|
%I #27 Dec 30 2020 01:41:52
%S 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,
%T 18,18,19,20,20,21,21,22,23,23,24,24,25,26,26,27,28,28,29,29,30,31,31,
%U 32,32,33,34,34,35,36,36,37,37,38,39,39,40,41,41,42,42,43
%N Take Zeckendorf representation of n (A014417(n)), drop least significant bit, take inverse Zeckendorf representation.
%C In other words, the Zeckendorf representation of a(n) is obtained from the Zeckendorf representation of n by deleting the least significant bit.
%C Theorem: The first differences (1,1,0,1,0,1,1,0,...) form the Fibonacci word A005614. (The proof is straightforward.)
%C The difference sequence agrees with A005614 after the first two terms of A005614. - _Clark Kimberling_, Dec 29 2020
%H Rémy Sigrist, <a href="/A319433/b319433.txt">Table of n, a(n) for n = 2..25000</a>
%F a(n) = floor((n-1)/tau)+2, where tau = golden ratio. - _Clark Kimberling_, Dec 29 2020
%e 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.
%t r = (1 + Sqrt[5])/2; t = Table[Floor[(n - 1)/r] + 2, {n, 0, 150}] (* A319433 *)
%t Differences[t] (* A005614 after the 1st 2 terms *)
%t (* _Clark Kimberling_, Dec 29 2020 *)
%o (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
%Y Cf. A003714, A005614, A014417.
%K nonn
%O 2,2
%A _N. J. A. Sloane_, Sep 30 2018
%E More terms from _Rémy Sigrist_, Oct 04 2018
|
