%I #13 Jul 13 2023 08:37:40
%S 0,1,1,8,11,45,80,251,517,1432,3195,8317,19360,48827,116213,288360,
%T 694331,1708397,4138480,10138363,24636645,60217912,146570491,
%U 357833309,871703360,2126857275,5183425493,12642971912,30819571387,75160150861
%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)/2)^n + ((x-d)/2)^n, where d = sqrt(x^2+8). For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232.
%H G. C. Greubel, <a href="/A192427/b192427.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 (1,5,-2,-4).
%F From _Colin Barker_, May 12 2014: (Start)
%F a(n) = a(n-1) + 5*a(n-2) - 2*a(n-3) - 4*a(n-4).
%F G.f.: x*(1+2*x^2)/(1-x-5*x^2+2*x^3+4*x^4). (End)
%F a(n) = Sum_{k=0..n} T(n, k)*Fibonacci(k), where T(n, k) = [x^k] ((x + sqrt(x^2+8))^n + (x - sqrt(x^2+8))^n)/2^n. - _G. C. Greubel_, Jul 13 2023
%e The first five polynomials p(n,x) and their reductions are as follows:
%e p(0,x) = 2 -> 2
%e p(1,x) = x -> x
%e p(2,x) = 4 + x^2 -> 5 + x
%e p(3,x) = 6*x + x^3 -> 1 + 8*x
%e p(4,x) = 8 + 8*x^2 + x^4 -> 18 + 11*x.
%e From these, read A192426 = (2, 0, 5, 1, 18, ...) and a(n) = (0, 1, 1, 8, 11, ...).
%t (See A192426.)
%t LinearRecurrence[{1,5,-2,-4}, {0,1,1,8}, 40] (* _G. C. Greubel_, Jul 13 2023 *)
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1+2*x^2)/(1-x-5*x^2+2*x^3+4*x^4) )); // _G. C. Greubel_, Jul 13 2023
%o (SageMath)
%o @CachedFunction
%o def a(n): # a = A192427
%o if (n<4): return (0,1,1,8)[n]
%o else: return a(n-1) + 5*a(n-2) - 2*a(n-3) - 4*a(n-4)
%o [a(n) for n in range(41)] # _G. C. Greubel_, Jul 13 2023
%Y Cf. A000045, A192232, A192426.
%K nonn,easy
%O 0,4
%A _Clark Kimberling_, Jun 30 2011
%E Typo in name corrected by _G. C. Greubel_, Jul 13 2023