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A098614 Product of Fibonacci and Catalan numbers: a(n) = A000045(n+1)*A000108(n). 12
1, 1, 4, 15, 70, 336, 1716, 9009, 48620, 267410, 1494844, 8465184, 48466796, 280073300, 1631408400, 9568812015, 56466198990, 335002137360, 1997007404700, 11955535480350, 71850862117320, 433322055191220, 2621615826231480, 15906988165723200, 96775058652983100 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
Radius of convergence: r = (sqrt(5)-1)/8; A(r) = sqrt(2+2/sqrt(5)). More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.
LINKS
Paul Barry and Arnauld Mesinga Mwafise, Classical and Semi-Classical Orthogonal Polynomials Defined by Riordan Arrays, and Their Moment Sequences, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.5.
FORMULA
G.f.: A(x) = sqrt( (1-2*x - sqrt(1-4*x-16*x^2))/10 )/x.
G.f. satisfies: A(x) = sqrt( 1 + 2*x*A(x)^2 + 5*x^2*A(x)^4 ).
a(n) == 1 (mod 2) iff n = 2^k - 1 for k>=0.
n*(n+1)*a(n) -2*n*(2*n-1)*a(n-1) -4*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 17 2018
Sum_{n>=0} a(n)/8^n = 2*sqrt(2/5). - Amiram Eldar, May 06 2023
EXAMPLE
Begins: {1*1, 1*1, 2*2, 3*5, 5*14, 8*42, 13*132, 21*429,...}.
MATHEMATICA
With[{nn=30}, Times@@@Thread[{Fibonacci[Range[nn]], CatalanNumber[ Range[ 0, nn-1]]}]] (* Harvey P. Dale, Nov 14 2011 *)
PROG
(PARI) {a(n)=local(X=x+O(x^(n+3)), A); A = sqrt( (1-2*x - sqrt(1-4*X-16*x^2)) / (10*x^2)); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) {a(n)=binomial(2*n, n)/(n+1)*round(((1+sqrt(5))^(n+1)-(1-sqrt(5))^(n+1))/(2^(n+1)*sqrt(5)))}
CROSSREFS
Sequence in context: A020020 A000882 A364908 * A367040 A356407 A366034
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Oct 09 2004
STATUS
approved

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Last modified April 25 04:42 EDT 2024. Contains 371964 sequences. (Running on oeis4.)