A247239 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.
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, 64, 1, 8, 84, 784, 3611, 5600, 2080, 128, 1, 9, 120, 1680, 13328, 42065, 40000, 8256, 256, 1, 10, 165, 3312, 42048, 241472, 499955, 288000, 32896, 512, 1
Offset: 0
Examples
Array a(n,m) begins: 1, 1, 1, 1, 1, 1, 1, 1, ... 1 (A000012) 2, 4, 8, 16, 32, 64, 128, 256, ... 2^(m+1) (A000079) 3, 10, 36, 136, 528, 2080, 8256, 32896, ... A007582 4, 20, 120, 800, 5600, 40000, 288000, 2080000, ... A093123 5, 35, 329, 3611, 42065, 499955, 5980889, 71698571, ... not in the OEIS ... 1st column is n+1 (A000027). 2nd column is A000292. 3rd column is not in the OEIS.
Links
- I. M. Gessel, Generating functions and generalized Dedekind sums, Electron. J. Combin.4 (1997), no. 2, Research Paper 11, 17 pp.
- Les Mathematiques, Somme des 1/sin^2, Sketch of a proof [in French].
- MilesB, How to prove this sum involving powers of cosec is an integer?, MathOverflow 444094.
- R. P. Stanley, Invariants of finite groups and their applications to combinatorics, Bull. Amer. Math. Soc. 1 (1979), 475-511.
Programs
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Mathematica
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
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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
Formula
First formulas for rows:
a(0,m) = 1.
a(1,m) = 2^(m + 1).
a(2,m) = 2^m + 2^(2*m + 1).
a(3,m) = 2*((5 - sqrt(5))^m + (5 + sqrt(5))^m).
a(4,m) = 2^(2*m + 1) + 3^m + 2^(2*m + 1)*3^m.
First formulas for columns:
a(n,0) = n + 1.
a(n,1) = (n + 1)*(n + 2)*(n + 3)/6.
a(n,2) = coefficient of x^n in the expansion of (1 - x^4)/(1 - x)^8.
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
Comments