# A mystery sequence

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?

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 $$\sum _{k=n}^\infty\frac{q^{k^2}}{(q;q)_{k-n}(q^{n+1};q)_{k}}=\frac{q^{n^2}}{(q^{n+1};q)_n}\sum _{k=0}^\infty\frac{q^{k^2+2kn}}{(q;q)_{k}(q^{2n+1};q)_{k}}=\frac{q^{n^2}}{(q^{n+1};q)_\infty}.$$
• Miraculous! It can be added that the series in $(1\mathrm d)$ is the third order mock theta function, it appears in OEIS A000025 where it is also said that it is the generating function for difference between the number of partitions with even rank and with odd rank. – მამუკა ჯიბლაძე Feb 21 '17 at 20:09
• Presumably $(4)$ can be used to also prove the second identity (for $f_t$)... – მამუკა ჯიბლაძე Feb 21 '17 at 20:28
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.
• $\sum_{\substack{m,n\geqslant0\\n\ne1}}(-1)^mq^{\frac{(m+n)(3m+n+1)}2}$ is a Hecke type double sum, so it can be expressed in terms of Appell-Lerch sums and theta functions. The function$g(x)$ also can be expressed in similar form. So the proof should be straightforward. – Nemo Feb 20 '17 at 21:26