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a(n) = [(bn+c)r]-b[nr]-[cr], where (r,b,c)=(sqrt(2),4,3) and []=floor.
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%I #10 Jul 04 2017 18:28:12

%S 1,3,1,2,0,2,3,1,3,0,2,4,1,3,1,2,0,2,3,1,3,0,2,4,1,3,0,2,0,1,3,1,2,0,

%T 2,3,1,3,0,2,4,1,3,1,2,0,2,3,1,3,0,2,4,1,3,1,2,0,1,3,1,2,0,2,3,1,3,0,

%U 2,4,1,3,1,2,0,2,3,1,3,0,2,4,1,3,1,2,0,2,3,1,3,0,2,3,1,3,0,2,0,1,3,1,2,0,2,3,1,3,0,2,4,1,3,1,2,0,2,3,1,3,0,2,4,1,3,1,2,0,1,3,1,2,0,2,3,1,3,0

%N a(n) = [(bn+c)r]-b[nr]-[cr], where (r,b,c)=(sqrt(2),4,3) 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

%H G. C. Greubel, <a href="/A190561/b190561.txt">Table of n, a(n) for n = 1..1000</a>

%t r = Sqrt[2]; b = 4; c = 3;

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

%t t = Table[f[n], {n, 1, 200}] (* A190561 *)

%t Flatten[Position[t, 0]] (* A190562 *)

%t Flatten[Position[t, 1]] (* A190563 *)

%t Flatten[Position[t, 2]] (* A190564 *)

%t Flatten[Position[t, 3]] (* A190565 *)

%t Flatten[Position[t, 4]] (* A190566 *)

%Y Cf. A190562 to A190566.

%K nonn

%O 1,2

%A _Clark Kimberling_, May 12 2011