cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A002114 Glaisher's H' numbers.

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%I A002114 M4810 N2057 #70 Aug 18 2025 15:43:02
%S A002114 1,11,301,15371,1261501,151846331,25201039501,5515342166891,
%T A002114 1538993024478301,533289474412481051,224671379367784281901,
%U A002114 113091403397683832932811,67032545884354589043714301,46211522130188693681603906171
%N A002114 Glaisher's H' numbers.
%C A002114 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 A002114 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 A002114 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D A002114 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H A002114 Vincenzo Librandi, <a href="/A002114/b002114.txt">Table of n, a(n) for n = 1..100</a>
%H A002114 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 A002114 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 A002114 <a href="/index/Ge#Glaisher">Index entries for sequences related to Glaisher's numbers</a>
%F A002114 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 A002114 H'(n) = A000436(n)/2^(2n+1). - _Philippe Deléham_, Jan 17 2004
%F A002114 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 A002114 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 A002114 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 A002114 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 A002114 a(n) ~ (2*n)! * 3^(2*n+1/2) / Pi^(2*n+1). - _Vaclav Kotesovec_, Feb 26 2014
%F A002114 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 A002114 From _Vaclav Kotesovec_, May 05 2020: (Start)
%F A002114 a(n) = (2*n)! * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (sqrt(3)*(2*Pi)^(2*n+1)).
%F A002114 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 A002114 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 A002114 a := n -> (-1)^n*6^(2*n)*(Zeta(0,-n*2,1/3)-Zeta(0,-n*2, 5/6)):
%p A002114 seq(a(n), n=1..14);
%t A002114 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 A002114 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 A002114 (Maxima)
%o A002114 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 A002114 nice,easy,nonn
%O A002114 1,2
%A A002114 _N. J. A. Sloane_