|
%I #40 Sep 08 2022 08:46:07
%S 1,12,68,504,3355,23256,158717,1089648,7463884,51170460,350695511,
%T 2403786672,16475579353,112925875764,774003961940,5305106018016,
%U 36361727272627,249227013404808,1708227291909269,11708364225400920,80250321774226396,550043889533755332,3770056901455017263
%N A fourth-order linear divisibility sequence related to the Fibonacci numbers: a(n) = (1/2)*Fibonacci(3*n)*Lucas(n).
%C 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.
%D S. Koshkin, Non-classical linear divisibility sequences ..., Fib. Q., 57 (No. 1, 2019), 68-80.
%H G. C. Greubel, <a href="/A238536/b238536.txt">Table of n, a(n) for n = 1..1185</a>
%H Peter Bala, <a href="/A238536/a238536.pdf">A family of linear divisibility sequences of order four</a>
%H E. L. Roettger and H. C. Williams, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL24/Roettger/roettger12.html">Appearance of Primes in Fourth-Order Odd Divisibility Sequences</a>, J. Int. Seq., Vol. 24 (2021), Article 21.7.5.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Divisibility_sequence">Divisibility sequence</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Lucas_sequence">Lucas sequence</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,19,4,-1).
%F a(n) = (1/2)*Fibonacci(3*n)*Lucas(n) = (1/2)*A000045(3*n)*A000032(n).
%F a(n) = (1/2)*Fibonacci(2*n)*Fibonacci(3*n)/Fibonacci(n).
%F 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 ).
%F The sequence can be extended to negative indices by setting a(-n) = -a(n).
%F O.g.f. x*(1 + 8*x + x^2)/( (1 + 3*x + x^2)*(1 - 7*x + x^2) ).
%F Recurrence equation: a(n) = 4*a(n-1) + 19*a(n-2) + 4*a(n-3) - a(n-4).
%F a(n) = (1/2) * (Fibonacci(4*n) + (-1)^n*Fibonacci(2*n)). - _Ralf Stephan_, Mar 01 2014
%p with(combinat): lucas:= n->fibonacci(n+1)+ fibonacci(n-1):
%p seq(1/2*lucas(n)*fibonacci(3*n), n = 1..24);
%t 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 *)
%o (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
%o (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
%Y Cf. A000032, A000045, A127595, A215466, A238537, A238538.
%K nonn,easy
%O 1,2
%A _Peter Bala_, Feb 28 2014
|
