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A109262
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A Catalan transform of the Fibonacci numbers.
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10
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0, 1, 2, 6, 19, 63, 215, 749, 2650, 9490, 34318, 125104, 459152, 1694914, 6287896, 23429158, 87635243, 328917615, 1238303243, 4674847097, 17692789741, 67114622451, 255120892105, 971649360211, 3707176155659, 14167390221873
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OFFSET
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0,3
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COMMENTS
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A column of A109267.
Hankel transform is -Fibonacci(2*n). a(n+1) has Hankel transform Fibonacci(2*n+1). - Paul Barry, Nov 22 2007
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..200
Paul Barry, Chebyshev moments and Riordan involutions, arXiv:1912.11845 [math.CO], 2019.
Stoyan Dimitrov, On permutation patterns with constrained gap sizes, arXiv:2002.12322 [math.CO], 2020.
Sergio Falcon, Catalan transform of the K-Fibonacci sequence, Commun. Korean Math. Soc. 28 (2013), No. 4, pp. 827-832.
Guo-Niu Han, Enumeration of Standard Puzzles
Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]
Merve Taştan and Engin Özkan, Catalan transform of the k-Jacobsthal sequence, Electronic Journal of Mathematical Analysis and Applications (2020) Vol. 8, No. 2, 70-74.
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FORMULA
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G.f.: x*c(x)/(1 - x*c(x) - x^2*c(x)^2) = (1 - sqrt(1-4*x))/(2*(x + sqrt(1-4*x))) where c(x) is the g.f. of A000108.
a(n) = Sum_{k=0..n} (k/(2*n-k))*binomial(2*n-k, n-k)*Fibonacci(k).
a(n) = Sum_{k=0..n} A106566(n,k)*A000045(k). - Philippe Deléham, Oct 28 2008
a(n) = Sum_{k=0..n} A039599(n,k)*(-1)^(k+1)*A000045(k). - Philippe Deléham, Oct 28 2008
n*a(n) - (7*n-4)*a(n-1) + (7*n-2)*a(n-2) + (19*n-60)*a(n-3) + 2*(2*n-7)*a(n-4) = 0. - R. J. Mathar, Nov 26 2012
Recurrence: n*(5*n-11)*a(n) = 2*(20*n^2 - 59*n + 30)*a(n-1) - 15*(5*n^2 - 19*n + 16)*a(n-2) - 2*(2*n-5)*(5*n-6)*a(n-3). - Vaclav Kotesovec, Feb 13 2014
a(n) ~ 5*4^n/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Feb 13 2014
a(n) = (1/(2*sqrt(5)))*Catalan(n-1)*Sum_{j=0..1} ((-1)^j + sqrt(5)) * Hypergeometric2F1([2,1-n], [2*(1-n)], (1+(-1)^j*sqrt(5))/2). - G. C. Greubel, May 30 2022
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MATHEMATICA
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CoefficientList[Series[(1-Sqrt[1-4*x])/(2*(Sqrt[1-4*x]+x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)
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PROG
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(Magma) [n eq 0 select 0 else (&+[k*Binomial(2*n-k-1, n-1)*Fibonacci(k): k in [0..n]])/n: n in [0..30]]; // G. C. Greubel, May 30 2022
(SageMath) [0]+[(1/n)*sum(k*binomial(2*n-k-1, n-1)*fibonacci(k) for k in (1..n)) for n in (1..30)] # G. C. Greubel, May 30 2022
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CROSSREFS
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Cf. A000045, A000108, A039599, A081696, A106566, A109267.
Sequence in context: A284216 A059712 A059713 * A006724 A057409 A346157
Adjacent sequences: A109259 A109260 A109261 * A109263 A109264 A109265
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KEYWORD
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easy,nonn
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AUTHOR
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Paul Barry, Jun 24 2005
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STATUS
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approved
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