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Aug 14, 2023 at 12:26 comment added isnet Thanks for the suggested paper.
Aug 11, 2023 at 14:18 comment added Steven Creech I added an edit that links to a paper that I think will answer your question in general
Aug 11, 2023 at 14:18 history edited Steven Creech CC BY-SA 4.0
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Aug 11, 2023 at 6:37 comment added isnet What if $p_i$ are not distinct? For example, lets say $n=p^3$.
Aug 10, 2023 at 21:39 comment added Steven Creech Good catch, so the general formula for square-free should be $\frac{\sigma(n)}{2^k}-1$
Aug 10, 2023 at 19:30 comment added Erick Wong Since this is the squarefree case, it happens that $\prod (p+1)$ can be identified with the sum of divisors function $\sigma(n)$, though this isn't true in general.
Aug 10, 2023 at 19:27 vote accept isnet
Aug 10, 2023 at 18:55 history answered Steven Creech CC BY-SA 4.0