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).