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Consider the following two infinite series $$\sum_{n\geq0}a(n)q^n=\prod_{k\geq1}\frac1{(1-q^k)^2(1-q^{5k})^2} \,\,\,\, \text{and} \,\,\, \sum_{n\geq0}b(n)q^n=\prod_{k\geq1}\frac1{(1-q^k)^2(1-q^{7k})^2}.$$ Denote the triangular numbers and pentagonal numbers by $t_n:=\binom{n+1}2$ and $\omega_n:=\frac{n(3n+1)}2$, respectively.

QUESTION. Are these two congruences true modulo $4$? $$\sum_{n\geq0}a(4n+3)q^n\equiv \sum_{n\in\mathbb{Z}}q^{5t_n} \qquad \text{and} \qquad \sum_{n\geq0}b(4n+3)q^n\equiv 2\sum_{n\in\mathbb{Z}}q^{14\omega_n}.$$

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  • $\begingroup$ It seems that $\omega_n$ does not appear in your question - would one of $t_n$ be $\omega_n$ instead? $\endgroup$
    – Seewoo Lee
    Commented Jan 23 at 18:36
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    $\begingroup$ Ouch. You are right. $\endgroup$
    – T. Amdeberhan
    Commented Jan 23 at 18:38
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    $\begingroup$ Isn't that just a matter of q-series multisection? $\endgroup$
    – Max Alekseyev
    Commented Jan 23 at 22:19
  • $\begingroup$ True, this is 4-sectioning but the point is always to determine each section explicitly. Is it not? After all Ramanujan's famous $\sum p(5n+4)q^n=5\frac{(q^5)_{\infty}^5}{(q)_{\infty}^6}$ is also multi-sectioning. $\endgroup$
    – T. Amdeberhan
    Commented Jan 24 at 1:24

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