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A141260
a(n) = 1 if n == {0,1,3,4,5,7,9,11} mod 12, otherwise a(n) = 0.
3
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
1,1
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
EXAMPLE
a(16) = 1 since 16 == 4 (mod 12).
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 *)
CROSSREFS
Cf. A141259. Note that A035263 has a similar definition, but is a different sequence.
Sequence in context: A323045 A104106 A349167 * A029883 A035263 A089045
KEYWORD
nonn
AUTHOR
Gary W. Adamson, Jun 18 2008
EXTENSIONS
Edited by N. J. A. Sloane, Jun 28 2008, Jan 14 2009
STATUS
approved