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.

A238536 A fourth-order linear divisibility sequence related to the Fibonacci numbers: a(n) = (1/2)*Fibonacci(3*n)*Lucas(n).

Original entry on oeis.org

1, 12, 68, 504, 3355, 23256, 158717, 1089648, 7463884, 51170460, 350695511, 2403786672, 16475579353, 112925875764, 774003961940, 5305106018016, 36361727272627, 249227013404808, 1708227291909269, 11708364225400920, 80250321774226396, 550043889533755332, 3770056901455017263
Offset: 1

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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 linear divisibility sequence of order 2, i.e., U(n) divides U(m) whenever n divides m and U(n) <> 0. In general, 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 linear divisibility sequence of order 4. For a proof and a generalization of this result see the Bala link. Here we take p = 3 with P = 1 and Q = -1, for which U(n) is the sequence of Fibonacci numbers, A000045, V(n) is the sequence of Lucas numbers, A000032, and normalize the sequence to have the initial term 1. For other sequences of this type see A238537 and A238538.

References

  • S. Koshkin, Non-classical linear divisibility sequences ..., Fib. Q., 57 (No. 1, 2019), 68-80.

Crossrefs

Programs

  • Magma
    I:=[12, 68, 504, 3355]; [1] cat [n le 4 select I[n] else 4*Self(n-1) + 19*Self(n-2) + 4*Self(n-3) - Self(n-4): n in [1..30]]; // G. C. Greubel, Dec 25 2017
  • Maple
    with(combinat): lucas:= n->fibonacci(n+1)+ fibonacci(n-1):
    seq(1/2*lucas(n)*fibonacci(3*n), n = 1..24);
  • Mathematica
    Table[Fibonacci(3*n)*Lucas(n)/2, {n,1,30}] (* or *) Join[{1}, LinearRecurrence[{4,19,4,-1}, {12, 68, 504, 3355}, 30]] (* G. C. Greubel, Dec 25 2017 *)
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,4,19,4]^(n-1)*[1;12;68;504])[1,1] \\ Charles R Greathouse IV, Oct 07 2016
    

Formula

a(n) = (1/2)*Fibonacci(3*n)*Lucas(n) = (1/2)*A000045(3*n)*A000032(n).
a(n) = (1/2)*Fibonacci(2*n)*Fibonacci(3*n)/Fibonacci(n).
a(n) = (1/(2*sqrt(5)))*( ((7 + 3*sqrt(5))/2)^n - ((7 - 3*sqrt(5))/2)^n + (-1)^n*((3 + sqrt(5))/2)^n - (-1)^n*((3 - sqrt(5))/2)^n ).
The sequence can be extended to negative indices by setting a(-n) = -a(n).
O.g.f. x*(1 + 8*x + x^2)/( (1 + 3*x + x^2)*(1 - 7*x + x^2) ).
Recurrence equation: a(n) = 4*a(n-1) + 19*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = (1/2) * (Fibonacci(4*n) + (-1)^n*Fibonacci(2*n)). - Ralf Stephan, Mar 01 2014