cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A238537 A fourth-order linear divisibility sequence related to the Pell numbers.

Original entry on oeis.org

1, 42, 1379, 47124, 1599205, 54335358, 1845747527, 62701403688, 2130000094537, 72357312787410, 2458018570699691, 83500274463891516, 2836551311028252973, 96359244313163973414, 3273377755262716618895, 111198484435049515150416, 3777475093033912744231057
Offset: 1

Views

Author

Peter Bala, Feb 28 2014

Keywords

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

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{28, 202, 28, -1}, {1, 42, 1379, 47124}, 17] (* Jean-François Alcover, Nov 02 2019 *)

Formula

a(n) = (1/5)*A000129(3*n)*A001333(n).
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