A mystery sequence

The conjectured identity
$$
f(q)=(q;q)_inftyleft(1+sum_{k=1}^infty q^k(-q;q)^2_{k-1}right)=sum_{substack{m,ngeqslant0\nne1}}(-1)^mq^{frac{(m+n)(3m+n+1)}2},tag{1}
$$
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
$$
(q;q)_inftysum_{k=1}^infty q^k(-q;q)^2_{k-1}=sum_{substack{mge0}}sum_{substack{nge 1}}(-1)^mq^{frac{(m+n)(3m+n+1)}2}.tag{1a}
$$
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{mge0}}sum_{substack{nge 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
$$
sum _{k=n}^inftyfrac{q^{k^2}}{(q;q)_{k-n}(q^{n+1};q)_{k}}=frac{q^{n^2}}{(q^{n+1};q)_n}sum _{k=0}^inftyfrac{q^{k^2+2kn}}{(q;q)_{k}(q^{2n+1};q)_{k}}=frac{q^{n^2}}{(q^{n+1};q)_infty}.
$$

Found the following purely empirically, have no idea how to prove it:
$$
sum_{k=0}^infty a_kq^k=sum_{substack{m,ngeqslant0\nne1}}(-1)^mq^{frac{(m+n)(3m+n+1)}2};
$$
also, in case this might be useful,
$$
f_t(q)=sum_{substack{m,ngeqslant0\nne1}}(-1)^mfrac{1+t^{2n-1}}{(1+t)t^{n-1}}q^{frac{(m+n)(3m+n+1)}2}
$$
but I don’t have a proof of it either.

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