In the 1988 Narosa edition of Ramanujan's The Lost Notebook and Other Unpublished Papers, on the first line of page 1 is the following: $$ \Big(1+\frac1a\Big) \Bigg\{\frac{1}{(1-aq)(1-q/a)}+\frac{q(1+q)(1+q^2)}{(1-aq)(1-aq^3)(1-q/a)(1-q^3/a)}+\frac{q^2(1+q)..(1+q^4)}{..}+..\Bigg\} .$$ The infinite sum in braces with $a=-1$ has a $q$-series expansion$$ A(q) := 1 - q + 3q^2 - 2q^3 + 3q^4 - 3q^5 + 4q^6 - 3q^7 + 6q^8 - 4q^9 +\dots$$ which is the generating function of OEIS sequence A292511 and I conjecture that $A292511(n-1) = -(-1)^n A260195(n)$ where $A260195(n)$ is the number of integer triples $[x,y,z]$ such that $1\le\min(x,z),\max(x,z)\le y$ and $y^2-(x^2-x+z^2-z)/2=n$.

Is this conjecture true?

It may help that G. E. Andrews and B. C. Berndt in Ramanujan's lost notebook, Part I, page 277, in discussing Ramanujan's equation (Entry 12.4.5), have a result in equation (12.4.23) that leads to $$ q A(q) = \prod_{n>0} \frac{1+q^n}{1-q^n} \sum_{n>0} -(-1)^n q^{n^2}\frac{1-q^{2n-1}}{(1+q^{2n-1})^2} $$ which gives an alternate way of getting the $q$-series expansion.

  • $\begingroup$ @domotorp you could take the opportunity of your massive tagging "oeis" to also add the tag "sequences-and-series". $\endgroup$
    – YCor
    Mar 14, 2018 at 10:10

1 Answer 1


This conjecture is equivalent to the following $$\frac{q}{(1-q)^2}\sum_{n=0}^\infty(-q)^n \frac{(q;q^2){}_n(-q^2;q^2){}_n}{(q^3;q^2){}_n^2}=\sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)},\tag{1}$$ and after using the identity pointed out by M. Somos (with $q$ relaced by $-q$) $$ \frac{(q;-q)_\infty}{(-q;-q)_\infty}\sum_{n=-\infty}^\infty\frac{q^{n^2}}{(1-q^{2 n-1})^2}=\sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)}.\tag{1a} $$

Initial proof of $(1a)$ utilized the identity $$ \sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)-1}=\frac13\sum_{r,s,t\ge 0}(2q^{rs+st+tr+r+t+s}+(-q)^{rs+st+tr+r+t+s}),\tag{2} $$ which turned out to be another conjecture due to Michael Somos. So that initial proof was not self sufficient. Self sufficient proofs of $(1a)$ and $(2)$ are given below.

Proof of $(1a)$. By shifting the summation of variables so that all summations start from $0$ and by elementary rearrangements of terms we get \begin{align} \sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)-1}&=2\sum_{r,s,t\ge 0}q^{\frac{r^2+3r}{2}+s^2+2s+2rs+t(r+1+2s)}-\sum_{1\le r\le t}q^{t^2-(r^2-r)-1}\\ &=\left(\sum_{r,s,t\ge 0}+\sum_{r,s,t< 0}\right)q^{\frac{r^2+3r}{2}+s^2+2s+2rs+t(r+1+2s)}\\ &=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-q^{r+1+2s}}. \end{align}

It turns out that it is easier to work with a more general sum $$ g(z,q)=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z\,q^{r+2s}}. $$

Following the general method outlined in the article by E. Mortenson we derive the functional equation satisfied by $g(z,q)$: \begin{align} 0&=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right){q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}\\ &=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z\,q^{r+1+2s}}({1-z\,q^{r+1+2s}})\\ &=g(qz,q)-qz\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+5r}{2}+s^2+4s+2rs}}{1-z\,q^{r+1+2s}}\\ &=g(qz,q)-\frac{z}{q}\left(\sum_{r\ge 1,s\ge 0}-\sum_{r\le 0,s<0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z\,q^{r+2s}}\\ &=g(qz,q)-\frac{z}{q}g(z,q)+\frac{z}{q^2}j\left(-q,q^2\right) m\left(-{z }/{q},q^2,-q\right),\tag{3} \end{align} where in the last line we used notations $$ J_m=(q^m;q^m)_\infty,\quad j(x,q)=(q;q)_\infty(x;q)_\infty(q/x;q)_\infty,\quad m(x,q,z)=\frac1{j(z,q)}\sum_{n=-\infty}^\infty\frac{(-z)^nq^{n(n-1)/2}}{1-xzq^{n-1}}.\tag{4} $$

