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

Author

Clark Kimberling, Jul 13 2011

Keywords

• nonn
• easy

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.

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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).

Crossrefs

Cf. A000045, A192232, A192744.

Programs

GAP

F:=Fibonacci;; List([0..40], n-> F(n+4)+2*F(n+2)-(3*n+5)); # G. C. Greubel, Jul 12 2019

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

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 *)

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

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