A192981 - OEIS

%I #20 Sep 08 2022 08:45:58

%S 1,0,2,5,12,24,45,80,138,233,388,640,1049,1712,2786,4525,7340,11896,

%T 19269,31200,50506,81745,132292,214080,346417,560544,907010,1467605,

%U 2374668,3842328,6217053,10059440,16276554,26336057,42612676,68948800

%N Constant term of 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) + (n-1)^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="/A192981/b192981.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 (3,-2,-1,1).

%F a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4).

%F G.f.: (1 - 3*x + 4*x^2)/((1 - x)^2*(1 - x - x^2)). - _Colin Barker_, May 11 2014

%F a(n) = Lucas(n+2) + Fibonacci(n+1) - (2*n+3). - _G. C. Greubel_, Jul 25 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-1)^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}] (* A192981 *)

%t u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192982 *)

%t (* Additional programs *)

%t Table[LucasL[n+2]_Fibonacci[n+1]-(2*n+3), {n,0,40}] (* _G. C. Greubel_, Jul 25 2019 *)

%o (PARI) vector(40, n, n--; f=fibonacci; f(n+3)+2*f(n+1) -(2*n+3)) \\ _G. C. Greubel_, Jul 25 2019

%o (Magma) F:=Fibonacci; [F(n+3)+2*F(n+1) -(2*n+3): n in [0..40]]; // _G. C. Greubel_, Jul 25 2019

%o (Sage) f=fibonacci; [f(n+3)+2*f(n+1) -(2*n+3) for n in (0..40)] # _G. C. Greubel_, Jul 25 2019

%o (GAP) F:=Fibonacci;; List([0..40], n-> F(n+3)+2*F(n+1) -(2*n+3)); # _G. C. Greubel_, Jul 25 2019

%Y Cf. A000032, A000045, A192232, A192744, A192951, A192982.

%K nonn,easy

%O 0,3

%A _Clark Kimberling_, Jul 14 2011