A conjecture by the late Romanian mathematician Alexandru Lupas. Posted in sci.math in 2005, but no proof was found. Physicist Alan Sokal just reminded me of it, saying it was related to something he is working on.

Let $P_n(z)$ be the Legendre polynomials, defined by the generating function $$ \big(1-2tz+t^2\big)^{-1/2} = \sum_{k=0}^\infty t^k P_k(z) . $$ Let $g(\alpha,\beta) = 4\cos(2\alpha)+8\sin(\beta)\cos(\alpha)+5$ be defined for $(\alpha,\beta) \in (-\pi,\pi)\times (-\pi,\pi)$ . Let $A_n$ be these Apéry numbers $$ A_n = \sum_{k=0}^n \binom{n}{k}\binom{n+k}{k}\sum_{j=0}^k \binom{k}{j}^3 . $$

$$ P_0(z)=1;\qquad P_1(z)=z;\qquad P_2(z)=\frac{3}{2}\;z^2-\frac{1}{2};\qquad P_3(z)=\frac{5}{2}\;z^3-\frac{3}{2}\;z; \\ A_0 = 1;\qquad A_1=5;\qquad A_2=73;\qquad A_3=1445;\qquad A_4=33001 . $$

Prove or disprove:
$A_n$ is the average of $P_n\circ g$. More explicitly: for all natural numbers $n$, $$ A_n = \frac{1}{4\pi^2}\int_{-\pi}^\pi \int_{-\pi}^\pi P_n\big(g(\alpha,\beta)\big)\;d\beta\;d\alpha $$

This is surely true (Sokal says he has checked it through $n=123$). But can you prove it?

additional notes

Also $$ A_n = \sum_{k=0}^n \binom{k}{n}^2\binom{n+k}{k}^2 $$ The conjecture should be equivalent to $$ \frac{1}{4\pi^2}\int_{-\pi}^\pi\int_{-\pi}^\pi\frac{d\alpha\;d\beta}{\sqrt{ t^2-2t(4\cos(2\alpha)+8\sin\beta\cos\alpha+5)+1}\;} =\sum_{k=0}^\infty A_k t^k $$ In the integral we can change variables to get $$ \frac{1}{\pi^2}\int_{-1}^1\int_{-1}^1\frac{dp\;dq}{\sqrt{1-p^2} \sqrt{1-q^2}\sqrt{1-2t(8pq+8q^2+1)+t^2}\;} $$

  • $\begingroup$ I believe the formula should read $$A_n = \frac{1}{4\pi^2}\int_{-\pi}^\pi \int_{\color{red}-\pi}^\pi P_n\big(g(\alpha,\beta)\big)\;d\beta\;d\alpha$$ $\endgroup$
    – heropup
    Commented Aug 18, 2014 at 19:43
  • $\begingroup$ Just a suggestion, Wadim Zudilin quit MO some time ago; he is a CARMA at Newcastle, Australia. His sort of thing. I write to him with questions occasionally, he sends brief answers or references. Has a new book with Shallit and Borwein cambridge.org/us/academic/subjects/mathematics/number-theory/… $\endgroup$
    – Will Jagy
    Commented Aug 18, 2014 at 23:30
  • 3
    $\begingroup$ @Will, my late colleague, Alf van der Poorten, is also credited as an author of that book. $\endgroup$ Commented Aug 19, 2014 at 1:53
  • $\begingroup$ @Gerry, yes, I was going to mention him and found that I could not be sure of the spelling. It seems quite a good book, I will probably buy it, I don't buy many books these days. $\endgroup$
    – Will Jagy
    Commented Aug 19, 2014 at 1:59
  • $\begingroup$ See the follow-up question here: mathoverflow.net/questions/179108 $\endgroup$ Commented Aug 22, 2014 at 14:01

1 Answer 1


Here as an approach. It is known that: $P_{n}(x) = \sum_{k=0}^{n}\binom{n}{k}\binom{-n-1}{k}\left(\frac{1-x}{2}\right)^{k}$ (see for example http://en.wikipedia.org/wiki/Legendre_polynomials) Therefore these expressions are equal $$ \sum_{k=0}^{n}\binom{n}{k}\binom{n+k}{k}\sum_{j=0}^{k}\binom{k}{j}^{3}=\sum_{k=0}^{n}\binom{n}{k}\binom{-n-1}{k}\frac{1}{4 \pi^{2}}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\left(\frac{1-g(x,y)}{2}\right)^{k}dydx $$ if $$ \sum_{j=0}^{k}\binom{k}{j}^{3}=\frac{1}{4 \pi^{2}}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\left(\frac{g(x,y)-1}{2}\right)^{k}dydx $$ One can see that $$ \left(\frac{g(x,y)-1}{2}\right)^{k} = 4^{k} (\cos^{2}x + \cos x \sin y )^k $$ Therefore we get $$ \frac{1}{4 \pi^{2}}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\left(\frac{g(x,y)-1}{2}\right)^{k}dydx = \frac{4^k}{4\pi^{2}}\int_{0}^{2\pi}\int_{0}^{2\pi}\sum_{j=0}^{k}\binom{k}{j}\cos^{2k-j}x \sin^{j} y dxdy $$

Since \begin{align*} \int_{0}^{2\pi} \cos^{m} x dx=\int_{0}^{2\pi} \sin^{m} x dx = \frac{2\pi} {2^m}\binom{m}{m/2} \quad \text{for} \quad m \quad \text{even, otherwise =0} \end{align*} We get $$ \frac{1}{4 \pi^{2}}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\left(\frac{g(x,y)-1}{2}\right)^{k}dydx = \sum_{\ell=0}^{k/2}\binom{k}{2\ell}\binom{2\ell}{\ell}\binom{2(k-\ell)}{k-\ell} $$

and now we want to show that $$ \sum_{j=0}^{k}\binom{k}{j}^{3}=\sum_{\ell=0}^{k/2}\binom{k}{2\ell}\binom{2\ell}{\ell}\binom{2(k-\ell)}{k-\ell} $$ The last equality follows from the identity: $$ \sum_{j=0}^{k}\binom{k}{j}^{3}=\sum_{\ell=k/2}^{k}\binom{k}{\ell}^{2}\binom{2\ell}{k} $$ See for example http://arxiv.org/pdf/math/0311195v1.pdf formula (2). because $$ \binom{k}{2\ell}\binom{2\ell}{\ell}\binom{2k-2\ell}{k-\ell}=\binom{k}{k-\ell}^{2}\binom{2k-2\ell}{k} $$

  • 8
    $\begingroup$ The identity involving $\sum_j {k \choose j}^3$ can also be proved by observing that both sides satisfy the recursion $(8k^2+32k+32)f(k+1) + (7k^2+35k+44)f(k+2) -(k^2+6k+9)f(k+3)$ and have the same initial values, where the recursion can be found by standard techniques (e.g., $\mathtt{listtorec}$ in Maple's $\mathtt{gfun}$ package). $\endgroup$ Commented Aug 18, 2014 at 18:13
  • 8
    $\begingroup$ Another proof : $\sum_{j} \binom{k}{j}^3 = \sum_r \binom{k}{r} \binom{k}{k-r} \binom{k}{r} = \sum_r \sum_t \binom{k}{r}\binom{k}{k-r}\binom{r}{t}\binom{k-r}{t} = \sum_r\sum_t \binom{k}{t}\binom{k-t}{r-t}\binom{k}{t}\binom{k-t}{k-r-t}= \sum_{t} \binom{k}{t}^2 \sum_r \binom{k-t}{r-t}\binom{k-t}{k-r-t} = \sum_{u} \binom{k}{u}^2 \sum_r \binom{u}{k-r}\binom{u}{r} = \sum_u \binom{k}{u}^2 \binom{2u}{k} = \sum_u \binom{k}{u} \binom{2u}{u} \binom{u}{k-u} = \sum_\ell \binom{k}{\ell}\binom{2(k-\ell)}{k-\ell} \binom{k-\ell}{\ell} = \sum_\ell \binom{k}{2\ell}\binom{2\ell}{\ell}\binom{2(k-\ell)}{k-\ell}.$ $\endgroup$ Commented Aug 18, 2014 at 19:09

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