Regarding the $\sigma (n)$ function. This question relates to Robin's Inequality. Is $\sigma{(n^2)}$ < (2 n) $\sigma{(n)}$ ? For what integer values of n is this satisfied?
 A: 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 $1.368432778$.
A: 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.
