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Expansion of 1/((1+x+x^2)*(1+5*x+x^2)).
5

%I #14 Jan 03 2023 03:57:15

%S 1,-6,29,-138,660,-3162,15151,-72594,347819,-1666500,7984680,

%T -38256900,183299821,-878242206,4207911209,-20161313838,96598657980,

%U -462831976062,2217561222331,-10624974135594,50907309455639,-243911573142600,1168650556257360,-5599341208144200

%N Expansion of 1/((1+x+x^2)*(1+5*x+x^2)).

%C In reference to the program code, A004254(n+1) = 1ibaseiseq[A*B](n).

%C Superseeker finds: a(n) + a(n+1) + a(n+2) = (-1)^n*A004254(n+3).

%H Colin Barker, <a href="/A110311/b110311.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-6,-7,-6,-1).

%F a(n+2) = - 5*a(n+1) - a(n) + ((-1)^n)*A109265(n+1)/2.

%F a(n) = -6*a(n-1) - 7*a(n-2) - 6*a(n-3) - a(n-4) for n>3. - _Colin Barker_, May 14 2019

%F a(n) = (1/4)*(5*U(n, -5/2) + U(n-1, -5/2) - U(n, -1/2) - U(n-1, -1/2)), where U(n, x) = ChebyshevU(n, x). - _G. C. Greubel_, Jan 02 2023

%p seriestolist(series(1/((x^2+5*x+1)*(x^2+x+1)), x=0,25));

%t LinearRecurrence[{-6,-7,-6,-1}, {1,-6,29,-138}, 40] (* _G. C. Greubel_, Jan 02 2023 *)

%o (PARI) Vec(1/((1+x+x^2)*(1+5*x+x^2)) + O(x^25)) \\ _Colin Barker_, May 14 2019

%o (Magma) R<x>:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/((1+x+x^2)*(1+5*x+x^2)) )); // _G. C. Greubel_, Jan 02 2023

%o (SageMath)

%o def U(n,x): return chebyshev_U(n,x)

%o def A110311(n): return (1/4)*(5*U(n, -5/2) + U(n-1, -5/2) - U(n, -1/2) - U(n-1, -1/2))

%o [A110311(n) for n in range(41)] # _G. C. Greubel_, Jan 02 2023

%Y Cf. A004253, A004254, A110307, A110308, A110309, A110310.

%K easy,sign

%O 0,2

%A _Creighton Dement_, Jul 19 2005