Denote $\phi(q):=\prod_{j\geq1}(1-q^j)$ and let $\xi=e^{\frac{2\pi i}3}$ be a cube root of unity.

Define the sequence $u(n)$ by $$\prod_{n\geq1}\prod_{s=1}^2(1-q^n\xi^{ns})(1-q^{2n}\xi^{ns})=\sum_{n\geq0}u(n)\,q^n.$$

QUESTION. Is this true? $$\sum_{n\geq0} u(3n+2)q^{3n+2}=3q^2\,\phi^2(q^9)\,\phi^2(q^{18}).$$

Denote $\phi(q):=\prod_{j\geq1}(1-q^j)$ and let $\xi=e^{\frac{2\pi i}3}$ be a cube root of unity.

Define the sequence $u(n)$ by $$\prod_{n\geq1}\prod_{s=1}^2(1-q^n\xi^{ns})(1-q^{2n}\xi^{ns}) =\sum_{n\geq0}u(n)\,q^n.$$

QUESTION. Is this true? $$\sum_{n\geq0} u(3n+2)q^{3n+2}=3q^2\,\phi^2(q^9)\,\phi^2(q^{18}).$$

Addendum. To help readers, if we let $$f_n(q)=(1 + q^{3n - 1} + q^{2(3n - 1)})(1 + q^{3n - 2} + q^{2(3n-2)}) (1 - q^{3n})^2,$$ then we have $$\prod_{n\geq1}\prod_{s=1}^2(1-q^n\xi^{ns})(1-q^{2n}\xi^{ns}) =\prod_{n\geq1}f_n(q) f_n(q^2).$$

Denote $\phi(q):=\prod_{j\geq1}(1-q^j)$ and let $\xi=e^{\frac{2\pi i}3}$ be a cube root of unity.

Define the sequences $u(n)$ by $$\prod_{n\geq1}\prod_{s=1}^2(1-q^n\xi^{ns})(1-q^{2n}\xi^{ns})=\sum_{n\geq0}u(n)\,q^n.$$

QUESTION. Is this true? $$\sum_{n\geq0} u(3n+2)q^{3n+2}=3q^2\,\phi^2(q^9)\,\phi^2(q^{18}).$$

Denote $\phi(q):=\prod_{j\geq1}(1-q^j)$ and let $\xi=e^{\frac{2\pi i}3}$ be a cube root of unity.

Define the sequence $u(n)$ by $$\prod_{n\geq1}\prod_{s=1}^2(1-q^n\xi^{ns})(1-q^{2n}\xi^{ns})=\sum_{n\geq0}u(n)\,q^n.$$

QUESTION. Is this true? $$\sum_{n\geq0} u(3n+2)q^{3n+2}=3q^2\,\phi^2(q^9)\,\phi^2(q^{18}).$$