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
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Programs
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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 *) -
Maxima
a(n):=sum(4^(j)*binomial((n-1)/2,j),j,0,(n-1)/2); /* Vladimir Kruchinin, May 31 2014 */
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)*Fibonacci(n-2*k).
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
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