A190427 a(n) = [(b*n+c)*r] - b*[n*r] - [c*r], where (r,b,c)=(golden ratio,2,1) and []=floor.

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

Offset: 1

Author

Clark Kimberling, May 10 2011

Keywords

• nonn

Comments

Write a(n) = [(b*n+c)*r] - b*[n*r] - [c*r]. 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,2,0): A078588, A005653, A005652

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

(golden ratio,3,0): A140397 - A190400

(golden ratio,3,1): A140431 - A190435

(golden ratio,3,2): A140436 - A190439

Examples

			a(1)=[3r]-2[r]-1=4-3-1=1.
a(2)=[5r]-2[2r]-1=8-6-1=1.
a(3)=[7r]-2[3r]-1=11-8-1=2.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001622, A078588, A190428, A190429, A190430.

Programs

Magma

[Floor((2*n+1)*(1+Sqrt(5))/2) - 2*Floor(n*(1+Sqrt(5))/2) - 1: n in [1..100]]; // G. C. Greubel, Apr 06 2018

Mathematica

r = GoldenRatio; b = 2; c = 1;
f[n_] := Floor[(b*n + c)*r] - b*Floor[n*r] - Floor[c*r];
t = Table[f[n], {n, 1, 320}] (* A190427 *)
Flatten[Position[t, 0]] (* A190428 *)
Flatten[Position[t, 1]] (* A190429 *)
Flatten[Position[t, 2]] (* A190430 *)
Table[Floor[(2*n+1)*GoldenRatio] - 2*Floor[n*GoldenRatio] -1, {n,1,100}] (* G. C. Greubel, Apr 06 2018 *)

PARI

for(n=1,100, print1(floor((2*n+1)*(1+sqrt(5))/2) - 2*floor(n*(1+sqrt(5))/2) - 1, ", ")) \\ G. C. Greubel, Apr 06 2018

Python

from mpmath import mp, phi
from sympy import floor
mp.dps=100
def a(n): return floor((2*n + 1)*phi) - 2*floor(n*phi) - 1
print([a(n) for n in range(1, 132)]) # Indranil Ghosh, Jul 02 2017

Formula

a(n) = [(2*n+1)*r] - 2*[n*r] - 1, where r=(1+sqrt(5))/2.