Up to $10^6$: $\sigma(8n+1) \mod 4 = OEIS A001935(n) \mod 4$ (Number of partitions with no even part repeated ) Up to $10^6$: 
$\sigma(8n+1) \mod 4 = OEIS A001935(n) \mod 4$ 
A001935 Number of partitions with no even part repeated
Is this true in general?
It would mean relation between restricted partitions of $n$ and divisors of $8n+1$.
Another one up to $10^6$ is:
$\sigma(4n+1) \mod 4 = A001936(n) \mod 4$
A001936 Expansion of q^(-1/4) (eta(q^4) / eta(q))^2 in powers of q
$\sigma(n)$ is sum of divisors of $n$.

sigma(8n+1) mod 4 starts: 1, 1, 2, 3, 0, 2, 1, 0, 0, 2, 1, 2, 2, 0, 2, 1, 0, 2, 0, 2, 0, 3, 0, 0, 2, 0, 0, 0, 3, 2
sigma(4n+1) mod 4 starts: 1, 2, 1, 2, 2, 0, 3, 2, 0, 2, 2, 2, 1, 2, 0, 2, 0, 0, 2, 0, 1, 0, 2, 0, 2, 2

Update
Up to 10^7
A001935 mod 4 is zero for n = 9m+4 or 9m+7
A001936 mod 4 is zero for n = 9m+5 or 9m+8
Question about computability

Let's call A001936(n) by $a(n)$. Here is a sketch of why $$a(n)\equiv \sigma(4n+1)\pmod{4}$$
Firs note that the generating function of $a(n)$ is 
$$A(x)=\sum_{n\geq 0}a(k)x^n=\prod_{k\geq 1}\left(\frac{1-x^{4k}}{1-x^k}\right)^2$$ for $\sigma(2n+1)$ the generating function is $$B(x)=\sum_{k\geq 0}\sigma(2k+1)x^k=\prod_{k\geq 0}(1-x^k)^4(1+x^k)^8$$
So $$B(x)\equiv \prod_{k\geq 1}(1+x^{2k})^2(1+x^{4k})^2\equiv \prod_{k\geq 1}(\frac{1-x^{8k}}{1-x^{2k}})^2\equiv A(x^2)\pmod{4}$$
Now the proof is complete once we know that $$B(x)\equiv \sum_{k\geq 0} \sigma(4n+1)x^{2n}\pmod{4}$$ this is an other way of saying $\sigma(4n-1)$ is divisible by $4$, which can be shown by pairing up the divisors $d+\frac{4n-1}{d}\equiv 0\pmod{4}$.
The proof for the other congruence is similar, but slightly longer, I might update this post later to include it. 

Let's prove that $\sigma(8n+1)\equiv q(n)\pmod{4}$, where $q(n)$ is the number of partitions with no even part repeated. The generating function is $$Q(x)=\sum_{n\geq 0}q(n)x^n=\prod_{k\geq 1}\frac{1-x^{4k}}{1-x^k}$$
Since we know from above that $$\sum_{n\geq 0}\sigma(4n+1)x^{2n}\equiv \prod_{k\geq 1}(1+x^{2k})^2(1+x^{4k})^2 \pmod{4}$$ we conclude that $$L(x)=\sum_{n\geq 0}\sigma(4n+1)x^n\equiv Q(x)^2 \pmod{4}$$
so that $$\sum_{n\geq 0} \sigma(8n+1)x^{2n}\equiv \frac{L(x)+L(-x)}{2}\pmod{4}$$
So to finish off the proof we need the following
$$\frac{Q(x)^2+Q(-x)^2}{2}\equiv Q(x^2)\pmod{4}$$  which I will leave as an exercise Actually let me write the proof, just to make sure I didn't mess up calculations. This reduces to proving
$$\frac{\prod_{k\geq 1}(1+x^{2k})^4(1+x^{2k-1})^2+\prod_{k\geq 1}(1+x^{2k})^4(1-x^{2k-1})^2}{2}$$ $$\equiv \prod_{k\geq 1}(1+x^{4k-2})(1+x^{4k})^2 \pmod{4}$$ and since $$(1+x^{2k})^4\equiv (1+x^{4k})^2 \pmod{4}$$ this reduces to
$$\frac{\prod_{k\geq 1}(1+x^{2k-1})^2+\prod_{k\geq 1}(1-x^{2k-1})^2}{2}\equiv \prod_{k\geq 1} (1+x^{4k-2})\pmod{4}$$ but we can write $$\prod_{k\geq 1}(1-x^{2k-1})^2\equiv \left(\prod_{k\ geq 1}(1+x^{2k-1})^2\right) \left(1-4\sum_{k\geq 1}\frac{x^{2k-1}}{(1+x^{2k-1})^2}\right)\pmod{8}$$ therefore now we have to show
$$\prod_{k\geq 1}(1+x^{2k-1})^2\left(1-2\sum_{k\geq 1}\frac{x^{2k-1}}{(1+x^{2k-1})^2}\right)\equiv \prod_{k\geq 1}(1+x^{4k-2})\pmod{4}$$ Now everything is clear since $$\prod_{k\geq 1}(1+x^{2k-1})^2\equiv \prod_{k\geq 1}(1+x^{4k-2})\left(1+2\sum_{k\geq 1}\frac{x^{2k-1}}{(1+x^{2k-1})^2}\right)\pmod{4}$$
