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%I #23 Oct 14 2016 08:07:28
%S 0,1,1,5,7,25,41,125,225,625,1195,3125,6227,15625,32059,78125,163727,
%T 390625,831505,1953125,4206145,9765625,21215481,48828125,106782837,
%U 244140625,536618341,1220703125,2693492305,6103515625,13507578125
%N An inverse Chebyshev transform of the Fibonacci numbers.
%C 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)).
%C Hankel transform is (-1)^n*(2^n-0^n)/2. Hankel transform of a(n+1) is A141125. - _Paul Barry_, Jun 05 2008
%C Basically A000351 interleaved with A144635. - _Peter Luschny_, May 31 2014
%H Vincenzo Librandi, <a href="/A100095/b100095.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: (x^2*sqrt(1-4*x^2)+x*(1-4*x^2))/((1-4*x^2)*(1-5*x^2)).
%F a(n) = sum( k=0..floor(n/2), binomial(n, k)Fib(n-2k) ).
%F 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
%F Recurrence: (n-2)*a(n) = (9*n-22)*a(n-2) - 20*(n-3)*a(n-4). - _Vaclav Kotesovec_, Feb 12 2014
%F a(n) ~ 5^((n-1)/2). - _Vaclav Kotesovec_, Feb 12 2014
%F a(n) = sum(j=0..(n-1)/2, 4^(j)*binomial((n-1)/2,j)). - _Vladimir Kruchinin_, May 31 2014
%F 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
%t 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 *)
%o (Maxima)
%o a(n):=sum(4^(j)*binomial((n-1)/2,j),j,0,(n-1)/2); /* _Vladimir Kruchinin_, May 31 2014 */
%Y Cf. A000351, A144635, A000045, A100096, A100097.
%K nonn,easy
%O 0,4
%A _Paul Barry_, Nov 03 2004
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