A190893 a(n) = [3en] - 3[en], where [ ] = floor.

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, 1, 0, 2, 2, 1, 0, 2, 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

Offset: 1

Author

Clark Kimberling, May 26 2011

Keywords

• nonn

Comments

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.

Crossrefs

Cf. A191103, A191104, A191105 (3-way partition of N, from this sequence).

Programs

Mathematica

f[n_] := Floor[3 n*E] - 3*Floor[n*E];
t = Table[f[n], {n, 1, 220}] (* A190893 *)
Flatten[Position[t, 0]]      (* A191103 *)
Flatten[Position[t, 1]]      (* A191104 *)
Flatten[Position[t, 1]]      (* A191105 *)
f[n_]:=Module[{c=E*n},Floor[3*c]-3*Floor[c]]; Array[f,150] (* Harvey P. Dale, Feb 08 2013 *)