A conjecture by the late Romanian mathematician Alexandru Lupas.
Posted in [sci.math](http://mathforum.org/kb/message.jspa?messageID=3704516) in 2005, but no proof was found.
Physicist [Alan Sokal](http://www.physics.nyu.edu/faculty/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](http://oeis.org/A005259)
$$
A\_n = \sum\_{k=0}^n \binom{n}{k}\binom{n+k}{k}\sum\_{j=0}^k \binom{k}{j}^3 .
$$


**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
$$
\frac{1}{\pi^2}\int\_{-1}^1\int\_{-1}^1\frac{dp\;dq}{\sqrt{1-p^2}
\sqrt{1-q^2}\sqrt{1-2t(8pq+8q^2+1)+t^2}\;}
$$



A conjectured formula for Apéry numbers
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 
$$
\binom{k}{2\ell}\binom{2\ell}{\ell}\binom{2k-2\ell}{k-\ell}=\binom{k}{k-\ell}^{2}\binom{2k-2\ell}{k}
$$


