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%I #16 Sep 08 2022 08:45:58
%S 0,1,0,3,10,27,60,121,228,411,718,1227,2064,3433,5664,9291,15178,
%T 24723,40188,65233,105780,171411,277630,449523,727680,1177777,1906080,
%U 3084531,4991338,8076651,13068828,21146377,34216164,55363563,89580814
%N Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.
%C The titular polynomials are defined recursively: p(n,x) = x*p(n-1,x) - 2 + n^2, with p(0,x)=1. For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232 and A192744.
%H G. C. Greubel, <a href="/A192959/b192959.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (4,-5,1,2,-1).
%F a(n) = 4*a(n-1) - 5*a(n-2) + a(n-3) + 2*a(n-4) - a(n-5).
%F From _R. J. Mathar_, May 09 2014: (Start)
%F G.f.: x*(1 -4*x +8*x^2 -3*x^3)/((1-x-x^2)*(1-x)^3).
%F a(n) - a(n-1) = A192958(n-1). (End)
%F a(n) = 6*Fibonacci(n+2) - (n^2 + 4*n + 6). - _G. C. Greubel_, Jul 12 2019
%t (* First program *)
%t q = x^2; s = x + 1; z = 40;
%t p[0, x]:= 1;
%t p[n_, x_]:= x*p[n-1, x] + n^2 - 2;
%t Table[Expand[p[n, x]], {n, 0, 7}]
%t reduce[{p1_, q_, s_, x_}]:= FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
%t t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
%t u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192958 *)
%t u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192959 *)
%t (* Second program *)
%t With[{F=Fibonacci}, Table[6*F[n+2]-(n^2+4*n+6), {n,0,40}]] (* _G. C. Greubel_, Jul 12 2019 *)
%o (PARI) vector(40, n, n--; f=fibonacci; 6*f(n+2)-(n^2+4*n+6)) \\ _G. C. Greubel_, Jul 12 2019
%o (Magma) F:=Fibonacci; [6*F(n+2)-(n^2+4*n+6): n in [0..40]]; // _G. C. Greubel_, Jul 12 2019
%o (Sage) f=fibonacci; [6*f(n+2)-(n^2+4*n+6) for n in (0..40)] # _G. C. Greubel_, Jul 12 2019
%o (GAP) F:=Fibonacci;; List([0..40], n-> 6*F(n+2)-(n^2+4*n+6)); # _G. C. Greubel_, Jul 12 2019
%Y Cf. A000045, A192232, A192744, A192951, A192958.
%K nonn
%O 0,4
%A _Clark Kimberling_, Jul 13 2011