A189480 [4rn]-4[rn], where r=sqrt(5) and [ ]=floor.

0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 2, 3, 0, 1, 2, 3, 0

Offset: 1

Author

Clark Kimberling, Apr 23 2011

Keywords

• nonn

Comments

Suppose, in general, 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. These b+1 (or b) position sequences comprise a partition of the positive integers.

Links

• Ivan Panchenko, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A190813, A190883, A190884, A190885.

Programs

Mathematica

r=Sqrt[5];
f[n_]:=Floor[4 n*r]-4*Floor[n*r];
t=Table[f[n],{n,1,320}] (*A189480*)
Flatten[Position[t,0]]  (*A190813*)
Flatten[Position[t,1]]  (*A190883*)
Flatten[Position[t,2]]  (*A190884*)
Flatten[Position[t,3]]  (*A190885*)