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A238536
A fourth-order linear divisibility sequence related to the Fibonacci numbers: a(n) = (1/2)*Fibonacci(3*n)*Lucas(n).
11
1, 12, 68, 504, 3355, 23256, 158717, 1089648, 7463884, 51170460, 350695511, 2403786672, 16475579353, 112925875764, 774003961940, 5305106018016, 36361727272627, 249227013404808, 1708227291909269, 11708364225400920, 80250321774226396, 550043889533755332, 3770056901455017263
OFFSET
1,2
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 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.
LINKS
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
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 *)
PROG
(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
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Feb 28 2014
STATUS
approved