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A100095
An inverse Chebyshev transform of the Fibonacci numbers.
7
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
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
KEYWORD
nonn,easy
AUTHOR
Paul Barry, Nov 03 2004
STATUS
approved