For sum of divisors of 4n+1 and 4n+3 and is there a known inequality with n? I was trying to understand the relationship between $S(n)$ and $S(4n+1)$ and between $S(n)$ and $S(4n+3)$.
It seems that up to $10^5$, the ratios $S(4n+1)/S(n)$ and $S(4n+3)/S(n)$ are always $<1$, while for up to $10^7$ the ratio $S(4n+3)/S(n)$ remains to be less than one. I would be interested in these ratios. Don't know if there's any concrete results.
 A: I assume you mean that the ratios are greater than $1$.
The reason that you need to go quite high to see counterexamples is fairly simple. In order to have $S(n)>S(4n+1)$, you certainly need $S(n)>4n$. But this can only happen if $n$ has a lot of relatively small factors. In particular,
$$\frac{S(n)}{n}=\sum_{d\mid n}\frac{1}{d},$$
and you would need to add together lots of different numbers of the form $1/d$ to get a sum exceeding $4$. (Even if you just added together $1/1+1/2+1/3+\cdots$, you need at least $31$ terms to get to $4$, and so you certainly need at least $31$ factors, but actually you will need to use significantly more factors than this.)
Actually, the first number for which $S(n)>4n$ is $27720$, which has $96$ factors. As it happens, this also satisfies $S(n)>S(4n+1)$, because $4n+1$ is prime. So essentially the only difficulty in finding numbers where $S(n)>S(4n+1)$ is in making $S(n)>4n$.
The reason it's even harder to get a value where $S(n)>S(4n+3)$ is that if $n$ is a multiple of $3$ then so is $4n+3$, meaning the ratio $S(n)/n$ actually needs to be greater than $16/3$, whereas if $n$ is not a multiple of $3$ that makes it harder to get the ratio even above $4$.
edit to answer question in comments: We can make the ratio $S(n)/S(4n+1)$ arbitrarily large. Fix $k$, and let $n$ be a multiple of $k!$ such that $4n+1$ is prime; this exists by Dirichlet's theorem. In particular it is divisible by $1,...,k$ so $S(n)/n>1/1+1/2+...+1/k>\log_2 k$. But $S(4n+1)=4n+2$ and so $S(n)/S(4n+1)>0.24\log_2 k$, say, which can be made arbitrarily large by choosing $k$ appropriately. The same argument works for $4n+3$, except you need to make $\frac{4n+3}{3}$ prime instead.
