OFFSET
0,1
COMMENTS
Also, positions of 3's in A139764, the smallest term in Zeckendorf representation of n. - John W. Layman, Aug 25 2011
The formula a(n) = 3*A003622(n)-n+1 = 3AA(n)-n+1 conjectured by Layman below is correct, since it is well known that AA(n)+1 = B(n) = A(n)+n, where B = A001950, and so 3AA(n)-n+1 = 3B(n)-n-2 = 3A(n)+2n-2. - Michel Dekking, Aug 31 2017
From Amiram Eldar, Mar 21 2022: (Start)
Numbers k for which the Zeckendorf representation A014417(k) ends with 1, 0, 0.
The asymptotic density of this sequence is 1/phi^4 = 2/(7+3*sqrt(5)), where phi is the golden ratio (A001622). (End)
LINKS
Amiram Eldar, Table of n, a(n) for n = 0..10000
J. H. Conway and N. J. A. Sloane, Notes on the Para-Fibonacci and related sequences.
Clark Kimberling, Complementary equations and Wythoff Sequences, JIS, Vol. 11 (2008), Article 08.3.3.
N. J. A. Sloane, Classic Sequences.
FORMULA
a(n) = F(4)A(n)+F(3)(n-1) = 3A(n)+2n-2, where A = A000201 and F = A000045. - Michel Dekking, Aug 31 2017
It appears that a(n) = 3*A003622(n) - n + 1. - John W. Layman, Aug 25 2011
MAPLE
t := (1+sqrt(5))/2 ; [ seq(3*floor((n+1)*t)+2*n, n=0..80) ];
MATHEMATICA
Table[3 Floor[n GoldenRatio] + 2 n - 2, {n, 46}] (* Michael De Vlieger, Aug 31 2017 *)
PROG
(Python)
from sympy import floor
from mpmath import phi
def a(n): return 3*floor((n + 1)*phi) + 2*n # Indranil Ghosh, Jun 10 2017
(Python)
from math import isqrt
def A035337(n): return 3*(n+isqrt(5*n**2)>>1)+(n-1<<1) # Chai Wah Wu, Aug 11 2022
(PARI) a(n) = 2*n + 3*floor((1+sqrt(5))*(n+1)/2); \\ Altug Alkan, Sep 18 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved