A100095 An inverse Chebyshev transform of the Fibonacci numbers.

0, 1, 1, 5, 7, 25, 41, 125, 225, 625, 1195, 3125, 6227, 15625, 32059, 78125, 163727, 390625, 831505, 1953125, 4206145, 9765625, 21215481, 48828125, 106782837, 244140625, 536618341, 1220703125, 2693492305, 6103515625, 13507578125

Offset

0,4

Comments

Image of x/(1-x-x^2) under the transform g(x)->(1/sqrt(1-4xx^2)g(xc(x^2)), where c(x) is the g.f. of the Catalan numbers A000108. This is the inverse of the Chebyshev transform which takes A(x) to ((1-x^2)/(1+x^2))A(x/(1+x^2)).

Hankel transform is (-1)^n*(2^n-0^n)/2. Hankel transform of a(n+1) is A141125. - Paul Barry, Jun 05 2008

Basically A000351 interleaved with A144635. - Peter Luschny, May 31 2014

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Formula

G.f.: (x^2*sqrt(1-4*x^2)+x*(1-4*x^2))/((1-4*x^2)*(1-5*x^2)).

a(n) = sum( k=0..floor(n/2), binomial(n, k)Fib(n-2k) ).

Conjecture: (-n+2)*a(n) +(-n+3)*a(n-1) +(9*n-22)*a(n-2) +(9*n-31)*a(n-3) +20*(-n+3)*a(n-4) +20*(-n+4)*a(n-5)=0. - R. J. Mathar, Nov 24 2012

Recurrence: (n-2)*a(n) = (9*n-22)*a(n-2) - 20*(n-3)*a(n-4). - Vaclav Kotesovec, Feb 12 2014

a(n) ~ 5^((n-1)/2). - Vaclav Kotesovec, Feb 12 2014

a(n) = sum(j=0..(n-1)/2, 4^(j)*binomial((n-1)/2,j)). - Vladimir Kruchinin, May 31 2014

a(2*n) = 5^n/sqrt(5) - 2^n * (2*n-1)!! * hypergeom([1, n+1/2], [n+1], 4/5)/(5*n!), a(2*n+1) = 5^n. - Vladimir Reshetnikov, Oct 13 2016

Mathematica

CoefficientList[Series[(x^2*Sqrt[1-4*x^2]+x*(1-4*x^2))/((1-4*x^2)*(1-5*x^2)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 12 2014 *)

Prog

(Maxima)
a(n):=sum(4^(j)*binomial((n-1)/2, j), j, 0, (n-1)/2); /* Vladimir Kruchinin, May 31 2014 */

Crossrefs

Cf. A000351, A144635, A000045, A100096, A100097.

Sequence in context: A003595 A018697 A018319 * A227922 A179782 A013626

Adjacent sequences: A100092 A100093 A100094 * A100096 A100097 A100098

Keyword

nonn,easy

Author

Paul Barry, Nov 03 2004

Status

approved