Proofs of some conjectures of Chan and Wang on congruences for \((q;q)_\infty^{\frac{a}{b}}\) (Q829768)

In 2003, \textit{S.T.~Ng} [The Ramanujan's partition congruences. Singapore: National University of Singapore (Undergratuate Thesis) (2003)] investigated the congruence properties for \(p_{a/b}(n)\), where \(a/b\) is a negative rational number. Using the theory of modular forms, he proved that for any \(n\geq0\), \[ p_{-2/3}(19n+9)\equiv0\pmod{19}. \] Here, we say that \(c/d\equiv0\pmod{p^k}\) if \(\nu_p(c)-\nu_p(d)\geq k\), where \(\nu_p(x)\) denote the highest power of \(p\) dividing \(x\). In 2019, \textit{H.H.~Chan} and \textit{L.~Wang} [Acta Arith. 187, No. 1, 59--80 (2019; Zbl 1435.11135)] further studied congruences properties satisfied by \(p_{a/b}(n)\), where \(a/b\in\mathbb{Q}\backslash\mathbb{Z}\). They derived a number of congruences for \(p_{a/b}(n)\). For instance, they proved that for any \(n\geq0\), \begin{align*} p_{1/3}(41n+37) &\equiv0\pmod{41},\\ p_{-1/5}(71n+29) &\equiv0\pmod{71}. \end{align*} At the end of their paper, Chan and Wang posed many conjectural congruences modulo powers of \(5\) and \(7\) enjoyed by \(p_{a/b}(n)\). For instance, they conjectured that for any \(n\geq0\), \begin{align*} p_{-3/4}(125n+r) &\equiv0\pmod{5^5},\qquad r\in\{93,118\},\tag{1}\\ p_{-1/2}(2401n+r) &\equiv0\pmod{7^4},\qquad r\in\{979,1665,2008,2351\}.\tag{2} \end{align*} Applying Ramanujan's modular equations of fifth, seventh and thirteenth orders, the authors not only confirm these conjectures due to Chan and Wang, but establish many new infinite families of congruences modulo powers of \(5\), \(7\), and \(13\) for \(p_{a/b}(n)\). For instance, they prove that for any \(a\in\mathbb{Z}\) and \(b\in\mathbb{N}_+\), \begin{itemize} \item If \(\gcd(5,b)=1\) and \(3125|(2343b-a)\), then \[ p_{a/b}(125n+r)\equiv0\pmod{5^5},\qquad r\in\{93,118\}.\tag{3} \] \item If \(\gcd(7,b)=1\) and \(2401|(1200b-a)\), then \[ p_{a/b}(2401n+r)\equiv0\pmod{7^4},\qquad r\in\{979,1665,2008,2351\}.\tag{4} \] \item If \(\gcd(13,b)=1\), \(1\leq j\leq7\), and \(13^j|(16b-a)\), \[ p_{a/b}{\left(13^jn+\dfrac{(3r+2)\times13^{j-1}-2}{3}\right)}\equiv0\pmod{13^j},\qquad r\in\{4,5,7,8,9,11,12\}. \] \end{itemize} Obviously, (1) and (2) are the special cases of (3) and (4), respectively.