OFFSET
1,2
COMMENTS
Numbers n such that Fibonacci(n) mod 4 = 1. - Gary Detlefs, Jun 01 2012
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).
FORMULA
From Colin Barker, Mar 13 2012
G.f.: x*(1+x+3*x^2+x^3)/((1-x)^2*(1+x+x^2)).
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4. (End)
a(n) = 2*n-1-floor((n mod 3)/2). - Gary Detlefs, Jun 01 2012
From Wesley Ivan Hurt, Jun 10 2016: (Start)
a(n) = (6*n-4+cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/3.
a(3k) = 6k-1, a(3k-1) = 6k-4, a(3k-2) = 6k-5. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = (6-sqrt(3))*Pi/18 - log(2)/6. - Amiram Eldar, Dec 16 2021
MAPLE
A047268:=n->(6*n-4+cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/3: seq(A047268(n), n=1..100); # Wesley Ivan Hurt, Jun 10 2016
MATHEMATICA
Select[Range[0, 120], MemberQ[{1, 2, 5}, Mod[#, 6]]&] (* Vincenzo Librandi, Apr 26 2012 *)
PROG
(Magma) I:=[1, 2, 5, 7]; [n le 4 select I[n] else Self(n-1)+Self(n-3)-Self(n-4): n in [1..70]]; // Vincenzo Librandi, Apr 26 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved