A189921 - OEIS

%I #35 Jul 01 2023 08:30:05

%S 1,0,1,0,1,1,0,0,0,1,1,1,0,0,1,0,1,0,0,1,1,1,1,0,0,1,1,0,1,0,1,0,1,1,

%T 0,0,0,0,0,1,1,1,1,1,0,0,1,1,1,0,1,0,1,1,0,1,1,0,1,0,0,0,1,0,1,1,1,0,

%U 0,0,1,0,0,1,0,0,0,1,1,1,1,1,1,0,0,1,1,1,1,0,1,0,1,1,1,0,1,1,0,1,1,0

%N Wythoff representation of natural numbers.

%C The row lengths sequence of this array is A135817.

%C For the Wythoff representation of n see the W. Lang reference.

%C The Wythoff complementary sequences are A(n):=A000201(n) and B(n)=A001950(n), n>=1. The Wythoff representation of n=1 is A(1) and for n>=2 there is a unique representation as composition of A- or B-sequence applied to B(1)=2. E.g. n=4 is A(A(B(1))), written as AAB or as `110` or here as 1,1,0, i.e., 1 for A and 0 for B.

%C The Wythoff orbit of 1 (starting always with B(1), applying any number of A- or B-sequences) produces every number n>1 just once. This produces a binary Wythoff code for n>1, ending always in 0 (for B(1)).

%D Wolfdieter Lang, The Wythoff and the Zeckendorf representations of numbers are equivalent, in G. E. Bergum et al. (edts.) Application of Fibonacci numbers vol. 6, Kluwer, Dordrecht, 1996, pp. 319-337. [See A317208 for a link.]

%H Amiram Eldar, <a href="/A189921/b189921.txt">Table of n, a(n) for n = 1..9415</a> (first 1000 rows)

%H Clark Kimberling, <a href="http://www.fq.math.ca/Scanned/33-1/kimberling.pdf">The Zeckendorf array equals the Wythoff array</a>, Fibonacci Quarterly 33 (February, 1995) 3-8.

%H Wolfdieter Lang, <a href="/A135817/a135817.txt"> Wythoff representations for n=1...150. </a>

%F The entries of row n, n>=1, are given by W(n), computed with the algorithm given on p. 335 of the W. Lang reference. 1 is used for Wythoff's A sequence and 0 for the B sequence.

%e n=1:  1;

%e n=2:  0;

%e n=3:  1, 0;

%e n=4:  1, 1, 0;

%e n=5:  0, 0;

%e n=6:  1, 1, 1, 0;

%e n=7:  0, 1, 0;

%e n=8:  1, 0, 0;

%e ...

%e 1 = A(1); 2 = B(1), 3 = A(B(1)), 4 = A(A(B(1))),

%e 5 = B(B(1)), 6 = A(A(A(B(1)))), 7 = B(A(B(1))),

%e 8 = A(B(B(1))), ...

%t z[n_] := Floor[(n + 1)*GoldenRatio] - n - 1; h[n_] := z[n] - z[n - 1]; w[n_] := Module[{m = n, zm = 0, hm, s = {}}, While[zm != 1, hm = h[m]; AppendTo[s, hm]; If[hm == 1, zm = z[m], zm = z[z[m]]]; m = zm]; s]; w[0] = 0; Table[w[n], {n, 1, 25}] // Flatten (* _Amiram Eldar_, Jul 01 2023 *)

%Y See A317208 for another encoding; also for the link to the scanned W. Lang article with corrections.

%Y Cf. A135817 (length), A189920 (Zeckendorf).

%K nonn,easy,tabf,base

%O 1

%A _Wolfdieter Lang_, Jun 12 2011