A042941 Convolution of Catalan numbers A000108 with A038845.

1, 13, 110, 765, 4746, 27314, 149052, 781725, 3975730, 19730150, 95973956, 459145778, 2165937060, 10095323460, 46566906872, 212857023069, 965208806082, 4345780250270, 19442667426420, 86489687956518

Offset: 0

Author

Wolfdieter Lang

Keywords

• easy
• nonn

Comments

Also convolution of A018218(n+1), n >= 0, with A000302 (powers of 4); also convolution of A000346 with A002697.

Links

• Fung Lam, Table of n, a(n) for n = 0..1000

Programs

Mathematica

CoefficientList[Series[(1-Sqrt[1-4*x])/(2*x*(1-4*x)^3), {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 16 2014 *)

Formula

a(n) = binomial(n+3, 2)*(4^(n+1) - A000984(n+3)/A000984(2)) / 2.

G.f.: c(x)/(1-4*x)^3, where c(x) is the g.f. for Catalan numbers.

Recurrence: (n+1)*a(n) = 128*(1-2*n)*a(n-4) + 32*(8*n-1)*a(n-3) - 24*(4*n+1)*a(n-2) + 2*(8*n+5)*a(n-1). - Fung Lam, Apr 13 2014

a(n) ~ 2^(2*n)*(n^2 - 8*n^(3/2)/(3*sqrt(Pi))). - Fung Lam, Apr 13 2014

Recurrence: n*(n+1)*a(n) = 2*n*(4*n+9)*a(n-1) - 8*(n+2)*(2*n+3)*a(n-2). - Vaclav Kotesovec, Apr 16 2014