A076026 Expansion of g.f.: (1-4*x*C)/(1-5*x*C) where C = (1/2-1/2*(1-4*x)^(1/2))/x = g.f. for Catalan numbers A000108.
1, 1, 6, 37, 230, 1434, 8952, 55917, 349374, 2183230, 13643972, 85270626, 532926716, 3330739972, 20816939100, 130105200765, 813155081070, 5082210417270, 31763782696740, 198523522444950, 1240771573465140, 7754820693127020, 48467623215477120, 302922622226091090
Offset: 0
References
- L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Richard Ehrenborg, Gábor Hetyei, and Margaret Readdy, Catalan-Spitzer permutations, arXiv:2310.06288 [math.CO], 2023. See p. 20.
Programs
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Magma
R
:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( (2- 4*Sqrt(1-4*x))/(3-5*Sqrt(1-4*x)) )); // G. C. Greubel, May 04 2019 -
Mathematica
CoefficientList[Series[(2-4*Sqrt[1-4*x])/(3-5*Sqrt[1-4*x]), {x, 0, 30}], x] (* Vaclav Kotesovec, Dec 09 2013 *) Flatten[{1,Table[FullSimplify[(2*n)!*Hypergeometric2F1Regularized[1, n+1/2, n+2, 16/25] / (25*n!) + 3*5^(2*n-1)/4^(n+1)], {n,1,30}]}] (* Vaclav Kotesovec, Dec 09 2013 *)
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PARI
my(x='x+O('x^30)); Vec((2-4*sqrt(1-4*x))/(3-5*sqrt(1-4*x))) \\ G. C. Greubel, May 04 2019
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Sage
((2-4*sqrt(1-4*x))/(3-5*sqrt(1-4*x))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 04 2019
Formula
a(n+1) = Sum_{k=0..n} A039598(n,k)*4^k. - Philippe Deléham, Mar 21 2007
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j]=Catalan(j-i), (i<=j), and A[i,j]=0, otherwise. Then, for n >= 1, a(n+1)=(-1)^n*charpoly(A,-5). - Milan Janjic, Jul 08 2010
From Gary W. Adamson, Jul 25 2011: (Start)
a(n) = upper left term in M^(n-1), M = an infinite square production matrix as follows:
6, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, ...
... (End)
D-finite with recurrence: 4*n*a(n) = (41*n-24)*a(n-1) - 50*(2*n-3)*a(n-2). - Vaclav Kotesovec, Dec 09 2013
a(n) ~ 3*5^(2*n-1)/4^(n+1). - Vaclav Kotesovec, Dec 09 2013
O.g.f. A(x) = (1 - *Sum_{n >= 1} binomial(2*n,n)*x^n)/(1 - (3/2)*Sum_{n >= 1} binomial(2*n,n)*x^n). - Peter Bala, Sep 01 2016
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