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%I #11 Sep 09 2019 12:49:35
%S 2,1,0,2,1,0,0,2,1,0,2,1,1,0,2,1,0,2,1,1,0,2,1,0,2,2,1,0,2,1,0,2,2,1,
%T 0,2,1,0,0,2,1,0,2,1,0,0,2,1,0,2,1,1,0,2,1,0,2,1,1,0,2,1,0,2,2,1,0,2,
%U 1,0,2,2,1,0,2,1,0,0,2,1,0,2,1,1,0,2,1,0,2,1,1,0,2,1,0,2,2,1,0,2,1,0,2,2,1,0,2,1,0,0,2,1,0,2,1,0,0,2,1,0,2,1,1,0,2,1,0,2,1,1,0,2
%N a(n) = [3en] - 3[en], where [ ] = floor.
%C Suppose, in general, that a(n) = [(bn+c)r] - b[nr] - [cr]. If r > 0 and b and c are integers satisfying b >= 2 and 0 <= c <= b-1, then 0 <= a(n) <= b. The positions of 0 in the sequence a are of interest, as are the position sequences for 1,2,...,b. These b+1 (or b) position sequences comprise a partition of the positive integers.
%t f[n_] := Floor[3 n*E] - 3*Floor[n*E];
%t t = Table[f[n], {n, 1, 220}] (* A190893 *)
%t Flatten[Position[t, 0]] (* A191103 *)
%t Flatten[Position[t, 1]] (* A191104 *)
%t Flatten[Position[t, 1]] (* A191105 *)
%t f[n_]:=Module[{c=E*n},Floor[3*c]-3*Floor[c]]; Array[f,150] (* _Harvey P. Dale_, Feb 08 2013 *)
%Y Cf. A191103, A191104, A191105 (3-way partition of N, from this sequence).
%K nonn
%O 1,1
%A _Clark Kimberling_, May 26 2011
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