A141260 a(n) = 1 if n == {0,1,3,4,5,7,9,11} mod 12, otherwise a(n) = 0.

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

Offset: 1

Author

Gary W. Adamson, Jun 18 2008

Keywords

• nonn

Comments

Also characteristic function of A141259.

Let S be the period-3 sequence (1,0,1,1,0,1,1,0,1,...); create a hole after every (1,0,1) segment getting 1,0,1__1,0,1__1,0,1__1,0,1,_1,0,1__,... Then insert successive terms of S into the holes.

In more detail: define S to be 1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1,0,1___...

If we fill the holes with S we get A141260:

1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,

........1.........0.........1.........1.........0.......1.........1.........0...

- the result is

1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.0.1.... = A141260

But instead, if we define T recursively by filling the holes in S with the terms of T itself, we get A035263:

1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0, 1___1, 0,

........1.........0.........1.........1.........1.......0.........1.........0...

- the result is

1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.1.1.0.1.0.1..0..1.1.1..0..1.0.1.. = A035263

Examples

			a(16) = 1 since 16 == 4 (mod 12).
		

Links

• Index entries for characteristic functions

Crossrefs

Cf. A141259. Note that A035263 has a similar definition, but is a different sequence.

Programs

Mathematica

Table[If[MemberQ[{0,1,3,4,5,7,9,11},Mod[n,12]],1,0],{n,110}] (* or *) PadRight[{},110,{1,0,1,1,1,0,1,0,1,0,1,1}] (* Harvey P. Dale, Mar 29 2015 *)

Extensions

Edited by N. J. A. Sloane, Jun 28 2008, Jan 14 2009