This question arose from the recent one, roots of a polynomial linked to mock theta function?. Let $$ g(x):=\sum_{k=0}^\infty x^k\prod_{j=1}^{k-1}(1 + x^j)^2\\=1+x+x^2+3 x^3+4 x^4+6 x^5+10 x^6+15 x^7+21 x^8+30 x^9+43 x^{10}+59 x^{11}+...; $$ the sequence $1,1,1,3,4,6,10,15,21,30,43,59,...$ with the generating function $g(x)$ is A059618 on OEIS, it is the sequence of numbers of strongly unimodal partitions.

Now let $$ f(q):=g(q)\prod_{n=1}^\infty(1-q^n), $$ and let $a_k$ be the $k$th coefficient in the Maclaurin series for $f$, $$ f(x)=\sum_{k=0}^\infty a_kx^k\\=1-x^2+x^3+x^6+x^7-x^9+x^{10}-x^{14}+x^{18}-x^{20}+x^{21}+x^{25}+x^{26}-x^{27}\\+x^{28}-x^{30}+x^{33}-x^{35}+x^{36}-x^{39}-x^{40}+x^{42}-x^{44}+2x^{45}-x^{49}+x^{52}-x^{54}\\+x^{55}+x^{56}+x^{57}-x^{60}-x^{65}+... $$ The sequence of $a_k$, starting with

1,0,-1,1,0,0,1,1,0,-1,1,0,0,0,-1,0,0,0,1,0,-1,1,0,0,0,1,1,-1,1,0,-1,0,0,1,0,-1,1,0,0,-1,-1,0,1,0,-1,2,0,0,...

is not on OEIS. Among the first 1000 terms of the sequence, there are 609 zeroes, 182 ones, 161 -1s, 19 of them are 2 ($a_{45},a_{150},a_{210},a_{221},a_{273},a_{300},...$), 22 are -2 ($a_{77},a_{90},a_{165},a_{225},...$), and two of them ($a_{525}$ and $a_{825}$) are 3; seems like $a_k$ are zero for $k=2^j$ ($j>0$), for $k=p$ or $k=2p$, with $p$ prime $>7$, $k=3p$ and $k=4p$ with $p$ prime $\geqslant23$, $k=5p$ with $p$ prime $>31$, $6p$ for $p>37$, $7p$ and $8p$ for $p>43$, $9p$ for $p>47$, $10p$ for $p>61$, $11p$ for $p>67$,...

What may (or may not) be relevant is another sequence obtained from introducing new variable in the way I learned from a paper by Rhoades linked to from the above OEIS page for $g$.

Let $$ g_t(q):=\sum_{k=0}^\infty q^k\prod_{j=1}^{k-1}(1 + q^jt)(1+q^j/t), $$ and let $$ f_t(q)=g_t(q)\prod_{n=1}^\infty(1-q^n), $$ so that $g_1(q)=g(q)$ and $f_1(q)=f(q)$. Then $$ f_t(q)=1-q^2+\frac{1+t^3}{(1+t)t}q^3+\frac{1+t^5}{(1+t)t^2}q^6+q^7-\frac{1+t^3}{(1+t)t}q^9+\frac{1+t^7}{(1+t)t^3}q^{10}+...; $$ most coefficients have form $\pm\frac{1+t^{2j+1}}{(1+t)t^j}$, except that I cannot figure out how $j$ depends on the number of the coefficient. Exceptions here start from the $15$th coefficient, which is $\frac{1+t^9}{(1+t)t^4}-1$ and the $45$th one which is $\frac{1+t^{17}}{(1+t)t^8}+\frac{1+t^3}{(1+t)t}$.

Despite all these clues, to my shame I've given up searching for an explicit formula for $a_k$. Is there one? I am pretty sure there is, but what is it?