A047268 Numbers that are congruent to {1, 2, 5} mod 6.

1, 2, 5, 7, 8, 11, 13, 14, 17, 19, 20, 23, 25, 26, 29, 31, 32, 35, 37, 38, 41, 43, 44, 47, 49, 50, 53, 55, 56, 59, 61, 62, 65, 67, 68, 71, 73, 74, 77, 79, 80, 83, 85, 86, 89, 91, 92, 95, 97, 98, 101, 103, 104, 107, 109, 110, 113, 115, 116, 119, 121, 122, 125

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

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).

Programs

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

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 *)

Formula

From Colin Barker, Mar 13 2012: (Start)

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