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
$$
A_n = \sum_{k=0}^n \binom{n}{k}\binom{n+k}{k}\sum_{j=0}^k \binom{k}{j}^3 .
$$
examples
$$
P_0(z)=1;\qquad
P_1(z)=z;\qquad
P_2(z)=\frac{3}{2}\;z^2-\frac{1}{2};\qquad
P_3(z)=\frac{5}{2}\;z^3-\frac{3}{2}\;z;
\\
A_0 = 1;\qquad A_1=5;\qquad A_2=73;\qquad A_3=1445;\qquad A_4=33001 .
$$
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
$$
\frac{1}{\pi^2}\int_{-1}^1\int_{-1}^1\frac{dp\;dq}{\sqrt{1-p^2}
\sqrt{1-q^2}\sqrt{1-2t(8pq+8q^2+1)+t^2}\;}
$$
Best Answer
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 $$ \binom{k}{2\ell}\binom{2\ell}{\ell}\binom{2k-2\ell}{k-\ell}=\binom{k}{k-\ell}^{2}\binom{2k-2\ell}{k} $$