A192951 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

0, 1, 3, 9, 20, 40, 74, 131, 225, 379, 630, 1038, 1700, 2773, 4511, 7325, 11880, 19252, 31182, 50487, 81725, 132271, 214058, 346394, 560520, 906985, 1467579, 2374641, 3842300, 6217024, 10059410, 16276523, 26336025, 42612643, 68948766

Offset

0,3

Comments

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.

...

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:

...

If v(n)= n, then u1=A001595, u2=A104161.

If v(n)= n-1, then u1=A001610, u2=A066982.

If v(n)= 3n-1, then u1=A171516, u2=A192951.

If v(n)= 3n-2, then u1=A192746, u2=A192952.

If v(n)= 2n-1, then u1=A111314, u2=A192953.

If v(n)= n^2, then u1=A192954, u2=A192955.

If v(n)= -1+n^2, then u1=A192956, u2=A192957.

If v(n)= 1+n^2, then u1=A192953, u2=A192389.

If v(n)= -2+n^2, then u1=A192958, u2=A192959.

If v(n)= 2+n^2, then u1=A192960, u2=A192961.

If v(n)= n+n^2, then u1=A192962, u2=A192963.

If v(n)= -n+n^2, then u1=A192964, u2=A192965.

If v(n)= n(n+1)/2, then u1=A030119, u2=A192966.

If v(n)= n(n-1)/2, then u1=A192967, u2=A192968.

If v(n)= n(n+3)/2, then u1=A192969, u2=A192970.

If v(n)= 2n^2, then u1=A192971, u2=A192972.

If v(n)= 1+2n^2, then u1=A192973, u2=A192974.

If v(n)= -1+2n^2, then u1=A192975, u2=A192976.

If v(n)= 1+n+n^2, then u1=A027181, u2=A192978.

If v(n)= 1-n+n^2, then u1=A192979, u2=A192980.

If v(n)= (n+1)^2, then u1=A001891, u2=A053808.

If v(n)= (n-1)^2, then u1=A192981, u2=A192982.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..400

Index entries for linear recurrences with constant coefficients, signature (3, -2, -1, 1).

Formula

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

From Bruno Berselli, Nov 16 2011: (Start)

G.f.: x*(1+2*x^2)/((1-x)^2*(1 - x - x^2)).

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)

a(n) = Fibonacci(n+4) + 2*Fibonacci(n+2) - (3*n+5). - G. C. Greubel, Jul 12 2019

Mathematica

(* First program *)
q = x^2; s = x + 1; z = 40;
p[0, x]:= 1;
p[n_, x_]:= x*p[n-1, x] + 3n - 1;
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}]:= FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]
t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A171516 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192951 *)
(* Additional programs *)
LinearRecurrence[{3, -2, -1, 1}, {0, 1, 3, 9}, 40] (* Vincenzo Librandi, Nov 16 2011 *)
With[{F=Fibonacci}, Table[F[n+4]+2*F[n+2]-(3*n+5), {n, 0, 40}]] (* G. C. Greubel, Jul 12 2019 *)

Prog

(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
(Magma) F:=Fibonacci; [F(n+4)+2*F(n+2)-(3*n+5): n in [0..40]]; // G. C. Greubel, Jul 12 2019
(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
(PARI) vector(40, n, n--; f=fibonacci; f(n+4)+2*f(n+2)-(3*n+5)) \\ G. C. Greubel, Jul 12 2019
(Sage) f=fibonacci; [f(n+4)+2*f(n+2)-(3*n+5) for n in (0..40)] # G. C. Greubel, Jul 12 2019
(GAP) F:=Fibonacci;; List([0..40], n-> F(n+4)+2*F(n+2)-(3*n+5)); # G. C. Greubel, Jul 12 2019

Crossrefs

Cf. A000045, A192232, A192744.

Sequence in context: A145068 A293357 A202349 * A027114 A145070 A011796

Adjacent sequences: A192948 A192949 A192950 * A192952 A192953 A192954

Keyword

nonn,easy

Author

Clark Kimberling, Jul 13 2011

Status

approved