OFFSET
1,2
COMMENTS
Let P and Q be integers. The Lucas sequences U(n) and V(n) (which depend on P and Q) are a pair of integer sequences that satisfy the recurrence equation a(n) = P*a(n-1) - Q*a(n-2) with the initial conditions U(0) = 0, U(1) = 1 and V(0) = 2, V(1) = P, respectively. The sequence {U(n)}n>=1 is a divisibility sequence, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0. In general the sequence V(n) is not a divisibility sequence. However, it can be shown that if p >= 3 is an odd integer then the sequence {U(p*n)*V(n)}n>=1 is a divisibility sequence satisfying a linear recurrence of order 4. For a proof and a generalization of this result see the Bala link. Here we take p = 3 with P = 2 and Q = -1, for which U(n) is the sequence of Pell numbers A000129, and consider the normalized divisibility sequence with initial term equal to 1. For other sequences of this type see A238536 and A238538
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..654
E. L. Roettger and H. C. Williams, Appearance of Primes in Fourth-Order Odd Divisibility Sequences, J. Int. Seq., Vol. 24 (2021), Article 21.7.5.
Wikipedia, Divisibility sequence
Wikipedia, Pell number
Index entries for linear recurrences with constant coefficients, signature (28,202,28,-1)
FORMULA
a(n) = (1/(20*sqrt(2)))*((1 + sqrt(2))^(3*n) - (1 - sqrt(2))^(3*n))*( (1 + sqrt(2))^n + (1 - sqrt(2))^n ).
O.g.f.: x*(1 + 14*x + x^2)/( (1 + 6*x + x^2)*(1 - 34*x + x^2) ).
Recurrence equation: a(n) = 28*a(n-1) + 202*a(n-2) + 28*a(n-3) - a(n-4).
a(n) = (1/10) * (Pell(4n) + (-1)^n*Pell(2n)), with Pell(n) = A000129(n). - Ralf Stephan, Mar 01 2014
MATHEMATICA
LinearRecurrence[{28, 202, 28, -1}, {1, 42, 1379, 47124}, 17] (* Jean-François Alcover, Nov 02 2019 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Feb 28 2014
STATUS
approved