### 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 = β^{2} + α^{2}. 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.

I think that the sequence T(n) = a (b/a)^n does not satisfy the product rule unless a = b = 1/2, in which case T(n) = 1/2 holds for all n. [The argument given in the post correctly establishes that T(m) T(n) + T(m-1) T(n-1) = T(m + n) holds for any sequence of the given form— but this is not the product rule, which would have T(mn) on the right hand side.]

Clearly the constant sequence T(n) = 1/2 satisfies the product rule. Conversely, fix any a and b, with a nonzero, and assume that the sequence T(n) = a (b/a)^n satisfies the product rule. As established in the post we must have

(*) b = b^2 + a^2.

Since a is nonzero, we deduce from (*) that b is nonzero.

As a special case (m = n = 2) of the product rule, we have

T(2*2) = T(2) T(2) + T(1) T(1).

Using the definition of T we it follows that a (b/a)^4 = a (b/a)^2 a (b/a)^2 + a (b/a)^1 a (b/a)^1, and simplifying this equality we learn that

a (b/a)^4 = b^2 (b/a)^2 + a^2 (b/a)^2

Cancelling the nonzero factor (b/a)^2 from both sides, we deduce that a (b/a)^2 = b^2 + a^2; from (*) it follows that a (b/a)^2 = b and hence (b/a)^2 = b/a.

Since the number x = b/a is a nonzero solution to the equation x^2 = x, it follows that x = 1, and hence a = b. The equation (*) then implies that b = 2b^2, and as b is nonzero we deduce that 1 = 2b, so b = 1/2.

You are quite right. I confused T(mn) with T(m+n). Thanks for noting the error.