A141212 a(n) = 1, if n == {1,3,4} mod 6; otherwise 0.

1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1

Offset: 1

Author

Gary W. Adamson, Jun 14 2008

Keywords

• nonn
• easy

Examples

			a(7) = 1 since 7 == 1 mod 6.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Crossrefs

Cf. A029739, A049347.

Programs

Mathematica

Table[If[MemberQ[{1,3,4},Mod[n,6]],1,0],{n,120}] (* or *) PadRight[{},120,{1,0,1,1,0,0}] (* Harvey P. Dale, May 03 2013 *)

Formula

a(n) = 1 if n is in A029739, otherwise 0.

Begin with the sequence S: (1,0,1,0,1,0,...) and create a hole every 3n-th place: 1,0_1,0_1,0_1,0_,... Then insert terms of the sequence S in the holes.

From R. J. Mathar, Jun 17 2008: (Start)

O.g.f.: x(1+x^2+x^3)/((1-x)(1+x)(x^2+x+1)(x^2-x+1)).

a(n) = 1/2+A049347(n-1)/2-(-1)^n/6-A087204(n)/6 = a(n-6). (End)

a(n) = ((Fibonacci(n+4) mod 4) mod 3) mod 2. - Gary Detlefs, Dec 29 2010