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%I #14 Jul 03 2017 22:37:18
%S 1,3,0,2,0,1,3,1,2,0,2,3,1,3,0,2,0,1,3,1,2,0,2,3,1,3,0,2,0,1,3,1,2,0,
%T 1,3,1,2,0,2,3,1,3,0,2,0,1,3,1,2,0,2,3,1,3,0,2,0,1,3,1,2,0,2,3,1,3,0,
%U 2,3,1,3,0,2,0,1,3,1,2,0,2,3,1,3,0,2,0,1,3,1,2,0,2,3,1,3,0,2,0,1,3,0,2,0,1,3,1,2,0,2,3,1,3,0,2,0,1,3,1,2,0,2,3,1,3,0,2,0
%N a(n) = [(bn+c)r]-b[nr]-[cr], where (r,b,c)=(sqrt(2),4,0) 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
%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="/A190544/b190544.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = [4nr] - 4*[nr], where r=sqrt(2).
%t r = Sqrt[2]; b = 4; c = 0;
%t f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];
%t t = Table[f[n], {n, 1, 200}] (* A190544 *)
%t Flatten[Position[t, 0]] (* A190545 *)
%t Flatten[Position[t, 1]] (* A190546 *)
%t Flatten[Position[t, 2]] (* A190547 *)
%t Flatten[Position[t, 3]] (* A190548 *)
%Y Cf. A190545, A190546, A190547, A190548.
%K nonn
%O 1,2
%A _Clark Kimberling_, May 12 2011
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