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 $d+\frac{4n-1}{d}\equiv 0\pmod{4}$.
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 $$\frac{Q(x)^2+Q(-x)^2}{2}\equiv Q(x)\pmod{4}$$ Q(x^2)\pmod{4}$$which I will leave as an exercise Actually let me write the proof, just to make sure I didn't mess up calculations. This reduces to proving$$\prod_{k\geq 1} $\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 \left(\frac{1-x^{4k}}{1-x^k}\right)\left(1-2\sum_{k\geq 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 \frac{x^{2k+1}}{(1+x^{2k+1})^2}\right)\pmod{4}$$which I will leave as an exercise :left(\prod_{k\ geq 1}(1+x^{2k-1})^2\right) \left(1-4\sum_{k\geq 1}\frac{x^{2k-1}}{(1+x^{2k-1})^2}\right)\pmod{8}$$ 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 $$\prod_{k\geq 1}(1+x^{2k-1})^2\equiv \prod_{k\geq 1}(1+x^{4k-2})\left(1+2\sum_{k\geq 1}\frac{x^{2k-1}}{(1+x^{2k-1})^2}\right)\pmod{4}$$ 3 added 9 characters in body 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 $$Q(n)=\sum_{n\geq Q(x)=\sum_{n\geq 0}q(n)x^n=\prod_{k\geq 1}\frac{1-x^{4k}}{1-x^k}$$ Since we know from above that $$\sum_{n\geq 0}\sigma(4n+1)x^{2n}\equiv \prod_{k\geq 1}(1+x^{2k})^2(1+x^{4k})^2$$ 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)\pmod{4}$$which reduces to$$\prod_{k\geq 1} \left(\frac{1-x^{4k}}{1-x^k}\right)\left(1-2\sum_{k\geq 1} \frac{x^{2k+1}}{(1+x^{2k+1})^2}\right)\pmod{4}$$which I will leave as an exercise :) 2 added 782 characters in body Let's call A001936(n) by a(n). Here is a sketch of why$$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(n)=\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$$we conclude that$$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)\pmod{4}$$which reduces to$$\prod_{k\geq 1} \left(\frac{1-x^{4k}}{1-x^k}\right)\left(1-2\sum_{k\geq 1} \frac{x^{2k+1}}{(1+x^{2k+1})^2}\right)\pmod{4}$\$ which I will leave as an exercise :)