[seqfan] Re: a question to the experts regarding the product of tau-values for consecutive integers (A092517)

[William Keith]

Well, I can give you one direction.  Suppose (n+1)^2 = b^{c_1} + ... +
b^{c_{n+1}}, with all c_i \geq 0.  Then

n^2 + n = (b^{c_1} - 1) + ... + (b^{c_{n+1}} - 1)

and the right-hand side is divisible by b-1, so the left-hand side is as
well.  Hence b-1 is a divisor of n(n+1).

So this reduces the question to why a divisor of n(n+1) always affords such
a representation.

Best,
William Keith