A190483 a(n) = [(bn+c)r]-b[nr]-[cr], where (r,b,c)=(sqrt(2),2,1) and []=floor.

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

Offset: 1

Author

Clark Kimberling, May 11 2011

Keywords

• nonn

Comments

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.

Examples:

(golden ratio,2,1): A190427-A190430

(sqrt(2),2,0): A190480

(sqrt(2),2,1): A190483-A190486

(sqrt(2),3,0): A190487-A190490

(sqrt(2),3,1): A190491-A190495

(sqrt(2),3,2): A190496-A190500

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A190484, A190485, A190486.

Programs

Mathematica

r = Sqrt[2]; b = 2; c = 1;
f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];
t = Table[f[n], {n, 1, 200}]  (* A190483 *)
Flatten[Position[t, 0]]   (* A190484 *)
Flatten[Position[t, 1]]   (* A190485 *)
Flatten[Position[t, 2]]   (* A190486 *)

Python

from sympy import sqrt, floor
r=sqrt(2)
def a(n): return floor((2*n + 1)*r) - 2*floor(n*r) - floor(r)
print([a(n) for n in range(1, 501)]) # Indranil Ghosh, Jul 02 2017