A119694 a(n) = Fibonacci(n) * Catalan(n).

0, 1, 2, 10, 42, 210, 1056, 5577, 30030, 165308, 923780, 5231954, 29953728, 173095700, 1008263880, 5913855450, 34898020290, 207042729630, 1234218400800, 7388927397390, 44406274641300, 267807758800920, 1620247684628040, 9831059348368050, 59810275503119232, 364767528768936300

Offset

0,3

Links

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Kenny B. Davenport, Problem B-1227, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 56, No. 2 (2018), p. 177; Generating Function of the Catalan Numbers, Solution to Problem B-1227 by Lauren G. Mcanany, ibid., Vol. 57, No. 2 (2019), pp. 178-179.

Vladimir V. Kruchinin and Maria Y. Perminova, Identities and Hadamard Product of the Generalized Fibonacci, Lucas, Catalan, and Harmonic Numbers, Journal of Integer Sequences, Vol. 28 (2025), Article 25.8.8. See p. 14.

Formula

a(n) = A000045(n) * A000108(n). - Alois P. Heinz, Aug 12 2017

Sum_{n>=0} a(n)/8^n = 4 - 6*sqrt(2/5) (Davenport, 2018). - Amiram Eldar, May 04 2023

G.f.: (1-sqrt((20*x+4*sqrt(-16*x^2-4*x+1)-12*x+6)/10))/(2*x). - Vladimir Kruchinin, Apr 12 2024

Maple

seq(combinat[fibonacci](n)*(binomial(2*n, n)/(n+1)), n=0..27);
# Alternative:
a:= proc(n) option remember; `if`(n<2, n,
      ((2*n-1)*(2*n*a(n-1)+(8*n-12)*a(n-2)))/(n*(n+1)))
    end:
seq(a(n), n=0..25);  # Alois P. Heinz, Aug 12 2017

Mathematica

Table[Fibonacci[n]CatalanNumber[n], {n, 0, 30}] (* Harvey P. Dale, Aug 27 2017 *)

Crossrefs

Cf. A000045, A000108, A000984, A371986.

Sequence in context: A084480 A309182 A099553 * A286760 A197048 A175613

Adjacent sequences: A119691 A119692 A119693 * A119695 A119696 A119697

Keyword

easy,nonn

Author

Zerinvary Lajos, Jun 09 2006

Extensions

Name edited by Alois P. Heinz, Aug 12 2017

Status

approved