login
Glaisher's H' numbers.
(Formerly M4810 N2057)
15

%I M4810 N2057 #66 Dec 12 2021 20:35:49

%S 1,11,301,15371,1261501,151846331,25201039501,5515342166891,

%T 1538993024478301,533289474412481051,224671379367784281901,

%U 113091403397683832932811,67032545884354589043714301,46211522130188693681603906171

%N Glaisher's H' numbers.

%C a(n) mod 9 = 1,2,4,8,7,5 repeated period 6 (A153130, see also A001370). a(n) mod 10 = 1. - _Paul Curtz_, Sep 10 2009

%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="/A002114/b002114.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 Vladimir Kruchinin, <a href="http://arxiv.org/abs/1009.2565">Composition of ordinary generating functions</a>, arXiv:1009.2565 [math.CO], 2010. [Author's named corrected by _N. J. A. Sloane_, Dec 12 2021]

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

%F H'(n) = H(n)/3, where H(n)=2^(2n+1)*I(n) (see A002112) and e.g.f. for (-1)^n*I(n) is (3/2)/(1+exp(x)+exp(-x)) (see A047788, A047789).

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

%F For n > 0, H'(n) = Sum{k = 0..n, T(n, k)*9^(n-k)*2^(k-1) }; where DELTA is the operator defined in A084938, T(n, k) is the triangle, read by rows, given by :[0, 1, 0, 4, 0, 9, 0, 16, 0, 25, ...] DELTA [1, 0, 10, 0, 28, 0, 55, 0, 90, ..]= {1}; {0, 1}; {0, 1, 1}; {0, 1, 12, 1}; {0, 1, 63, 123, 1}; {0, 1, 274, 2366, 1234, 1}; ... For 1, 10, 28, 55, 90, 136, ... see A060544 or A060544. - _Philippe Deléham_, Jan 17 2004

%F E.g.f. 1/2*1/(2*cos(x)-1). a(n)=sum(sum(binomial(k,j)*(-1)^(k-j+1)*1/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(2*i-j)^(2*n),i,0,floor((j-1)/2)),j,0,k)*(-2)^(k-1),k,1,2*n), n>0. - _Vladimir Kruchinin_, Aug 05 2010

%F E.g.f.: E(x)= 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=0,1,..., a(k+1)*x^(2k+2)), then A002114(k) = a(k+1)*(2*k+2)!. - _Sergei N. Gladkovskii_, Jan 09 2012

%F a(n) ~ (2*n)! * 3^(2*n+1/2) / Pi^(2*n+1). - _Vaclav Kotesovec_, Feb 26 2014

%F a(n) = (-1)^n*6^(2*n)*(zeta(-n*2,1/3)-zeta(-n*2,5/6)), where zeta(a, z) is the generalized Riemann zeta function.

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

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

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

%F Conjectural e.g.f.: Sum_{n >= 1} (-1)^n*Product_{k = 1..n} (1 - exp(A007310(k)*z) ) = z + 11*z^2/2! + 301*z^3/3! + .... - _Peter Bala_, Dec 09 2021

%p a := n -> (-1)^n*6^(2*n)*(Zeta(0,-n*2,1/3)-Zeta(0,-n*2, 5/6)):

%p seq(a(n), n=1..14);

%t Select[Rest[With[{nn=28},CoefficientList[Series[1/(2 (2Cos[x]-1)), {x,0,nn}], x]Range[0,nn]!]],#!=0&] (* _Harvey P. Dale_, Jul 27 2011 *)

%t FullSimplify[Table[(-1)^(s+1) * BernoulliB[2*s] * (Zeta[2*s + 1, 1/6] - Zeta[2*s + 1, 5/6]) / (4*Pi*Sqrt[3]*Zeta[2*s]), {s, 1, 20}]] (* _Vaclav Kotesovec_, May 05 2020 *)

%o (Maxima)

%o a(n) := sum(sum(binomial(k,j)*(-1)^(k-j+1)*1/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(2*i-j)^(2*n),i,0,floor((j-1)/2)),j,0,k)*(-2)^(k-1),k,1,2*n) (* _Vladimir Kruchinin_, Aug 05 2010 *)

%K nice,easy,nonn

%O 1,2

%A _N. J. A. Sloane_