A135594 a(n) = (1/2^n) * Sum_{i=0..n} (-1)^(n-i) * binomial(n,i) * A000364(i).
1, 0, 1, 6, 73, 1380, 37801, 1417626, 69802993, 4369750440, 339034806001, 31935510092046, 3590398569115513, 474937566660074700, 73024143791301120601, 12914495107705743175266, 2603190607000627341985633, 593297406341867021292734160
Offset: 0
References
- I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, Exercise 4.2.2.(b).
Programs
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Maple
A000364 := proc(n) option remember ; (2*n)!*coeftayl(sec(x),x=0,2*n) ; end: A135594 := proc(n) add((-1)^(n-i)*binomial(n,i)*A000364(i),i=0..n)/2^n ; end: seq(A135594(n),n=0..20) ; # R. J. Mathar, Mar 14 2008 f:=sec(z): fser:=series(f,z=0,63): for n from 0 to 60 do b[n]:=factorial(n)*coeff(fser,z,n) end do: a:= proc(n) options operator, arrow: add((-1)^(n-k)*binomial(n,k)*b[2*k],k=0..n)/2^n end proc: seq(a(n),n=0..16); # Emeric Deutsch, Mar 17 2008
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Mathematica
Table[(-1)^n*Sum[Binomial[n, k]*EulerE[2*k], {k, 0, n}]/2^n, {n, 0, 20}] (* Vaclav Kotesovec, Jun 08 2019 *)
Formula
G.f.: 1/Q(0), where Q(k)= 1 + x - x*(2*k+1)*(k+1)/(1 - x*(2*k+1)*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 04 2013
a(n) ~ 2^(3*n + 3) * n^(2*n + 1/2) / (Pi^(2*n + 1/2) * exp(2*n)). - Vaclav Kotesovec, Jun 08 2019
Conjecture: e.g.f. as a continued fraction: 2*exp(-t)/(2 - (1-exp(-4*t))/(2 - (1-exp(-8*t))/(2 - (1-exp(-12*t))/(2 - ... )))) = 1 + t^2/2! + 6*t^3/3! + 73*t^4/4! + .... Cf. A000657 and A005799. - Peter Bala, Dec 21 2019
Extensions
More terms from R. J. Mathar and Emeric Deutsch, Mar 03 2008
Comments