Since the Appell-Lerch sum $m(x,q,z)$ satisfies the equation (eq. 2.2c in Mortenson's paper) $$ m(qx,q,z)=1-xm(x,q,z),\tag{5} $$ one can show by elementary calculation that the function $$ g_1(z,q)=-\frac{z}{q^3}j\left(-q,q^2\right) m\left(-{z }/{q^2},q^2,-q\right) m\left(-{z}/{q},q^2,-q\right) $$ satisfies the same functional equation $(3)$ as $g(z,q)$. Now consider the function $$ h(z,q)=g(z,q)-g_1(z,q), $$ for which we have the functional equation $$ h(qz,q)=\frac{z}{q}h(z,q).\tag{6} $$ From the definition of $g(z,q)$ it is easy to see that it has simple poles at $z_0=q^n,~n\in\mathbb{Z}$ with residues being equal to finite sums of powers of $q$, i.e. polynomials in $q$. For example for $n\ge 1$ the residue is $$ \sum_{r+2s=n}q^{\frac{r^2-r}{2}+s^2+2rs}. $$

When $m\left(-{z }/{q^2},q^2,-q\right) $ is singular then $m\left(-{z }/{q},q^2,-q\right) $ is not and vice versa. So from the definition of $m(x,q,z)$ one can see that $g_1(z,q)$ has the same set of simple poles as $g(z,q)$. To find residues one can use the identity $m(-1,q^2,-q)=0$ and the functional equation $(5)$. The result is that $h(z,q)$ is analytic. Analytic function with functional equation $(6)$ is $0$ (for details one can consult Mortenson's paper, section 5). So we proved that $$ \left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z q^{r+2s}}=-\frac{z}{q^3}j\left(-q,q^2\right) m\left(-{z }/{q^2},q^2,-q\right) m\left(-{z}/{q},q^2,-q\right).\tag{7} $$

We want to calculate this expression when $z=q$. It is known that $m\left(-{z }/{q^2},q^2,-q\right)$ is singular at $z=q$ and $m\left(-{z }/{q},q^2,-q\right)=0$ at $z=q$. So one needs to resolve this ambiguity by applying L'Hôpital's rule: \begin{align} &\lim_{z\to q}m\left(-{z }/{q^2},q^2,-q\right)m\left(-{z }/{q},q^2,-q\right)\\ &=\frac{q}{j\left(-q,q^2\right) }\lim_{z\to q}\frac{m\left(-{z }/{q},q^2,-q\right)}{1-z/q}\\ &=\frac{q}{j\left(-q,q^2\right)}\frac{\frac d{dz}m\left(-{z }/{q},q^2,-q\right){\Large|_{\small{z=q}}}}{-1/q}\\ &=-\frac{q^2}{j^2\left(-q,q^2\right)}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}q^{2n-2}}{(1-q^{2n-1})^2}\\ &=-\frac{q}{j^2\left(-q,q^2\right)}\left(\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}-\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{1-q^{2n-1}}\right)\\ &=-\frac{q}{j^2\left(-q,q^2\right)}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}. \end{align}

So when $z\to q$ eq. $(7)$ becomes $$ \left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-q^{r+1+2s}}=\frac{1}{qj\left(-q,q^2\right)}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}\\ =\frac{(-q;-q)_\infty}{q(q;-q)_\infty}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}.\Box $$

Proof of $(2)$.

Here it is proved that $$ \frac13\sum_{r,s,t\ge 0}(2q^{rs+st+tr+r+t+s}+(-q)^{rs+st+tr+r+t+s})=\frac{(-q;-q)_\infty}{q(q;-q)_\infty}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2},(*) $$ which together with $(1a)$ will imply $(2)$.

