A025226 a(n) = a(1)*a(n-1) + a(2)*a(n-2) + ...+ a(n-1)*a(1) for n >= 2. Also a(n) = 3^n*C(n-1), where C = A000108 (Catalan numbers).
3, 9, 54, 405, 3402, 30618, 288684, 2814669, 28146690, 287096238, 2975361012, 31241290626, 331638315876, 3553267670100, 38375290837080, 417331287853245, 4566095267100210
Offset: 1
Keywords
Examples
a(3) = 3^3*C(2) = 27*2 = 54.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..900
- Volkan Yildiz, Counting with 3-valued truth tables of bracketed formulae connected by implication, arXiv:2010.10303 [math.GM], 2020.
- Volkan Yildiz, Notes on algebraic structure of truth tables of bracketed formulae connected by implications, arXiv:2106.04728 [math.CO], 2021.
Programs
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Magma
[3^n*Catalan(n-1): n in [1..30]]; // G. C. Greubel, May 20 2022
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Mathematica
Rest[CoefficientList[Series[(1-Sqrt[1-12x])/2,{x,0,20}],x]] (* Harvey P. Dale, Mar 09 2011 *) -
PARI
a(n)=polcoeff((1-sqrt(1-12*x+x*O(x^n)))/2,n)
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SageMath
[3^n*catalan_number(n-1) for n in (1..30)] # G. C. Greubel, May 20 2022
Formula
a(n) = Sum_{j=1..n-1} a(j)*a(n-j), with a(1) = 3.
a(n) = 3^n * A000108(n-1).
G.f.: (1-sqrt(1-12*x))/2. - Michael Somos, Jun 08 2000
Given g.f. C(x) and given A(x)= g.f. of A100239, then B(x) = A(x) - (1+2*x) satisfies B(x) = x - C(x*B(x)). - Michael Somos, Sep 07 2005
G.f.: (1 - U(0))/x where U(k)= 1 - 3*x/U(k+1) ; (continued fraction, 1-step). - Sergei N. Gladkovskii, Oct 30 2012
D-finite with recurrence: n*a(n) +6*(3-2*n)*a(n-1) = 0. - R. J. Mathar, Nov 12 2012
a(n) = 3^n/(4*n-2)*binomial(2*n,n). - Vaclav Kotesovec, Oct 11 2013
Comments