|
%I #8 Mar 30 2012 18:57:28
%S 2,4,1,3,1,2,4,2,3,1,3,4,2,4,1,3,0,2,4,1,3,1,2,4,2,3,1,3,0,2,4,1,3,1,
%T 2,4,2,3,1,3,4,2,4,1,3,1,2,4,2,3,1,2,4,2,3,1,3,0,2,4,1,3,1,2,4,2,3,1,
%U 3,4,2,4,1,3,1,2,4,2,3,1,3,4,2,4,1,3,0,2,4,1,3,1,2,4,2,3,1,3,0,2,4,1,3,1,2,4,2,3,1,3,4,2,4,1,3,1,2,4,1,3,1,2,4,2,3,1,3,0
%N [(bn+c)r]-b[nr]-[cr], where (r,b,c)=(sqrt(2),4,2) 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),2,1): A190483-A190486
%C (sqrt(2),3,0): A190487-A190490
%C (sqrt(2),3,1): A190491-A190495
%C (sqrt(2),3,2): A190496-A190500
%t r = Sqrt[2]; b = 4; c = 2;
%t f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];
%t t = Table[f[n], {n, 1, 200}] (* A190555 *)
%t Flatten[Position[t, 0]] (* A190556 *)
%t Flatten[Position[t, 1]] (* A190557 *)
%t Flatten[Position[t, 2]] (* A190558 *)
%t Flatten[Position[t, 3]] (* A190559 *)
%t Flatten[Position[t, 4]] (* A190486 *)
%Y Cf. A190556-A190559, A190486.
%K nonn
%O 1,1
%A _Clark Kimberling_, May 12 2011
|
