Timeline for Conjectural nonvanishing of some combinatorial sums (6j symbols)
Current License: CC BY-SA 4.0
14 events
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| Jul 20, 2018 at 17:31 | history | edited | Abdelmalek Abdesselam | CC BY-SA 4.0 |
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| Jul 10, 2018 at 3:30 | comment | added | Gerhard Paseman | Rather than write out an answer, I will suggest rewriting n=i+k+m, have the index i go from 0 to k/2 or just short of that, and then rewrite as (k choose i)^3 times a sum or difference of two binomial terms involving i. Now when m is large enough, the cubed term changes faster with i than the sum or difference, so that your alternating sum is nonzero by monotonicity. Gerhard "Overlooks Some Details In Comment" Paseman, 2018.07.09. | |
| Jul 5, 2018 at 19:38 | comment | added | Abdelmalek Abdesselam | @GerhardPaseman: I tried your mod 3 suggestion which helps a little. See edited question. | |
| Jul 5, 2018 at 19:37 | history | edited | Abdelmalek Abdesselam | CC BY-SA 4.0 |
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| Jul 5, 2018 at 18:50 | comment | added | Gerhard Paseman | 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. | |
| Jul 5, 2018 at 16:19 | comment | added | Abdelmalek Abdesselam | @Bullet51: The case $m=k+1$ is too easy anyway. | |
| Jul 5, 2018 at 11:44 | comment | added | LeechLattice | 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. | |
| Jul 4, 2018 at 22:30 | comment | added | Abdelmalek Abdesselam | @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? | |
| Jul 4, 2018 at 22:01 | comment | added | Gerhard Paseman | Do you know the behaviour of these sums mod 3 or mod 7? Gerhard "Prefers To Add Smaller Sums" Paseman, 2018.07.04 | |
| Jul 4, 2018 at 21:45 | history | edited | Abdelmalek Abdesselam |
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| Jul 4, 2018 at 19:43 | history | edited | Abdelmalek Abdesselam | CC BY-SA 4.0 |
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| Jul 4, 2018 at 17:49 | history | edited | Abdelmalek Abdesselam | CC BY-SA 4.0 |
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| Jul 4, 2018 at 17:11 | history | edited | Abdelmalek Abdesselam | CC BY-SA 4.0 |
fixed mistake in formula
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| Jul 4, 2018 at 16:01 | history | asked | Abdelmalek Abdesselam | CC BY-SA 4.0 |