Regarding the $\sigma (n)$ function.

Question

This question relates to Robin's Inequality. Is $\sigma{(n^2)}$ < (2 n) $\sigma{(n)}$ ? For what integer values of n is this satisfied?

Answer 1

Sorry: not sure about this. With the inclusion of the factor 2, your quantity is not multiplicative. Evidently the large (and increasing) values of $\sigma(n^2) / (n \sigma(n)),$ which really is multiplicative, occur at the primorials, that is the product of the consecutive primes. Here are the first few ratios:

    n       s(n)      s(n^2)  s(n^2) / n s(n)
    1         1         1   1                              1 =  1 
    2         3         7   1.166666666666667              2 = 2
    6        12        91   1.263888888888889              6 = 2 * 3
   30        72      2821   1.306018518518518             30 = 2 * 3 * 5
  210       576    160797   1.329340277777778            210 = 2 * 3 * 5 * 7
 2310      6912  21386001   1.339411037457912           2310 = 2 * 3 * 5 * 7 * 11

Should not be too hard to prove stuff and decide whether, at the primorials, the ratio is bounded. Simple analytic number theory bounds once you stick to primorials.

It appears that we always have $\sigma(n^2) < 2 n \sigma(n).$ As both sides are "multiplicative," it suffices to prove this for $1$ and all primes and prime powers. For example, prime $n=p,$ we get $\sigma(p^2) = p^2 + p + 1,$ while $n \sigma(n) = p (p+1) = p^2 + p,$ slightly smaller. But then double it and it becomes larger.

For $n=p^2,$ $\sigma(n^2) = \sigma(p^4) = p^4 + p^3 + p^2 + p + 1.$ Meanwhile, $n \sigma(n) = p^2 \sigma(p^2) = p^2 (p^2 + p + 1) = p^4 + p^3 + p^2,$ again slightly smaller, double it and it becomes bigger.

    n        s(n)     s(n^2)   s(n^2)-2ns(n)     n factored
    1         1         1        -1              1 =  1 
    2         3         7        -5              2 = 2
    3         4        13       -11              3 = 3
    4         7        31       -25              4 = 2^2
    5         6        31       -29              5 = 5
    6        12        91       -53              6 = 2 * 3
    7         8        57       -55              7 = 7
    8        15       127      -113              8 = 2^3
    9        13       121      -113              9 = 3^2
   10        18       217      -143             10 = 2 * 5
   11        12       133      -131             11 = 11
   12        28       403      -269             12 = 2^2 * 3
   13        14       183      -181             13 = 13
   14        24       399      -273             14 = 2 * 7
   15        24       403      -317             15 = 3 * 5
   16        31       511      -481             16 = 2^4
   17        18       307      -305             17 = 17
   18        39       847      -557             18 = 2 * 3^2
   19        20       381      -379             19 = 19
   20        42       961      -719             20 = 2^2 * 5
   21        32       741      -603             21 = 3 * 7
   22        36       931      -653             22 = 2 * 11
   23        24       553      -551             23 = 23
   24        60      1651     -1229             24 = 2^3 * 3
   25        31       781      -769             25 = 5^2
   26        42      1281      -903             26 = 2 * 13
   27        40      1093     -1067             27 = 3^3
   28        56      1767     -1369             28 = 2^2 * 7
   29        30       871      -869             29 = 29
   30        72      2821     -1499             30 = 2 * 3 * 5
   31        32       993      -991             31 = 31
   32        63      2047     -1985             32 = 2^5
   33        48      1729     -1439             33 = 3 * 11
   34        54      2149     -1523             34 = 2 * 17
   35        48      1767     -1593             35 = 5 * 7
jagy@phobeusjunior:~$

Answer 2

We can prove the inequality (actually a stronger inequality) with the explicit formula for $\sigma(n)$ in terms of the prime factorisation of $n$. For

$$n = \prod_{k=1}^r p_k^{\alpha_k},$$

we have

$$\sigma(n) = \prod_{k=1}^r \frac{p_k^{\alpha_k+1}-1}{p_k-1},$$

and so

$$\begin{align} \frac{\sigma(n^2)}{n\sigma(n)} &= \prod_{k=1}^r \frac{p_k^{2\alpha_k+1}-1}{p_k^{\alpha_k}(p_k^{\alpha_k+1}-1)}\\ &= \prod_{k=1}^r \frac{(p_k^{2\alpha_k+1}-p_k^{\alpha_k}) + (p_k^{\alpha_k}-1)}{p_k^{\alpha_k}(p_k^{\alpha_k+1}-1)}\\ &= \prod_{k=1}^r \left(1 + \frac{p_k^{\alpha_k}-1}{p_k^{\alpha_k}(p_k^{\alpha_k+1}-1)}\right)\\ &\leqslant \prod_{k=1}^r \left(1 + \frac{p_k-1}{p_k(p_k^2-1)}\right) \tag{$\ast$}\\ &= \prod_{k=1}^r \left(1 + \frac{1}{p_k(p_k+1)}\right)\\ &\leqslant \prod_{k=1}^r \left(1 + \frac{1}{p_k^2}\right)\\ &< \prod_{p\in\mathbb{P}} \left(1+\frac{1}{p^2}\right)\\ &= \prod_{p\in\mathbb{P}} \frac{1 - \frac{1}{p^4}}{1-\frac{1}{p^2}}\\ &= \frac{\zeta(2)}{\zeta(4)}\\ &= \frac{15}{\pi^2}\\ &< 1.52, \end{align}$$

where in $(\ast)$ we used that $\frac{m^\alpha-1}{m^\alpha(m^{\alpha+1}-1)} \leqslant \frac{m-1}{m(m^2-1)}$ for $m \geqslant 2$ and $\alpha \geqslant 1$, which can be elementarily verified. If we don't replace $\frac{1}{p(p+1)}$ with $\frac{1}{p^2}$ two lines below $(\ast)$, we get a sharp bound, which however I don't know how to evaluate explicitly. The sharp bound is approximately $1.368432778$.

Answer 3

To complete Daniel Fischer's answer:

A closed form for the sharp bound is $$\prod_{p \in \mathbb{P}}{\left(\dfrac{p^2 + p + 1}{p^2 + p}\right)} = \dfrac{\zeta(2)}{\zeta(3)} \approx 1.368432778.$$

See this MSE question for more.