A192422 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined below in Comments.

0, 1, 1, 5, 7, 20, 35, 83, 161, 355, 720, 1541, 3185, 6733, 14027, 29500, 61663, 129403, 270865, 567911, 1189440, 2492905, 5222449, 10943813, 22928815, 48044900, 100665083, 210927155, 441948689, 926020171, 1940274000, 4065458669, 8518311809

Offset: 0

Author

Clark Kimberling, Jun 30 2011

Keywords

• nonn

Comments

The polynomial p(n,x) is defined by ((x+d)/2)^n + ((x-d)/2)^n, where d=sqrt(x^2+4). For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232.

Assuming the o.g.f. given below, this sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. It is the case P1 = 1, P2 = -1, Q = -1 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. Cf. A100047. - Peter Bala, Aug 28 2019

Examples

			The first five polynomials p(n,x) and their reductions are as follows:
  p(0,x) = 2 -> 2
  p(1,x) = x -> x
  p(2,x) = 2 + x^2 -> 3 + x
  p(3,x) = 3*x + x^3 -> 1 + 5*x
  p(4,x) = 2 + 4*x^2 + x^4 -> 8 + 7*x.
From these, read A192421 = (2, 0, 3, 1, 8, ...) and a(n) = (0, 1, 1, 5, 7, ...).
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
• Index entries for linear recurrences with constant coefficients, signature (1,3,-1,-1).

Crossrefs

Cf. A000045, A100047, A192232, A192421.

Programs

Magma

R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x*(1+x^2)/(1-x-3*x^2+x^3+x^4) )); // G. C. Greubel, Jul 11 2023

Mathematica

(See A192421.)
LinearRecurrence[{1,3,-1,-1}, {0,1,1,5}, 40] (* G. C. Greubel, Jul 11 2023 *)

Maxima

a(n):=n*sum((binomial(n-i-1,i))/(n-2*i)*fib(n-2*i),i,0,(n-1)/2); /* Vladimir Kruchinin, Mar 20 2016 */

SageMath

@CachedFunction
def a(n): # a = A192422
    if (n<4): return (0,1,1,5)[n]
    else: return a(n-1) +3*a(n-2) -a(n-3) -a(n-4)
[a(n) for n in range(41)] # G. C. Greubel, Jul 11 2023

Formula

From Colin Barker, May 12 2014: (Start)

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

G.f.: x*(1 + x^2)/(1 - x - 3*x^2 + x^3 + x^4). (End)

From Vladimir Kruchinin, Mar 20 2016: (Start)

G.f.: ((1+x^2)/(1-x^2)) * F(x/(1-x^2)), where F(x) is g.f. of Fibonacci numbers (A000045).

a(n) = n*Sum_{i=0..floor((n-1)/2)} (binomial(n-i-1,i)/(n-2*i))*Fibonacci(n-2*i). (End)

a(n) = Sum_{j=0..n} T(n, j)*Fibonacci(j), where T(n, k) = [x^k] ((x + sqrt(x^2+4))^n + (x - sqrt(x^2+4))^n)/2^n. - G. C. Greubel, Jul 11 2023