This is my third question here.

For the past two days, I have asked two questions here, both of which have a bearing on my current question. I have been working on deriving an equation relating $\sigma(n^2)$ to $\sigma(n)$, if $n$ is odd.

Now, to begin with, we know that $\sigma(n^2)$ is odd, but we can't really be sure about the residue of $\sigma(n^2)$ modulo $4$, even if I know (by my current considerations) that $\sigma(n)$ is incongruent to $2$ modulo $4$. [Edit: As of today [June 22 2012], I am no longer sure about $\sigma(n) \not\equiv 2 \pmod 4$! :-(] I tried tackling the various cases that need to be considered to cover this problem in its full generality, but I think this mammoth task is too complex for me. Hence, I decided to ask this third question:

Is there a specific equation relating $\sigma(n^2)$ to $\sigma(n)$, if $n$ is odd?

Since $n \mid n^2$ and $n < n^2$, all the divisors of $n$ also divide $n^2$. Thus, intuitively, I know that $\sigma(n)$ forms part of the sum that is $\sigma(n^2)$. In other words, I have:

$$\sigma(n^2) = \sigma(n) + \sum_{\substack{d \mid n^2 \\ n < d \le n^2}}{d}.$$

(I am not so sure about that second addend, though.) Furthermore, since $I(n) < I(n^2) < (I(n))^2$, where $I(x) = \frac{\sigma(x)}{x}$ is the abundancy index of the positive integer $x$, then $n\sigma(n) < \sigma(n^2) < (\sigma(n))^2$. So then, I have:

$$\sigma(n^2) = n\sigma(n) + A, \hspace{0.10in} (**)$$


$$\sigma(n^2) = (\sigma(n))^2 - B. \hspace{0.10in} (***)$$

My obvious question at this point is: Is it possible to express $A$ or $B$ in terms of $n$, without using $(**)$ and $(***)$?

Appreciate any feedback/replies on this. Thanks!

  • 2
    $\begingroup$ There can be more terms in that sum $\sigma(n^2) = \sigma(n) + ...$. Take $n = 15$ and note that $15^2$ has $9$ as a divisor $\endgroup$ – Cocopuffs Jun 20 '12 at 22:48
  • 1
    $\begingroup$ Note that $\sigma(21)=\sigma(31)$ but $\sigma(21^2)\ne\sigma(31^2)$, so it is impossible for $\sigma(n^2)$ to be purely a function of $\sigma(n)$. On the other hand, allowing $\sigma(n^2)=f(n,\sigma(n))$, we should wonder how $f$ is being restricted (as $f(x,y)=\sigma(x^2)$ would be trivial): precisely what is or isn't allowed? $\endgroup$ – anon Jun 21 '12 at 1:44

Since $\sigma$ is multiplicative we can consider powers of primes as a base case. $$ \sigma(p^a) = \sum_{j=0}^a p^j = \frac{p^{a+1}-1}{p-1} \\ \sigma(p^{2a}) = \frac{p^{2a+1}-1}{p-1}=p^a \sigma(p^a)+\frac{p^a-1}{p-1} $$ So if $n=\prod p_i^{a_i}$ then $$ \sigma(n^2) = \prod \left(p_i^{a_i} \sigma(p_i^{a_i})+\frac{p_i^{a_i}-1}{p_i-1}\right) $$ The first term in the product gives $n\sigma(n)$, but the other terms rely on the factorization in a way that I don't think will allow a simple expression in terms of $n$ alone.

  • $\begingroup$ Thank you very much for this answer, Zander! $\endgroup$ – Jose Arnaldo Bebita-Dris Jun 21 '12 at 4:41
  • $\begingroup$ We can rewrite Zander's final expression for $\sigma(n^2)$ as: $$\sigma(n^2) = \prod\left({{p_i}\sigma({p_i}^{a_i}) + \sigma({p_i}^{a_i - 1})}\right).$$ $\endgroup$ – Jose Arnaldo Bebita-Dris Jun 21 '12 at 20:12

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