A002112 - OEIS

%I M3135 N1272 #53 Dec 24 2021 02:31:10

%S 3,33,903,46113,3784503,455538993,75603118503,16546026500673,

%T 4616979073434903,1599868423237443153,674014138103352845703,

%U 339274210193051498798433,201097637653063767131142903,138634566390566081044811718513

%N Glaisher's H numbers.

%D A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 76.

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H Vincenzo Librandi, <a href="/A002112/b002112.txt">Table of n, a(n) for n = 1..100</a>

%H J. W. L. Glaisher, <a href="https://doi.org/10.1112/plms/s1-31.1.216">On a set of coefficients analogous to the Eulerian numbers</a>, Proc. London Math. Soc., 31 (1899), 216-235.

%H Michael E. Hoffman, <a href="https://doi.org/10.37236/1453">Derivative polynomials, Euler polynomials, and associated integer sequences</a>, The Electronic Journal of Combinatorics 6.1 (1999).

%H <a href="/index/Ge#Glaisher">Index entries for sequences related to Glaisher's numbers</a>

%F H(n) = 2^(2n+1)*I(n), where e.g.f. for (-1)^n*I(n) is (3/2)/(1+exp(x)+exp(-x)) (see A047788, A047789).

%F H(n) = 3*A000436(n)/2^(2n+1) = 3*A002114(n). - _Philippe Deléham_, Jan 17 2004

%F E.g.f.: E(x) = 3*x^2/(G(0)-x^2); G(k) = 2*(2*k+1)*(k+1) - x^2 + 2*x^2*(2*k+1)*(k+1)/G(k+1); (continued fraction Euler's kind, 1-step ). - _Sergei N. Gladkovskii_, Jan 03 2012

%F If E(x) = Sum_{k>=0} a(k+1)*x^(2k+2), then A002112(k) = a(k+1)*(2*k+2)!. - _Sergei N. Gladkovskii_, Jan 09 2012

%F From _Vaclav Kotesovec_, May 05 2020: (Start)

%F a(n) = sqrt(3) * (2*n)! * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (2*Pi)^(2*n+1).

%F a(n) = (-1)^(n+1) * sqrt(3) * Bernoulli(2*n) * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (4*Pi*zeta(2*n)). (End)

%t e[0] = 1; e[n_] := e[n] = (-1)^n*(1 - Sum[(-1)^i*Binomial[2n, 2i]*3^(2n-2i)*e[i], {i, 0, n-1}]); a[n_] := 3*e[n]/2^(2n+1); Table[a[n], {n, 1, 14}] (* _Jean-François Alcover_, Jan 31 2012, after _Philippe Deléham_ *)

%K nonn,nice,easy

%O 1,1

%A _N. J. A. Sloane_