The sums of the type $\sum_{r,s,t\ge 0}q^{rs+st+tr+r+t+s}$ have been studied by Mortenson, Corollary 1.2, where he proved that \begin{align} \sum_{r,s,t\ge 0}q^{rs+st+tr}x^{r+t+s}+&\sum_{r,s,t< 0}q^{rs+st+tr}x^{r+t+s}=\\ &=3\frac{J_1^3j(x^2,q)}{j(x,q)^2}m(-q/x,q^2,qx^2)-2\frac{J_1^3J_2^3j(x^2,q^2)^3}{j(x,q)^3j(-x,q^2)^3}\tag{8} \end{align} with the notations $(4)$.

Since the difference of two Appel-Lerch functions $m(x,q,z)$ with different $z$ is a quotient of theta functions (eq. 2.6 in Mortenson's paper) we can write in $(8)$
$$ \small{m(-q/x,q^2,qx^2)=m\left(-{q}/{x},q^2,z\right)+z\frac{J_2^3 j\left({q x^2}/{z},q^2\right) j\left(-xzq^2,q^2\right)}{j\left(q x^2,q^2\right) j\left(-xq^2,q^2\right) j\left(z,q^2\right) j\left(-{q z}/{x},q^2\right)}.}\tag{8a} $$ We want to calculate the triple sum in $(8),(8a)$ when $x=q$. Note that $$ \sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}=\sum_{r,s,t< 0}q^{rs+st+tr}q^{r+t+s}. $$ However $j(q,q)=0$ and since $j(x,q)$ appears in the denominator at the RHS of $(3)$ one needs to use L'Hôpital's rule to calculate the limit $x\to q$ with the help of formulas $\frac d{dx}j(x/q,q){\Large|_{\small{x=q}}}=-\frac1q J_1^3$ and $j(q^nx,q)=(-1/x)^nq^{-n(n-1)/2}j(x,q)$.

First, we put $z=-q$ and take the limit $x\to q$ in $(8),(8a)$ to obtain \begin{align} &\sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}\\ &=\frac{3 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}-\frac{8 J_2^{12}}{j\left(-q,q^2\right)^3 J_1^6}-3\frac{j\left(-q^2,q^2\right) J_2^6}{q^2j\left(-q,q^2\right) j\left(q^3,q^2\right) j\left(-q^3,q^2\right) j\left(q,q^2\right)}. \end{align} One can show that $$ \frac{8 J_2^{12}}{j\left(-q,q^2\right)^3 J_1^6}+3\frac{j\left(-q^2,q^2\right) J_2^6}{q^2j\left(-q,q^2\right) j\left(q^3,q^2\right) j\left(-q^3,q^2\right) j\left(q,q^2\right)}=2\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3}. $$ So $$ \sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}=\frac{3 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}-2\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3}.\tag{9} $$ Similarly, the choice $z=q$ in $(8),(8a)$ leads to $$ \sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}=\frac{3 }{q j\left(q,q^2\right)}\sum _{r=-\infty}^\infty \frac{(-1)^rq^{r^2}}{\left(1+q^{2 r-1}\right)^2}+4\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3},\tag{10} $$ and after changing sign $q\to-q$: $$ \sum_{r,s,t\ge 0}(-q)^{rs+st+tr+r+t+s}=-\frac{3 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}+4\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3},\tag{10a} $$ Combine $(9)$ and $(10a)$ to get $$ \frac13\sum_{r,s,t\ge 0}(2q^{rs+st+tr+r+t+s}+(-q)^{rs+st+tr+r+t+s})=\frac{1 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}. $$ Now observe that $\frac{(q;-q)_{\infty } j\left(-q,q^2\right)}{(-q;-q)_{\infty }}=1$ to complete the proof of $(*)$.$~ \Box$

  • $\begingroup$ The equation (2) from A130695 is another of my conjectures. Do you happen to have a proof of it also? $\endgroup$
    – Somos
    Sep 19, 2017 at 15:02
  • $\begingroup$ @Somos , no. I thought it was proved. $\endgroup$
    – Nemo
    Sep 19, 2017 at 15:15
  • $\begingroup$ @Somos , is the relation of A292511 to Hurwitz class number also a conjecture? $\endgroup$
    – Nemo
    Sep 19, 2017 at 16:25
  • $\begingroup$ If you mean the formula in A260195 relating to Hurwitz class number, then yes. $\endgroup$
    – Somos
    Sep 19, 2017 at 16:30
  • 13
    $\begingroup$ You should write a joint paper. $\endgroup$
    – coudy
    Sep 28, 2017 at 7:10

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