OFFSET
1,2
COMMENTS
This is a divisibility sequence: if n divides m then a(n) divides a(m). The sequence satisfies a linear recurrence of order 6. In general, for integers r and s, the sequence Fibonacci(r*n) - 2*Fibonacci((r - 2*s)*n) + Fibonacci((r - 4*s)*n) is a divisibility sequence of the sixth order. This is the case r = 3, s = 1. See A127595 (case r = 4, s = 1).
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,4,-12,-4,4,1).
FORMULA
a(n) = Fibonacci(3*n) - 2*Fibonacci(n) + Fibonacci(-n).
a(2*n) = 5*Fibonacci(2*n)^3;
a(2n+1) = Fibonacci(2*n+1)*(5*Fibonacci(2*n+1)^2 - 4) = Fibonacci(2*n+1)*Lucas(2*n+1)^2.
O.g.f. x*(x^4 - x^3 + 8*x^2 + x + 1)/( (1 + x - x^2 )*(1 - x - x^2)*(1 - 4*x - x^2 ) ).
a(n) = 4*a(n-1) + 4*a(n-2) - 12*a(n-3) - 4*a(n-4) + 4*a(n-5) + a(n-6). - G. C. Greubel, Jun 02 2016
MAPLE
MATHEMATICA
LinearRecurrence[{4, 4, -12, -4, 4, 1}, {1, 5, 32, 135, 605, 2560}, 100] (* G. C. Greubel, Jun 02 2016 *)
Table[Fibonacci[3 n] - (2 + (-1)^n) Fibonacci[n], {n, 1, 30}] (* Vincenzo Librandi, Jun 02 2016 *)
PROG
(Magma) [Fibonacci(3*n)-(2+(-1)^n)*Fibonacci(n): n in [1..25]]; // Vincenzo Librandi, Jun 02 2016
(PARI) a(n)=fibonacci(3*n) - (2 + (-1)^n)*fibonacci(n) \\ Charles R Greathouse IV, Jun 08 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, May 29 2016
STATUS
approved