[seqfan] Re: a question to the experts regarding the product of tau-values for consecutive integers (A092517)
William Keith
william.keith at gmail.com
Thu Jan 6 05:20:02 CET 2022
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
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