A192240 Constant term in the reduction of the polynomial (x+3)^n by x^2 -> x+1.

1, 3, 10, 37, 149, 636, 2813, 12695, 57922, 265809, 1223521, 5640748, 26026505, 120137307, 554669594, 2561176781, 11826871933, 54615158940, 252210521317, 1164706900879, 5378632571666, 24838652091993, 114705606355625, 529714071477452

Offset: 0

Author

Clark Kimberling, Jun 26 2011

Keywords

• nonn

Comments

See A192232.

Links

• Robert Israel, Table of n, a(n) for n = 0..1498

Crossrefs

Cf. A192232.

Programs

Maple

seq(eval(rem((x+3)^n,x^2-x-1,x),x=0),n=0..50); # Robert Israel, Mar 14 2023

Mathematica

q[x_] := x + 1;
p[n_, x_] := (x + 3)^n;
reductionRules = {x^y_?EvenQ -> q[x]^(y/2),
   x^y_?OddQ -> x q[x]^((y - 1)/2)};
t = Table[
   Last[Most[
     FixedPointList[Expand[#1 /. reductionRules] &, p[n, x]]]], {n, 0,
     30}];
Table[Coefficient[Part[t, n], x, 0], {n, 30}]  (* A192240 *)
Table[Coefficient[Part[t, n], x, 1], {n, 30}]  (* A099453 *)
(* Peter J. C. Moses, Jun 26 2011 *)

Formula

Empirical g.f. and recurrence: (1-4*x)/(1-7*x+11*x^2). a(n) = 7*a(n-1) - 11*a(n-2). - Colin Barker, Feb 09 2012

Proof of recurrence: if r(n) == (x+3)^n mod (x^2-x-1), then r(n+j) == (x+1)^(n+j) mod (x^2 - x - 1). Now r(n+2) - 7*r(n+1) + 11*r(n) == ((x+3)^2 - 7*(x+3) + 11)*r(n) == 0 mod (x^2-x-1) since (x+3)^2 - 7*(x+3) + 11 = x^2 - x - 1. - Robert Israel, Mar 14 2023

a(n) = Sum_{i=0..n} (-1)^i*Fibonacci(i+1)*binomial(n,i)*4^(n-i) (conjecture). - Rigoberto Florez, Mar 25 2020

Extensions

Offset corrected by Robert Israel, Mar 14 2023