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%I #33 Sep 08 2022 08:45:58
%S 0,1,3,9,20,40,74,131,225,379,630,1038,1700,2773,4511,7325,11880,
%T 19252,31182,50487,81725,132271,214058,346394,560520,906985,1467579,
%U 2374641,3842300,6217024,10059410,16276523,26336025,42612643,68948766
%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) + 3n - 1, with p(0,x)=1. For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232 and A192744.
%C ...
%C The list of examples at A192744 is extended here; the recurrence is given by p(n,x) = x*p(n-1,x) + v(n), with p(0,x)=1, and the reduction of p(n,x) by x^2 -> x+1 is represented by u1 + u2*x:
%C ...
%C If v(n)= n, then u1=A001595, u2=A104161.
%C If v(n)= n-1, then u1=A001610, u2=A066982.
%C If v(n)= 3n-1, then u1=A171516, u2=A192951.
%C If v(n)= 3n-2, then u1=A192746, u2=A192952.
%C If v(n)= 2n-1, then u1=A111314, u2=A192953.
%C If v(n)= n^2, then u1=A192954, u2=A192955.
%C If v(n)= -1+n^2, then u1=A192956, u2=A192957.
%C If v(n)= 1+n^2, then u1=A192953, u2=A192389.
%C If v(n)= -2+n^2, then u1=A192958, u2=A192959.
%C If v(n)= 2+n^2, then u1=A192960, u2=A192961.
%C If v(n)= n+n^2, then u1=A192962, u2=A192963.
%C If v(n)= -n+n^2, then u1=A192964, u2=A192965.
%C If v(n)= n(n+1)/2, then u1=A030119, u2=A192966.
%C If v(n)= n(n-1)/2, then u1=A192967, u2=A192968.
%C If v(n)= n(n+3)/2, then u1=A192969, u2=A192970.
%C If v(n)= 2n^2, then u1=A192971, u2=A192972.
%C If v(n)= 1+2n^2, then u1=A192973, u2=A192974.
%C If v(n)= -1+2n^2, then u1=A192975, u2=A192976.
%C If v(n)= 1+n+n^2, then u1=A027181, u2=A192978.
%C If v(n)= 1-n+n^2, then u1=A192979, u2=A192980.
%C If v(n)= (n+1)^2, then u1=A001891, u2=A053808.
%C If v(n)= (n-1)^2, then u1=A192981, u2=A192982.
%H Vincenzo Librandi, <a href="/A192951/b192951.txt">Table of n, a(n) for n = 0..400</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 From _Bruno Berselli_, Nov 16 2011: (Start)
%F G.f.: x*(1+2*x^2)/((1-x)^2*(1 - x - x^2)).
%F a(n) = ((25+13*t)*(1+t)^n + (25-13*t)*(1-t)^n)/(10*2^n) - 3*n - 5 = A000285(n+2) - 3*n - 5 where t=sqrt(5). (End)
%F a(n) = Fibonacci(n+4) + 2*Fibonacci(n+2) - (3*n+5). - _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] + 3n - 1;
%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}] (* A171516 *)
%t u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192951 *)
%t (* Additional programs *)
%t LinearRecurrence[{3,-2,-1,1},{0,1,3,9},40] (* _Vincenzo Librandi_, Nov 16 2011 *)
%t With[{F=Fibonacci}, Table[F[n+4]+2*F[n+2]-(3*n+5), {n,0,40}]] (* _G. C. Greubel_, Jul 12 2019 *)
%o (Magma) I:=[0, 1, 3, 9]; [n le 4 select I[n] else 3*Self(n-1)-2*Self(n-2)-1*Self(n-3)+Self(n-4): n in [1..40]]; // _Vincenzo Librandi_, Nov 16 2011
%o (Magma) F:=Fibonacci; [F(n+4)+2*F(n+2)-(3*n+5): n in [0..40]]; // _G. C. Greubel_, Jul 12 2019
%o (PARI) a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 1,-1,-2,3]^n*[0;1;3;9])[1,1] \\ _Charles R Greathouse IV_, Mar 22 2016
%o (PARI) vector(40, n, n--; f=fibonacci; f(n+4)+2*f(n+2)-(3*n+5)) \\ _G. C. Greubel_, Jul 12 2019
%o (Sage) f=fibonacci; [f(n+4)+2*f(n+2)-(3*n+5) for n in (0..40)] # _G. C. Greubel_, Jul 12 2019
%o (GAP) F:=Fibonacci;; List([0..40], n-> F(n+4)+2*F(n+2)-(3*n+5)); # _G. C. Greubel_, Jul 12 2019
%Y Cf. A000045, A192232, A192744.
%K nonn,easy
%O 0,3
%A _Clark Kimberling_, Jul 13 2011
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