- $T(n) = \frac12 n(n+1)$ for all $n \ge 0$.

I wrote about this problem previously at MathBlag, but now I am trying to prepare the result for publication. I would appreciate feedback on my e-print.