OFFSET
1,2
COMMENTS
This is a divisibility sequence, that is, if n divides m then a(n) divides a(m). The sequence satisfies a sixth-order linear recurrence. More generally, the sequence s(n) := Fibonacci(2*n) + Fibonacci(4*n) + ... + Fibonacci(2*k*n) is a divisibility sequence for k = 1,2,3,.... See A215466 for the case k = 2. Cf. A273623, A273624.
From Peter Bala, Aug 05 2019: (Start)
Let U(n;P,Q), where P and Q are integer parameters, denote the Lucas sequence of the first kind. Then, excluding the cases P = -1 and P = 0, the sequence ( U(n;P,1) + U(2*n;P,1) + U(3*n;P,1))/(P^2 + P) is a sixth-order linear divisibility sequence with o.g.f. x*(1 - 2*(P^2 - 2)*x + (3*P^3 - 3*P^2 - 8*P + 10)*x^2 - 2*(P^2 - 2)*x^3 + x^4)/((1 - P*x + x^2)*(1 - (P^2 - 2)*x + x^2)*(1 - P*(P^2 - 3)*x + x^2)). This is the case P = 3.
More generally, the sequence U(n;P,1) + U(2*n;P,1) + ... + U(k*n;P,1) is a linear divisibility sequence of order 2*k. See, for example, A215466 with P = 3, k = 2. (End)
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..795
Index entries for linear recurrences with constant coefficients, signature (28,-204,434,-204,28,-1).
FORMULA
a(n) = -a(-n).
O.g.f.: x*(x^4 - 14*x^3 + 40*x^2 - 14*x + 1)/((x^2 - 3*x + 1)*(x^2 - 7*x + 1)*(x^2 - 18*x + 1)).
a(n) = 28*a(n-1) - 204*a(n-2) + 434*a(n-3) - 204*a(n-4) + 28*a(n-5) - a(n-6). - G. C. Greubel, Jun 02 2016
MAPLE
MATHEMATICA
LinearRecurrence[{28, -204, 434, -204, 28, -1}, {1, 14, 228, 3948, 69905, 1248072}, 100] (* G. C. Greubel, Jun 02 2016 *)
Table[1/12 (Fibonacci[2 n] + Fibonacci[4 n] + Fibonacci[6 n]), {n, 1, 30}] (* Vincenzo Librandi, Jun 02 2016 *)
PROG
(Magma) [1/12*(Fibonacci(2*n)+Fibonacci(4*n)+Fibonacci(6*n)): n in [1..25]]; // Vincenzo Librandi, Jun 02 2016
(PARI) A001906(n)=fibonacci(2*n)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, May 31 2016
STATUS
approved