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A100192
a(n) = Sum_{k=0..n} binomial(2n,n+k)*2^k.
5
1, 4, 18, 82, 374, 1704, 7752, 35214, 159750, 723880, 3276908, 14821668, 66991436, 302605528, 1366182276, 6165204102, 27811282374, 125415953208, 565408947756, 2548400193852, 11483706241044, 51739037228688, 233070330199296
OFFSET
0,2
COMMENTS
A transform of 2^n under the mapping g(x)->(1/sqrt(1-4x))g(xc(x)^2), where c(x) is the g.f. of the Catalan numbers A000108. A transform of 3^n under the mapping g(x)->(1/(c(x)*sqrt(1-4x))g(x*c(x)).
Hankel transform is A088138(n+1). - Paul Barry, Jan 11 2007
LINKS
FORMULA
G.f.: (sqrt(1-4*x)+1)/(sqrt(1-4*x)*(3*sqrt(1-4*x)-1)).
G.f.: sqrt(1-4*x)*(sqrt(1-4*x)-3*x+1)/((1-4*x)*(2-9*x)).
a(n) = sum{k=0..n, binomial(2n, n-k)2^k}.
a(n) = sum{k=0..n, C(2n,k)*2^(n-k)}; - Paul Barry, Jan 11 2007
a(n) = sum{k=0..n, C(n+k-1,k)3^(n-k)}; - Paul Barry, Sep 28 2007
Conjecture: 2*n*a(n) +(-23*n+16)*a(n-1) +3*(29*n-44)*a(n-2) +54*(-2*n+5)*a(n-3)=0. - R. J. Mathar, Nov 24 2012
a(n) ~ (9/2)^n. - Vaclav Kotesovec, Feb 12 2014
a(n) = [x^n] 1/((1 - x)^n*(1 - 3*x)). - Ilya Gutkovskiy, Oct 12 2017
MATHEMATICA
CoefficientList[Series[Sqrt[1-4*x]*(Sqrt[1-4*x]-3*x+1)/((1-4*x)*(2-9*x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)
CROSSREFS
Cf. A032443.
Sequence in context: A257059 A194460 A356289 * A052913 A279285 A129160
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Nov 08 2004
STATUS
approved