12
$\begingroup$

In the 1988 Narosa edition of Ramanujan's The Lost Notebook and Other Unpublished Papers, on the first line of page 1 is the following: $$ \Big(1+\frac1a\Big)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.

$\endgroup$
1
  • $\begingroup$ @domotorp you could take the opportunity of your massive tagging "oeis" to also add the tag "sequences-and-series". $\endgroup$
    – YCor
    Mar 14 '18 at 10:10
12
$\begingroup$

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 $(*)$.$~ \Box$

$\endgroup$
7
  • $\begingroup$ The equation (2) from A130695 is another of my conjectures. Do you happen to have a proof of it also? $\endgroup$
    – Somos
    Sep 19 '17 at 15:02
  • $\begingroup$ @Somos , no. I thought it was proved. $\endgroup$
    – Nemo
    Sep 19 '17 at 15:15
  • $\begingroup$ @Somos , is the relation of A292511 to Hurwitz class number also a conjecture? $\endgroup$
    – Nemo
    Sep 19 '17 at 16:25
  • $\begingroup$ If you mean the formula in A260195 relating to Hurwitz class number, then yes. $\endgroup$
    – Somos
    Sep 19 '17 at 16:30
  • 13
    $\begingroup$ You should write a joint paper. $\endgroup$
    – coudy
    Sep 28 '17 at 7:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.