A327542 A linear divisibility sequence of order 8.

1, 2, 16, 36, 171, 512, 2087, 6984, 26512, 92682, 341573, 1216512, 4429309, 15898766, 57595536, 207410832, 749793263, 2703799808, 9765692771, 35235657396, 127218945296, 459128080534, 1657436539337, 5982212358144, 21594204190521

Offset: 1

Author

Peter Bala, Sep 23 2019

Keywords

• nonn
• easy

Comments

Let f(x) = 1 + P*x + Q*x^2 + R*x^3 + x^4 be a monic quartic polynomial with integer coefficients. Let g(x) = x^4*f(1/x) = 1 + R*x + Q*x^2 + P*x^3 + x^4 be the reciprocal polynomial of f(x). Then the rational function x*d/dx( log(f(x)/g(-x)) ) is the generating function for a divisibility sequence satisfying a linear recurrence equation of order 8. Here we take f(x) = 1 + x - 2*x^2 + 3*x^3 + x^4 (and normalize the resulting divisibility sequence by removing a common factor of 4 from the terms of the sequence).

Roettger et al. constructed a 5-parameter family U_n(P1,P2,P3,P4,Q) of linear divisibility sequences of order 8. This sequence is a particular case of their result with parameters P1 = 2, P2 = -3, P3 = 0, P4 = -16 and Q = -1.

There are corresponding results for certain cubic polynomials - see A001945. See also A327541.

Links

• P. Bala, Some linear divisibility sequences of order 8
• E. L. Roettger, H. C. Williams, R. K. Guy, Some extensions of the Lucas functions, Number Theory and Related Fields: In Memory of Alf van der Poorten, Series: Springer Proceedings in Mathematics & Statistics 43, 271-311 (2013), chapter 5.
• Index entries for linear recurrences with constant coefficients, signature (2,7,-6,4,6,7,-2,-1).

Crossrefs

Cf. A001351, A001945, A327541.

Programs

Mathematica

a[n_] := With[{m = 1 - 2 Mod[n, 2]}, (m/4)(x^n - m/x^n) /. {Roots[1 + x - 2x^2 + 3x^3 + x^4 == 0, x] // ToRules} // Total // Round];
a /@ Range[25] (* Jean-François Alcover, Nov 11 2019 *)

Formula

a(2*n) = (1/4) * Sum_{i = 1..4} (alpha(i)^(2*n) - 1/alpha(i)^(2*n)), where alpha(i), 1 <= i <= 4, are the zeros of the quartic polynomial 1 + x - 2*x^2 + 3*x^3 + x^4.

a(2*n+1) = (-1/4) * Sum_{i = 1..4} (alpha(i)^(2*n+1) + 1/alpha(i)^(2*n+1)).

a(2*n)^2 = (-1/16) * Product_{i = 1..6} (1 - beta(i)^(2*n)), where beta(i), 1 <= i <= 6, are the zeros of the sextic polynomial x^6 + 2*x^5 + 2*x^4 - 14*x^3 + 2*x^2 + 2*x + 1.

a(2*n+1)^2 = (1/16) * Product_{i = 1..6} (1 + beta(i)^(2*n+1)).

a(n) = 2*a(n-1) + 7*a(n-2) - 6*a(n-3) + 4*a(n-4) + 6*a(n-5) + 7*a(n-6) - 2*a(n-7) - a(n-8).

O.g.f.: x*(1 + 5*x^2 - 4*x^3 - 5*x^4 - x^6)/((1 + x - 2*x^2 + 3*x^3 + x^4)*(1 - 3*x - 2*x^2 - x^3 + x^4)).