OFFSET
0,3
COMMENTS
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
LINKS
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.
FORMULA
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).
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
MATHEMATICA
CoefficientList[Series[(1-Sqrt[1-4*x])/(2*(Sqrt[1-4*x]+x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)
PROG
(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
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Jun 24 2005
STATUS
approved