A185020 a(n) = A000108(n)*A002605(n+1), where A000108 are the Catalan numbers.

1, 2, 12, 80, 616, 5040, 43296, 384384, 3500640, 32517056, 306896512, 2934597120, 28369508608, 276810483200, 2722537128960, 26963147796480, 268659456837120, 2691301381401600, 27089160416102400, 273833161582632960, 2778754123765002240, 28296326851107594240

Offset: 0

Author

Paul D. Hanna, Dec 26 2012

Keywords

• nonn

Comments

More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, S(0)=1, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.

Conjecture: +n*(n+1)*a(n) -4*n*(2*n-1)*a(n-1) -8*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Oct 08 2016

Examples

			G.f.: A(x) = 1 + 1*2*x + 2*6*x^2 + 5*16*x^3 + 14*44*x^4 + 42*120*x^5 + 132*328*x^6 +...+ A000108(n)*A002605(n+1)*x^n +...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..960

Crossrefs

Cf. A002605.

Programs

Mathematica

MapIndexed[CatalanNumber[#2 - 1] #1 &, Rest@ RecurrenceTable[{a[n] == 2 (a[n - 1] + a[n - 2]), a[0] == 0, a[1] == 1}, a, {n, 22}]] // Flatten (* or *)
CoefficientList[Series[Sqrt[(1 - 4 x - Sqrt[1 - 8 x - 32 x^2])/24]/x, {x, 0, 21}], x] (* Michael De Vlieger, Oct 08 2016 *)

PARI

{A000108(n) = binomial(2*n,n)/(n+1)}
{A002605(n) = polcoef(x/(1-2*x-2*x^2 +x*O(x^n)), n)}
{a(n)=A000108(n)*A002605(n+1)}
for(n=0,30,print1(a(n),", "))

Formula

G.f.: sqrt( (1-4*x - sqrt(1-8*x-32*x^2))/24 )/x.

G.f. A(x) satisfies A(x) = sqrt( 1 + 4*x*A(x)^2 + 12*x^2*A(x)^4 ). - Paul D. Hanna, Dec 14 2024