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A110311 Expansion of 1/((1+x+x^2)*(1+5*x+x^2)).

%I A110311 #14 Jan 03 2023 03:57:15
%S A110311 1,-6,29,-138,660,-3162,15151,-72594,347819,-1666500,7984680,
%T A110311 -38256900,183299821,-878242206,4207911209,-20161313838,96598657980,
%U A110311 -462831976062,2217561222331,-10624974135594,50907309455639,-243911573142600,1168650556257360,-5599341208144200
%N A110311 Expansion of 1/((1+x+x^2)*(1+5*x+x^2)).
%C A110311 In reference to the program code, A004254(n+1) = 1ibaseiseq[A*B](n).
%C A110311 Superseeker finds: a(n) + a(n+1) + a(n+2) = (-1)^n*A004254(n+3).
%H A110311 Colin Barker, <a href="/A110311/b110311.txt">Table of n, a(n) for n = 0..1000</a>
%H A110311 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-6,-7,-6,-1).
%F A110311 a(n+2) = - 5*a(n+1) - a(n) + ((-1)^n)*A109265(n+1)/2.
%F A110311 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 A110311 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 A110311 seriestolist(series(1/((x^2+5*x+1)*(x^2+x+1)), x=0,25));
%t A110311 LinearRecurrence[{-6,-7,-6,-1}, {1,-6,29,-138}, 40] (* _G. C. Greubel_, Jan 02 2023 *)
%o A110311 (PARI) Vec(1/((1+x+x^2)*(1+5*x+x^2)) + O(x^25)) \\ _Colin Barker_, May 14 2019
%o A110311 (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 A110311 (SageMath)
%o A110311 def U(n,x): return chebyshev_U(n,x)
%o A110311 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 [A110311(n) for n in range(41)] # _G. C. Greubel_, Jan 02 2023
%Y A110311 Cf. A004253, A004254, A110307, A110308, A110309, A110310.
%K A110311 easy,sign
%O A110311 0,2
%A A110311 _Creighton Dement_, Jul 19 2005