OFFSET
1,2
COMMENTS
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
REFERENCES
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.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..100
J. W. L. Glaisher, On a set of coefficients analogous to the Eulerian numbers, Proc. London Math. Soc., 31 (1899), 216-235.
Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010. [Author's named corrected by N. J. A. Sloane, Dec 12 2021]
FORMULA
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).
H'(n) = A000436(n)/2^(2n+1). - Philippe Deléham, Jan 17 2004
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
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
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
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
a(n) ~ (2*n)! * 3^(2*n+1/2) / Pi^(2*n+1). - Vaclav Kotesovec, Feb 26 2014
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.
From Vaclav Kotesovec, May 05 2020: (Start)
a(n) = (2*n)! * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (sqrt(3)*(2*Pi)^(2*n+1)).
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)
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
MAPLE
a := n -> (-1)^n*6^(2*n)*(Zeta(0, -n*2, 1/3)-Zeta(0, -n*2, 5/6)):
seq(a(n), n=1..14);
MATHEMATICA
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 *)
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 *)
PROG
(Maxima)
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 *)
CROSSREFS
KEYWORD
nice,easy,nonn
AUTHOR
STATUS
approved