From various considerations and with the help of J. Van der Jeugt, I was led to conjecture the following property of a class of Wigner 6j-symbols: for any integers $k,m$ with $m\ge k\ge 2$, $$ \left\{ \begin{array}{ccc} k & k & k \\ \frac{m}{2} & \frac{m}{2} & \frac{m}{2} \end{array} \right\}\ \neq\ 0 $$ except for the special case $k=2$, $m=3$.

Using Racah's single sum formula this amounts to the nonvanishing of the combinatorial sum $$ \sum_{n\in\mathbb{Z}} (-1)^n\ \left(\begin{array}{c} n+1 \\ 3k+1 \end{array}\right)\ \left(\begin{array}{c} k \\ n-k-m \end{array}\right)^3\ $$ with the usual convention that binomials are zero if the arguments are out of range.

Any ideas on how to prove this?

BTW, the exception $$ \left\{ \begin{array}{ccc} 2 & 2 & 2 \\ \frac{3}{2} & \frac{3}{2} & \frac{3}{2} \end{array} \right\}\ =\ 0 $$ has an interesting representation-theoretic interpretation (see this article by P. Minnaert).

Also note that the signs of these symbols look quite complicated. Using a matrix plot on Mathematica for the output of

Table[Sign[SixJSymbol[{2+i, 2+i, 2+i}, {(2+i +j)/2, (2+i+j)/2, (2+i+j)/2}]],

{i, 0, 200}, {j, 0, 200}]

produces this picture: enter image description here

The exceptional zero mentioned above is the tiny white square in the top left corner.

Note that following Gerhard's suggestion of looking at the problem mod 3 one can rule out zeros away from the white spots in: enter image description here

The above picture comes from using Fermat's Little Theorem to get rid of the third power mod 3 and the identity $$ \sum_{n\in\mathbb{Z}} (-1)^n\ \left(\begin{array}{c} n+1 \\ 3k+1 \end{array}\right)\ \left(\begin{array}{c} k \\ n-k-m \end{array}\right) \ =\ (-1)^m \left(\begin{array}{c} k+m+1 \\ 2k+1 \end{array}\right)\ . $$ This still leaves a piece of Swiss cheese.

  • $\begingroup$ Do you know the behaviour of these sums mod 3 or mod 7? Gerhard "Prefers To Add Smaller Sums" Paseman, 2018.07.04 $\endgroup$ – Gerhard Paseman Jul 4 '18 at 22:01
  • $\begingroup$ @GerhardPaseman: Indeed, looking at suitable congruences might be more promising than real-variable methods/inequalitiies, given how complicated the signs are. I didn't try mod p methods. Do you see interesting congruences? $\endgroup$ – Abdelmalek Abdesselam Jul 4 '18 at 22:30
  • $\begingroup$ Mod p methods may not work, as the $6j$ symbol with $k=p-1$ and $m=p$ is a multiple of $p$ when $p$ is prime, but the value itself is not 0. $\endgroup$ – LeechLattice Jul 5 '18 at 11:44
  • $\begingroup$ @Bullet51: The case $m=k+1$ is too easy anyway. $\endgroup$ – Abdelmalek Abdesselam Jul 5 '18 at 16:19
  • 1
    $\begingroup$ I have not computed these sums modulo any prime. I chose 3 and 7 because you can transform the sums to get simpler expressions (x^3 to x or 1 or -1 or 0). The hope is that you can say for which n the sum is not a multiple of 3 or 7. Gerhard "...And Then Further Inspiration Strikes... " Paseman, 2018.07.05. $\endgroup$ – Gerhard Paseman Jul 5 '18 at 18:50

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