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%I #17 Jul 11 2023 19:32:37
%S 0,2,4,32,96,512,1856,8576,33792,147456,602112,2566144,10637312,
%T 44892160,187269120,787087360,3292069888,13812760576,57837355008,
%U 242497880064,1015868817408,4258009186304,17841063460864,74771320537088,313317428035584
%N Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined below in Comments.
%C The polynomial p(n,x) is defined by ((x+d)^n - (x-d)^n)/(2*d), where d = sqrt(x+5). For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232.
%H G. C. Greubel, <a href="/A192387/b192387.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,12,-8,-16).
%F From _Colin Barker_, Dec 09 2012: (Start)
%F a(n) = 2*a(n-1) + 12*a(n-2) - 8*a(n-3) - 16*a(n-4).
%F G.f.: 2*x^2/(1-2*x-12*x^2+8*x^3+16*x^4). (End)
%F a(n) = 2^n*A112576(n). - _R. J. Mathar_, Mar 08 2021
%F From _G. C. Greubel_, Jul 10 2023: (Start)
%F T(n, k) = [x^k] ((x+sqrt(x+5))^n - (x-sqrt(x+5))^n)/(2*sqrt(x+5)).
%F a(n) = Sum_{k=0..n-1} T(n, k)*Fibonacci(k). (End)
%e The first five polynomials p(n,x) and their reductions are as follows:
%e p(0,x) = 1 -> 1
%e p(1,x) = 2*x -> 2*x
%e p(2,x) = 3 + x + 3*x^2 -> 8 + 4*x
%e p(3,x) = 12*x + 4*x^2 + 4*x^3 -> 8 + 32*x
%e p(4,x) = 9 + 6*x + 31*x^2 + 10*x^3 + 5*x^4 -> 96 + 96*x.
%e From these, read A192386 = (1, 0, 8, 8, 96, ...) and a(n) = (0, 2, 4, 32, 96, ...).
%t (See A192386.)
%t LinearRecurrence[{2,12,-8,-16}, {0,2,4,32}, 40] (* _G. C. Greubel_, Jul 10 2023 *)
%o (Magma)
%o R<x>:=PowerSeriesRing(Integers(), 41);
%o [0] cat Coefficients(R!( 2*x^2/(1-2*x-12*x^2+8*x^3+16*x^4) )); // _G. C. Greubel_, Jul 10 2023
%o (SageMath)
%o @CachedFunction
%o def a(n): # a = A192387
%o if (n<5): return (0,0,2,4,32)[n]
%o else: return 2*a(n-1) +12*a(n-2) -8*a(n-3) -16*a(n-4)
%o [a(n) for n in range(1,41)] # _G. C. Greubel_, Jul 10 2023
%Y Cf. A112576, A192232, A192386, A192388.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Jun 30 2011
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