A \(q\)-analogue for Euler's \(\zeta(6) = \pi^6/945\) (Q2299109)

A recent \(q\)-analogue for Euler's \(\zeta(6)=\pi^2/6\) is due to Z. W. Sun 2018 and states that if \(q\) is complex with \(|q|<1\), then \[ \sum_{k=0}^{\infty} \frac{q^k(1+q^{2k+1})}{(1-q^{2k+1})^2}=\prod_{n=1}^{\infty} \frac{(1-q^{2n})^4}{(1-q^{2n-1})^4}. \] Motivated by this the author proves the following \(q\)-analogues of Euler's \(\zeta(4)=\pi^4/90\) and \(\zeta(6)=\pi^6/945\): \[ \sum_{k=0}^{\infty} \frac{q^kP_2(q^{2k+1})}{(1-q^{2k+1})^4}=\prod_{n=1}^{\infty} \frac{(1-q^{2n})^8}{1-q^{2n-1})^8} \] and \[ \sum_{k=0}^{\infty} \frac{q^k(1+q^{2k+1})P_4(q^{2k+1})}{(1-q^{2k+1})^6}-\phi^{12}(q)=256 q\prod_{n=0}^{\infty}\frac{(1-q^{2n})^{12}}{(1-q^{2n-1})^{12}}, \] where \(P_2(x)=x^2+4x+1\), \(P_4(x)=x^4+236x^3+1446x^2+236x+1\), \(\phi(q)=\prod_{n=1}^{\infty} (1-q^n)\). The proofs use the \(q\)-analogue of Euler's Gamma function and a result of [\textit{K. Ono} et al., Aequationes Math. 50, No. 1--2, 73--94 (1995; Zbl 0828.11057)] relating the number of representations of \(n\) as a sum of \(12\)-triangular numbers with the sum of fifth powers of divisors of \(2n+3\) and the coefficient of the \(12\)th power of the Dedekind \(\eta\)-function.