A product rule for triangular numbers, part 2

by David Radcliffe

In the previous post, we showed that the sequence of triangular numbers satisfies the product rule

T(mn) = T(m) T(n) + T(m − 1) T(n − 1);    m, n ≥ 1.

Let α = T(0) and β = T(1). We will assume in this post that α ≠ 0. The case α = 0 is more difficult, and will be discussed in parts 3 and 4 of this series.

Substituting m = 1 and n = 1 into the product rule gives β = β2 + α2. Since α ≠ 0, it follows that β ≠ 1, so we can rewrite the above equation as

α/(1 – β) = β/α .

Substituting m = 1 into the product rule gives T(n) = βT(n) + αT(n-1), hence

T(n)/T(n-1) =  α/(1 – β) = β/α .

Therefore,

T(n) = α (β/α)n   for all n ≥ 0.

That is to say that T(n) is a geometric sequence. But a geometric sequence cannot satisfy the product rule unless it is a constant sequence (see the comments), in which case α = 1/2.

Therefore, the only solution to T(mn) = T(m) T(n) + T(m-1) T(n-1) satisfying T(0) ≠ 0 is the constant sequence T(n) = 1/2.

I am grateful to an anonymous commenter for pointing out a major error in a previous version of this post.

Advertisements