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%I #24 Dec 14 2023 05:17:43
%S 0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,
%T 1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,
%U -1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0,0,1,-1,0,0
%N Expansion of x/(1 + x + x^2 + x^3 + x^4).
%C Inverse binomial transform of A103311. A transform of the Fibonacci numbers: apply the Chebyshev transform (1/(1+x^2), x/(1+x^2)) followed by the binomial involution (1/(1-x),-x/(1-x)) followed by the inverse binomial transform (1/(1+x), x/(1+x)) (expressed as Riordan arrays) to the -F(n); equivalently, apply (1/(1+x^2),-x/(1+x^2)) to -F(n). Periodic {0,1,-1,0,0}.
%C Essentially the same as A010891. - _R. J. Mathar_, Apr 07 2008
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-1,-1,-1,-1).
%F Euler transform of length 5 sequence [ -1, 0, 0, 0, 1].
%F G.f.: x(1-x)/(1-x^5);
%F a(n) = -sqrt(1/5 + 2*sqrt(5)/25)*cos(4*Pi*n/5 + Pi/10) + sqrt(5)*sin(4*Pi*n/5 + Pi/10)/5 + sqrt(1/5 - 2*sqrt(5)/25)*cos(2*Pi*n/5 + 3*Pi/10) + sqrt(5)*sin(2*Pi*n/5 + 3*Pi/10)/5.
%F a(n) = A010891(n-1). - _R. J. Mathar_, Apr 07 2008
%F a(n) + a(n-1) = A092202(n). - _R. J. Mathar_, Jun 23 2021
%K easy,sign
%O 0,1
%A _Paul Barry_, Apr 02 2005
%E Corrected by _N. J. A. Sloane_, Nov 05 2005
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