A mystery sequence

The conjectured identity $$ f(q)=(q;q)_\infty\left(1+\sum_{k=1}^\infty q^k(-q;q)^2_{k-1}\right)=\sum_{\substack{m,n\geqslant0\\n\ne1}}(-1)^mq^{\frac{(m+n)(3m+n+1)}2},\tag{1} $$ using Euler's pentagonal number theorem $(q;q)_\infty=\sum _{m=-\infty}^\infty (-1)^m q^{\frac{1}{2} m (3 m+1)}$ can be brought to an equivalent form $$ (q;q)_\infty\sum_{k=1}^\infty q^k(-q;q)^2_{k-1}=\sum_{\substack{m\ge0}}\sum_{\substack{n\ge 1}}(-1)^mq^{\frac{(m+n)(3m+n+1)}2}.\tag{1a} $$ By identity $(4.1)$ in Rhoades' paper the sum on the LHS of $(1a)$ is $$ \sum_{k=1}^\infty q^k(-q;q)^2_{k-1}=\frac{1}{2(q;q)_{\infty }} \sum _{n=-\infty}^\infty \frac{q^{\frac{1}{2} n (n+1)}}{q^n+1}-\frac{1}{4}\sum _{k=0}^\infty \frac{q^{k^2}}{(-q;q)_k^2},\tag{2} $$ while the double sum in $(1a)$ \begin{align} \sum_{\substack{m\ge0}}\sum_{\substack{n\ge 1}}(-1)^mq^{\frac{(m+n)(3m+n+1)}2}&=\sum _{m=0}^\infty \sum _{k=m+1}^\infty (-1)^mq^{\frac{k(k+1+2m)}2}\\ &=\sum _{k=1}^\infty \sum _{m=0}^{k-1} (-1)^m q^{m k+\frac{1}{2} (k+1) k}\\ &=\sum _{k=1}^\infty q^{\frac{k^2}{2}+\frac{k}{2}} \frac{1-(-1)^k q^{k^2}}{1+q^k}.\tag{3} \end{align} Since $\sum _{n=-\infty}^\infty \frac{q^{\frac{1}{2} n (n+1)}}{q^n+1}=\frac{1}{2}+2 \sum _{n=1}^\infty \frac{q^{\frac{1}{2} n (n+1)}}{q^n+1}$ $(2)$ and $(3)$ contain the same sum $\sum _{n=1}^\infty \frac{q^{\frac{1}{2} n (n+1)}}{q^n+1}$. This sum cancels out after $(2)$ and $(3)$ are substituted in $(1a)$ resulting in $$ \frac14-\frac{(q;q)_{\infty }}{4}\sum _{k=0}^\infty \frac{q^{k^2}}{(-q;q)_k^2}=-\sum _{n=1}^\infty \frac{(-1)^n q^{\frac{3 n^2}{2}+\frac{n}{2}}}{1+q^n},\tag{1c} $$ and equivalently $$ \sum _{k=0}^\infty \frac{q^{k^2}}{(-q;q)_k^2}=\frac{2}{(q;q)_{\infty }}\sum _{n=-\infty}^\infty \frac{(-1)^n q^{\frac{3 n^2}{2}+\frac{n}{2}}}{1+q^n}.\tag{1d} $$ $(1d)$ corresponds to the special case $x=-1$ of the identity $$ \sum _{k=0}^\infty \frac{q^{k^2}}{(x q;q)_k(q/x;q)_k}=\frac{1-x}{(q;q)_{\infty }}\sum _{n=-\infty}^\infty \frac{(-1)^n q^{\frac{3 n^2}{2}+\frac{n}{2}}}{1-xq^n},\tag{4} $$ which can be obtained from Watson-Whipple transformation formula (see the paper "Modular transformations of Ramanujan’s fifth and seventh order mock theta functions", Ramanujan J. 7 (2003), 193–222. by Gordon and McIntosh). $(4)$ also can be proved directly by partial fractions expansion and the following limiting case of q-Gauss summation $$ \sum _{k=n}^\infty\frac{q^{k^2}}{(q;q)_{k-n}(q^{n+1};q)_{k}}=\frac{q^{n^2}}{(q^{n+1};q)_n}\sum _{k=0}^\infty\frac{q^{k^2+2kn}}{(q;q)_{k}(q^{2n+1};q)_{k}}=\frac{q^{n^2}}{(q^{n+1};q)_\infty}. $$

Found the following purely empirically, have no idea how to prove it: $$ \sum_{k=0}^\infty a_kq^k=\sum_{\substack{m,n\geqslant0\\n\ne1}}(-1)^mq^{\frac{(m+n)(3m+n+1)}2}; $$ also, in case this might be useful, $$ f_t(q)=\sum_{\substack{m,n\geqslant0\\n\ne1}}(-1)^m\frac{1+t^{2n-1}}{(1+t)t^{n-1}}q^{\frac{(m+n)(3m+n+1)}2} $$ but I don't have a proof of it either.