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 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, 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS Hankel transform is A157022(n+1). LINKS 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

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Last modified October 23 05:45 EDT 2018. Contains 316519 sequences. (Running on oeis4.)