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In general, very few prime factors in an odd perfect number can be of the form $n^2+1$. In particular, if N is an odd perfect number then $\frac{\sigma(N)}{N}=2$, and for any $m$ (perfect or not), $\... • 4,094 0 votes ### If$p^k m^2$is an odd perfect number with special prime$p$, then under what other conditions on$\sigma(p^k)/2$does$k=1$follow? This is a partial answer, in line with my first inquiry. It turns out that the converse $$k = 1 \implies \sigma(p^k)/2 \text{ is squarefree }$$ is true. The proof that follows is lifted from this ... • 1,326 1 vote Accepted ### Divisibility relation with a specific sum of divisors From the paper of Touchard that is linked in the question on we get the relation $$3nS_0(n)-\frac{n(n-1)\sigma(n)}{6}=\frac{10}{n}S_2(n) ....(1)$$ here$S_i(n)=\sum_{k=1}^{n-1}k^i\sigma(k)\sigma(n-k)\$....
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