A157021 - OEIS

%I #7 Oct 30 2017 07:39:46

%S 1,1,2,4,8,16,32,64,129,265,558,1200,2613,5721,12564,27702,61419,

%T 136987,307086,691012,1559430,3528310,8003808,18203788,41504967,

%U 94842031,217147258,498061096,1144296424,2633227232,6068715880,14006305208

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

%C Hankel transform is A157022(n+1).

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

%F 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

%F 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

%F a(n) ~ 3 * (1+sqrt(2))^(n + 3/2) / (2^(3/4) * sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, Oct 30 2017

%t 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 *)

%K easy,nonn

%O 0,3

%A _Paul Barry_, Feb 21 2009