Up to $2000$, $A145722(n-1) \equiv \sigma(4n-3) \pmod{5}$ A145722 is
Expansion of f(q) * f(q^5) / phi(-q^2)^2 in powers of q where f(), phi() are Ramanujan theta functions.
Using the pari program and offset 0, up to $2000$, 
$$A145722(n-1)  \equiv \sigma(4n-3) \pmod{5}$$

Q1 Is this congruence true?

@Gjergji Zaimi proved similar congruence in another answer.
Since sigma is multiplicative, the offsets bug me a bit.
I would expect if $a(f(n)) \equiv \sigma(g(n))$ and $a$
is defined in a "natural" way, then either $f(n) \mid g(n)$ or
$g(n) \mid f(n)$.

Q2 In case the congruence is true, where $n-1$ and $4n-3$ come from?

Certain sequences (like modular forms, sigma, totient) can be computed significantly
faster if the index is factored.

Q3 Does factoring $n-1$ helps in computing $A145722(n-1)$? (I expect no).

 A: Yes, the congruence is true. Here's a modular forms proof. Let $\eta(z) = q^{1/24} \prod_{n=1}^{\infty} (1-q^{n})$, where $q = e^{2 \pi i z}$. Define
$$
g(z) = \frac{\eta(16z) \eta(40z)^{3}}{\eta(4z) \eta(8z) \eta(20z) \eta(80z)} = \sum_{n=0}^{\infty} {\rm A145722}(n) q^{4n+1}.
$$
This is a modular form of weight zero for $\Gamma_{0}(80)$ and character $\chi_{5}$. Fermat's little theorem shows that $\eta^{5}(z) \equiv \eta(5z) \pmod{5}$ (in the sense that their Fourier coefficients are congruent modulo $5$). Applying $\eta(40z) \equiv \eta(8z)^{5}$, $\eta(20z) \equiv \eta(4z)^{5}$ and $\eta(80z) \equiv \eta(16z)^{5}$ shows that $g(z) \equiv h(z) \pmod{5}$ where
$$
h(z) = \frac{\eta(8z)^{14}}{\eta(4z)^{6} \eta(16z)^{4}}.
$$
This is a holomorphic modular form of weight $2$ for $\Gamma_{0}(16)$.
It is reasonably well-known that $F(z) = \frac{\eta(4z)^{8}}{\eta(2z)^{4}} = \sum_{n=0}^{\infty} \sigma(2n+1) q^{2n+1}$ is a modular form of weight $2$ for $\Gamma_{0}(4)$. Thus $F \otimes \chi_{-1} = \sum_{n=1}^{\infty} \chi_{-1}(2n+1) \sigma(2n+1) q^{2n+1}$ is a modular form of weight $2$ and level $16$.
The space of modular forms of weight $2$ and level $16$ is $5$-dimensional and by checking the coefficients of $q^{0}$ through $q^{4}$, we can see that
$$
h(z) = \frac{F + F \otimes \chi_{-1}}{2} = \sum_{n=0}^{\infty} \sigma(4n+1) q^{4n+1}.
$$
This proves the congruence. Hopefully this explains naturally why the shift in offset occurs.
The most efficient way to compute ${\rm A145722}(n)$ is probably using some analogue of the Rademacher formula for $p(n)$, and in no instance can this be computed more quickly by factoring $n$. In fact, since $\log({\rm A145722}(n)) \sim \frac{2 \pi \sqrt{n}}{\sqrt{5}}$ computing ${\rm A145722}(n)$ must take $\Omega(n^{1/2})$ time. Factoring $n$ is much faster than that. 
