OFFSET
1,2
COMMENTS
All prime Fibonacci numbers (A005478) except 2 and 3 are in this sequence. All terms equal 1 (mod 4). The indices of these Fibonacci numbers are 6k-1 or 6k+1.
This sequence can be thought of as two interlocking sequences, each of the form a(n) = 18a(n - 1) - a(n - 2).
Proof that all terms equal 1 (mod 4): From the Lucas 1876 identity Fib(2n+1) = Fib(n)^2 + Fib(n+1)^2 (see Weisstein, formula 60, or page 79 of Koshy), we see that odd-indexed Fibonacci numbers are the sum of two squares. Because a square is 0 or 1 (mod 4), the sum of two squares is 0, 1, or 2 (mod 4). All these terms are odd numbers. Hence, the only possibility is that they are 1 (mod 4). This can also be proved from the recursion formula.
REFERENCES
Thomas Koshy, "Fibonacci and Lucas Numbers and Applications", Wiley, New York, 2001.
LINKS
T. D. Noe, Table of n, a(n) for n = 1..200
Eric W. Weisstein, Fibonacci Number.
Index entries for linear recurrences with constant coefficients, signature (0,18,0,-1).
FORMULA
a(n) = 18*a(n-2) - a(n-4), with a(1)=1, a(2)=5, a(3)=13, and a(4)=89.
G.f.: x*(1-x)*(1+6*x+x^2)/((1+4*x-x^2)*(1-4*x-x^2)). - Colin Barker, Jun 19 2012
a(n) = (-(2-sqrt(5))^n + (-2-sqrt(5))^n*(-2+sqrt(5)) + 2*(-2+sqrt(5))^n + sqrt(5)*(-2+sqrt(5))^n + (2+sqrt(5))^n)/(2*sqrt(5)) for n>0. - Colin Barker, Mar 27 2016
MATHEMATICA
LinearRecurrence[{0, 18, 0, -1}, {1, 5, 13, 89}, 50]
PROG
(PARI) a(n)=fibonacci(n\2*6+if(n%2, 1, -1)) \\ Charles R Greathouse IV, Jun 08 2011
(PARI) Vec(x*(1-x)*(1+6*x+x^2)/((1+4*x-x^2)*(1-4*x-x^2)) + O(x^30)) \\ Colin Barker, Mar 27 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
T. D. Noe, May 24 2011
STATUS
approved