OFFSET
1,2
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 denote the reciprocal polynomial of f(x). Then the rational function x*d/dx( log(f(x)/g(x)) ) is the generating function of 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 2 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 the particular case of their result with parameters P1 = 4, P2 = -5, P3 = -16, P4 = 20 and Q = 1.
LINKS
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 (4,1,-4,-16,-4,1,4,-1).
FORMULA
a(n) = (1/2) * Sum_{i = 1..4} (alpha(i)^n - 1/alpha(i)^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(n)^2 = -(1/4) * Product_{i = 1..6} (1 - beta(i)^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(n) = 4*a(n-1) + a(n-2) - 4*a(n-3) - 16*a(n-4) - 4*a(n-5) + a(n-6) + 4*a(n-7) - a(n-8).
O.g.f.: x*(x^6 - x^4 + 8*x^3 - x^2 + 1)/((x^4 - x^3 - 2*x^2 - 3*x + 1)*(x^4 - 3*x^3 - 2*x^2 - x + 1)).
MATHEMATICA
LinearRecurrence[{4, 1, -4, -16, -4, 1, 4, -1}, {1, 4, 16, 72, 271, 1024, 3809, 13968}, 25] (* Jean-François Alcover, Nov 12 2019 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Sep 22 2019
STATUS
approved