

A319433


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


2



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
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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.)


LINKS



FORMULA



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 *)


PROG

(PARI) a(n) = my (f=2, v=0); while (fibonacci(f) < n, f++); while (n > 1, if (n >= fibonacci(f), v += fibonacci(f1); n = fibonacci(f); f); f); return (v) \\ Rémy Sigrist, Oct 04 2018


CROSSREFS



KEYWORD

nonn,base,changed


AUTHOR



EXTENSIONS



STATUS

approved



