A036281 Denominators in Taylor series for x * cosec(x).
1, 6, 360, 15120, 604800, 3421440, 653837184000, 37362124800, 762187345920000, 2554547108585472000, 401428831349145600000, 143888775912161280000, 846912068365871834726400000, 93067260259985915904000000, 2706661834818276108533760000000
Offset: 0
Examples
cosec(x) = x^(-1)+1/6*x+7/360*x^3+31/15120*x^5+... 1, 1/6, 7/360, 31/15120, 127/604800, 73/3421440, 1414477/653837184000, 8191/37362124800, ...
References
- M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 75 (4.3.68).
- G. W. Caunt, Infinitesimal Calculus, Oxford Univ. Press, 1914, p. 477.
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..224 (terms 0..100 from T. D. Noe)
- M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
- M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 75 (4.3.68).
- M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 30.
- J. Malenfant, Factorization of and Determinant Expressions for the Hypersums of Powers of Integers, arXiv preprint arXiv:1104.4332 [math.NT], 2011.
- Eric Weisstein's World of Mathematics, Hyperbolic Cosecant.
- Eric Weisstein's World of Mathematics, Cosecant.
- Herbert S. Wilf, Generatingfunctionology, Academic Press, NY, 1994. See p. 54.
- Index entries for Bernoulli numbers B(2n).
Programs
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Maple
series(csc(x),x,60);
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Mathematica
a[n_] := 2(2^(2n-1)-1) Abs[BernoulliB[2n]]/(2n)! // Denominator; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Jul 14 2018 *)
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Sage
def A036281_list(len): R, C = [1], [1]+[0]*(len-1) for n in (1..len-1): for k in range(n, 0, -1): C[k] = -C[k-1] / (k*(4*k+2)) C[0] = -sum(C[k] for k in (1..n)) R.append(C[0].denominator()) return R print(A036281_list(15)) # Peter Luschny, Feb 21 2016
Formula
A036280(n)/a(n)= 2 *(2^(2n-1) -1) *abs(B(2n)) / (2n)!.
From Arkadiusz Wesolowski, Oct 16 2013: (Start)
a(n) = A036280(n)*Pi^(2*n)/(zeta(2*n)*(2 - (2^(1-n))^2)).
a(n) = A230265(n)/2. (End)