A319433 Take Zeckendorf representation of n (A014417(n)), drop least significant bit, take inverse Zeckendorf representation.

0, 0, 1, 2, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 8, 9, 10, 10, 11, 11, 12, 13, 13, 14, 15, 15, 16, 16, 17, 18, 18, 19, 20, 20, 21, 21, 22, 23, 23, 24, 24, 25, 26, 26, 27, 28, 28, 29, 29, 30, 31, 31, 32, 32, 33, 34, 34, 35, 36, 36, 37, 37, 38, 39, 39, 40, 41, 41, 42, 42, 43

Offset: 0

Author

N. J. A. Sloane, Sep 30 2018

Keywords

• nonn
• base

Comments

In other words, the Zeckendorf representation of a(n) is obtained from the Zeckendorf representation of n by deleting the least significant bit.

Theorem: The first differences (1,1,0,1,0,1,1,0,...) form the Fibonacci word A005614. (The proof is straightforward.)

The difference sequence agrees with A005614 after the first two terms of A005614. - Clark Kimberling, Dec 29 2020

Examples

			n = 19 has Zeckendorf representation [1, 0, 1, 0, 0, 1], dropping last bit we get [1, 0, 1, 0, 0], which is the Zeckendorf representation of 11, so a(19) = 11.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 2..25000
• Johan Kok, Integer sequences with conjectured relation with certain graph parameters of the family of linear Jaco graphs, arXiv:2507.16500 [math.CO], 2025. See pp. 4-5.
• Jeffrey Shallit, The Hurt-Sada Array and Zeckendorf Representations, arXiv:2501.08823 [math.NT], 2025. See pp. 10-11.

Crossrefs

Cf. A003714, A005614, A014417.

Programs

Mathematica

r = (1 + Sqrt[5])/2; t = Table[Floor[(n - 1)/r] + 2, {n, 0, 150}] (* A319433 *)
Differences[t]  (* A005614 after the 1st 2 terms *)
(* Clark Kimberling, Dec 29 2020 *)

PARI

a(n) = my (f=2, v=0); while (fibonacci(f) < n, f++); while (n > 1, if (n >= fibonacci(f), v += fibonacci(f-1); n -= fibonacci(f); f--); f--); return (v) \\ Rémy Sigrist, Oct 04 2018

Formula

a(n) = floor((n+2)/tau)-1, where tau = golden ratio. - Clark Kimberling, Dec 29 2020; corrected by Harry Altman, Jun 06 2024

Extensions

More terms from Rémy Sigrist, Oct 04 2018