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

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

2,2

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

Links

Rémy Sigrist, Table of n, a(n) for n = 2..25000

Formula

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

Example

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.

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 *)

Prog

(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

Crossrefs

Cf. A003714, A005614, A014417.

Sequence in context: A347707 A029092 A280814 * A253671 A209081 A248610

Adjacent sequences: A319430 A319431 A319432 * A319434 A319435 A319436

Keyword

nonn,base

Author

N. J. A. Sloane, Sep 30 2018

Extensions

More terms from Rémy Sigrist, Oct 04 2018

Status

approved