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Array a(n,m) = ((n+2)/2)^m*Sum_{k=1..n+1} 1/sin(k*Pi/(n+2))^(2m), n>=0, k>=0, read by ascending antidiagonals.
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%I #46 Apr 04 2023 15:23:46

%S 1,2,1,3,4,1,4,10,8,1,5,20,36,16,1,6,35,120,136,32,1,7,56,329,800,528,

%T 64,1,8,84,784,3611,5600,2080,128,1,9,120,1680,13328,42065,40000,8256,

%U 256,1,10,165,3312,42048,241472,499955,288000,32896,512,1

%N Array a(n,m) = ((n+2)/2)^m*Sum_{k=1..n+1} 1/sin(k*Pi/(n+2))^(2m), n>=0, k>=0, read by ascending antidiagonals.

%C Unexpectedly, it is conjectured (proof wanted) that the expression ((n+2)/2)^m * Sum_{k=1..n+1} 1/sin(k*Pi/(n+2))^(2m), n>=0, k>=0, always gives an integer.

%C For example, a(3,1) = (5/2)*(1/sin(Pi/5)^2 + 1/sin((2*Pi)/5)^2 + 1/sin((3*Pi)/5)^2 + 1/sin((4*Pi)/5)^2) = (5/2)*(2/(5/8 - sqrt(5)/8) + 2/(5/8 + sqrt(5)/8)), which simplifies to 20.

%H I. M. Gessel, <a href="https://doi.org/10.37236/1326">Generating functions and generalized Dedekind sums</a>, Electron. J. Combin.4 (1997), no. 2, Research Paper 11, 17 pp.

%H Les Mathematiques, <a href="http://www.les-mathematiques.net/phorum/read.php?3,1024481">Somme des 1/sin^2</a>, Sketch of a proof [in French].

%H MilesB, <a href="https://mathoverflow.net/questions/444094/how-to-prove-this-sum-involving-powers-of-cosec-is-an-integer">How to prove this sum involving powers of cosec is an integer?</a>, MathOverflow 444094.

%H R. P. Stanley, <a href="https://doi.org/10.1090/S0273-0979-1979-14597-X">Invariants of finite groups and their applications to combinatorics</a>, Bull. Amer. Math. Soc. 1 (1979), 475-511.

%F First formulas for rows:

%F a(0,m) = 1.

%F a(1,m) = 2^(m + 1).

%F a(2,m) = 2^m + 2^(2*m + 1).

%F a(3,m) = 2*((5 - sqrt(5))^m + (5 + sqrt(5))^m).

%F a(4,m) = 2^(2*m + 1) + 3^m + 2^(2*m + 1)*3^m.

%F First formulas for columns:

%F a(n,0) = n + 1.

%F a(n,1) = (n + 1)*(n + 2)*(n + 3)/6.

%F a(n,2) = coefficient of x^n in the expansion of (1 - x^4)/(1 - x)^8.

%F Let b(N,m) be (N/2)^m times the coefficient of x^(2*m) in 1-N*x*cot(N*arcsin(x))/ sqrt(1-x^2). Then for m>0, a(n,m) = b(n+2,m). - _Ira M. Gessel_, Apr 04 2023

%e Array a(n,m) begins:

%e 1, 1, 1, 1, 1, 1, 1, 1, ... 1 (A000012)

%e 2, 4, 8, 16, 32, 64, 128, 256, ... 2^(m+1) (A000079)

%e 3, 10, 36, 136, 528, 2080, 8256, 32896, ... A007582

%e 4, 20, 120, 800, 5600, 40000, 288000, 2080000, ... A093123

%e 5, 35, 329, 3611, 42065, 499955, 5980889, 71698571, ... not in the OEIS

%e ...

%e 1st column is n+1 (A000027).

%e 2nd column is A000292.

%e 3rd column is not in the OEIS.

%t a[n_, m_] := ((n + 2)/2)^m*Sum[1/Sin[k*(Pi/(n + 2))]^(2*m), {k, 1, n + 1}]; Table[a[n - m, m] // FullSimplify, {n, 0, 10}, {m, 0, n}] // Flatten

%o (PARI) a(n,m)={t=Pi/(n+2);u=1+n/2;round(sum(k=1,n+1,(u/sin(k*t)^2)^m))} \\ _M. F. Hasler_, Dec 03 2014

%Y Cf. A000079, A000292, A007582, A093123.

%K nonn,tabl

%O 0,2

%A _Jean-François Alcover_, Nov 28 2014