A200312 a(n) = A000108(n)*A006130(n), where A000108 is the Catalan numbers and A006130(n) = A006130(n-1) + 3*A006130(n-2).
1, 1, 8, 35, 266, 1680, 12804, 93093, 726440, 5635058, 45063668, 362121760, 2955642508, 24284658100, 201428123040, 1680921310635, 14119413718770, 119205791509200, 1011387051005100, 8617021562542470, 73704123363739440, 632601537174078420
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 2*4*x^2 + 5*7*x^3 + 14*19*x^4 + 42*40*x^5 + 132*97*x^6 + 429*217*x^7 + ... + A000108(n)*A006130(n)*x^n + ... where the g.f. of A006130, F(x) = 1/(1-x-3*x^2), begins: F(x) = 1 + x + 4*x^2 + 7*x^3 + 19*x^4 + 40*x^5 + 97*x^6 + 217*x^7 + ...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..450
- S. B. Ekhad and M. Yang, Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences, (2017)
Programs
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Magma
m:=30; R
:=PowerSeriesRing(Rationals(), m); Coefficients(R!(Sqrt((1-2*x - Sqrt(1-4*x-48*x^2))/26)/x)); // G. C. Greubel, Jul 27 2018 -
Mathematica
CoefficientList[Series[Sqrt[(1 - 2*x - Sqrt[1 - 4*x - 48*x^2])/26]/x, {x, 0, 30}], x] (* G. C. Greubel, Jul 27 2018 *)
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PARI
{a(n)=binomial(2*n, n)/(n+1)*polcoeff(1/(1-x-3*x^2+x*O(x^n)),n)}
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PARI
{a(n)=polcoeff(sqrt((1-2*x - sqrt(1-4*x-48*x^2+x^3*O(x^n)))/26)/x,n)}
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PARI
{a(n)=polcoeff(serreverse(x*sqrt(1-12*x^2+x^2*O(x^n)) - x^2)/x,n)}
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PARI
{a(n)=polcoeff((1/x)*serreverse(x-x^2 - 6*x^3*sum(m=0,n\2,binomial(2*m,m)/(m+1)*3^m*x^(2*m))+x^3*O(x^n)),n)}
Formula
G.f.: sqrt( (1-2*x - sqrt(1-4*x-48*x^2))/26 )/x.
G.f.: (1/x)*Series_Reversion( x*sqrt(1-12*x^2) - x^2 ).
G.f.: (1/x)*Series_Reversion( x-x^2 - 6*x^3*Sum_{n>=0} A000108(n)*3^n*x^(2*n) ).
G.f. satisfies: A(x) = sqrt(1 + 2*x*A(x)^2 + 13*x^2*A(x)^4).
Conjecture: n*(n+1)*a(n) -2*n*(2*n-1)*a(n-1) -12*(2*n-1)*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 17 2011
a(n) = ( ((1+sqrt(13))/2)^(n+1) - ((1-sqrt(13))/2)^(n+1) )/sqrt(13) * binomial(2*n+1,n)/(2*n+1). - Paul D. Hanna, Sep 25 2012
0 = +a(n)*(+110592*a(n+3) -9216*a(n+4) -7392*a(n+5) +858*a(n+6)) +a(n+1)*(+6912*a(n+3) -1968*a(n+4) -910*a(n+5) +154*a(n+6)) +a(n+2)*(-240*a(n+3) -2*a(n+4) +41*a(n+5) -4*a(n+6)) +a(n+3)*(+6*a(n+3) +5*a(n+4) +3*a(n+5) -a(n+6)) for all n in Z. - Michael Somos, Jul 28 2018
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