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[(bn+c)r]-b[nr]-[cr], where (r,b,c)=(sqrt(3),2,1) and [ ]=floor.
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%I #7 Mar 30 2012 18:57:28

%S 2,1,1,2,2,1,0,2,1,1,0,2,1,1,2,2,1,1,2,2,1,0,2,1,1,0,2,1,1,2,2,1,1,2,

%T 1,1,0,2,1,1,0,2,1,1,2,2,1,1,2,1,1,0,2,1,1,2,2,1,1,2,2,1,0,2,1,1,0,2,

%U 1,1,2,2,1,1,2,2,1,0,2,1,1,0,2,1,1,2,2,1,1,2,1,1,0,2,1,1,0,2,1,1,2,2,1,0,2,1,1,0,2,1,1,2,2,1,1,2,2,1,0,2,1,1,0,2,1,1,2,2,1,1,2,1

%N [(bn+c)r]-b[nr]-[cr], where (r,b,c)=(sqrt(3),2,1) and [ ]=floor.

%C Write 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 position sequences comprise a partition of the positive integers.

%C Examples:

%C (golden ratio,2,1): A190427-A190430

%C (sqrt(2),2,0): A190480-A190482

%C (sqrt(2),2,1): A190483-A190486

%C (sqrt(2),3,0): A190487-A190490

%C (sqrt(2),3,1): A190491-A190495

%C (sqrt(2),3,2): A190496-A190500

%C (sqrt(2),4,c): A190544-A190566

%t r = Sqrt[3]; b = 2; c = 1;

%t f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];

%t t = Table[f[n], {n, 1, 200}] (* A190672 *)

%t Flatten[Position[t, 0]] (* A190673 *)

%t Flatten[Position[t, 1]] (* A190674 *)

%t Flatten[Position[t, 2]] (* A190675 *)

%Y Cf. A190673, A190674, A190675.

%K nonn

%O 1,1

%A _Clark Kimberling_, May 16 2011