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A141260
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a(n) = 1 if n == {0,1,3,4,5,7,9,11} mod 12, otherwise a(n) = 0.
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3
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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EXAMPLE
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a(16) = 1 since 16 == 4 (mod 12).
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MATHEMATICA
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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 *)
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CROSSREFS
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Cf. A141259. Note that A035263 has a similar definition, but is a different sequence.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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