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a(n) = [5nr]-5[nr], where r=sqrt(5).
6

%I #10 Jul 10 2020 03:51:10

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

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

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

%N a(n) = [5nr]-5[nr], where r=sqrt(5).

%C In general, suppose 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. For c=0, there are b of these position sequences, and they comprise a partition of the positive integers.

%F a(n) = [5nr]-5[nr], where r=sqrt(5).

%t r = Sqrt[5];

%t f[n_] := Floor[5n*r] - 5*Floor[n*r]

%t t = Table[f[n], {n, 1, 400}] (* A190886 *)

%t Flatten[Position[t, 0]] (* A190887 *)

%t Flatten[Position[t, 1]] (* A190888 *)

%t Flatten[Position[t, 2]] (* A190889 *)

%t Flatten[Position[t, 3]] (* A190890 *)

%t Flatten[Position[t, 4]] (* A190891 *)

%Y Cf. A190887, A190888, A190889, A190890, A190891.

%K nonn

%O 1,2

%A _Clark Kimberling_, May 26 2011