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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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%I #25 Mar 08 2024 11:57:35

%S 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,

%T 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,

%U 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

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

%C Also characteristic function of A141259.

%C 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.

%C 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___...

%C If we fill the holes with S we get A141260:

%C 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,

%C ........1.........0.........1.........1.........0.......1.........1.........0...

%C - the result is

%C 1..0..1.1.1..0..1.0.1..0..1.1.1..0..1.1.1..0..1.0.1.... = A141260

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

%C 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,

%C ........1.........0.........1.........1.........1.......0.........1.........0...

%C - the result is

%C 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

%H <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>

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

%t 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 *)

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

%K nonn

%O 1,1

%A _Gary W. Adamson_, Jun 18 2008

%E Edited by _N. J. A. Sloane_, Jun 28 2008, Jan 14 2009