A157021 a(n)=sum{k=0..floor(n/2), C(n,2k)*A000108(floor(k/2))}. Inverse binomial transform is aeration of doubled Catalan numbers.

1, 1, 2, 4, 8, 16, 32, 64, 129, 265, 558, 1200, 2613, 5721, 12564, 27702, 61419, 136987, 307086, 691012, 1559430, 3528310, 8003808, 18203788, 41504967, 94842031, 217147258, 498061096, 1144296424, 2633227232, 6068715880, 14006305208

Offset

0,3

Comments

Hankel transform is A157022(n+1).

Links

Table of n, a(n) for n=0..31.

Formula

G.f.: ((1-2x+2x^2)/(1-x)^3)*c(x^4/(1-x)^4) where c(x) is the g.f. of A000108.

Conjecture: (n+4)*a(n) +(-7*n-18)*a(n-1) +2*(11*n+12)*a(n-2) +8*(-5*n+2)*a(n-3) +(41*n-76)*a(n-4) +(-19*n+62)*a(n-5) +4*(-n+6)*a(n-6) +6*(n-6)*a(n-7)=0. - R. J. Mathar, Feb 05 2015

Conjecture:+(n+4)*(n^2-11*n+32)*a(n) +(-5*n^3+41*n^2-26*n-360)*a(n-1) +2*(5*n^3-47*n^2+112*n+64)*a(n-2) +2*(-5*n^3+53*n^2-182*n+168)*a(n-3) +(n-4)*(n^2-3*n-8)*a(n-4) +3*(n-4)*(n^2-9*n+22)*a(n-5)=0. - R. J. Mathar, Feb 05 2015

a(n) ~ 3 * (1+sqrt(2))^(n + 3/2) / (2^(3/4) * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Oct 30 2017

Mathematica

CoefficientList[Series[(1 - 2*x + 2*x^2) / (1-x)^3 * (1 - Sqrt[1 - 4*(x^4/(1-x)^4)]) / (2*(x^4/(1-x)^4)), {x, 0, 40}], x] (* Vaclav Kotesovec, Oct 30 2017 *)

Crossrefs

Sequence in context: A265407 A023422 A084638 * A210543 A006211 A180209

Adjacent sequences: A157018 A157019 A157020 * A157022 A157023 A157024

Keyword

easy,nonn

Author

Paul Barry, Feb 21 2009

Status

approved