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A200647 Number of equal bit-runs in Wythoff representation of n.

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%I A200647 #30 Jan 05 2025 19:51:39
%S A200647 1,1,2,2,1,2,3,2,2,3,4,2,1,2,3,4,4,3,2,3,2,2,3,4,4,3,4,5,4,2,3,4,2,1,
%T A200647 2,3,4,4,3,4,5,4,4,5,6,4,3,2,3,4,4,3,2,3,2,2,3,4,4,3,4,5,4,4,5,6,4,3,
%U A200647 4,5,6,6,5,4,5,4,2,3,4,4,3,4,5,4,2,3,4
%N A200647 Number of equal bit-runs in Wythoff representation of n.
%D A200647 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 A200647 Amiram Eldar, <a href="/A200647/b200647.txt">Table of n, a(n) for n = 1..10000</a>
%H A200647 Aviezri S. Fraenkel, <a href="http://www.jstor.org/stable/10.4169/000298910x496787">From Enmity to Amity</a>, American Mathematical Monthly, Vol. 117, No. 7 (2010) 646-648.
%H A200647 Clark Kimberling, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/33-1/kimberling.pdf">The Zeckendorf array equals the Wythoff array</a>, Fibonacci Quarterly 33 (1995) 3-8.
%e A200647 The Wythoff representation of 29 is '10110'. This has 4 equal bit-runs: '1', '0', '11' and '0'. So a(29) = 4.
%t A200647 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]; a[n_] := Length[Split[w[n]]]; Array[a, 100] (* _Amiram Eldar_, Jul 01 2023 *)
%Y A200647 Cf. A135817, A317208.
%K A200647 nonn
%O A200647 1,3
%A A200647 _Casey Mongoven_, Nov 19 2011
%E A200647 More terms from _Amiram Eldar_, Jul 01 2023