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%I #28 Jul 01 2023 08:30:00
%S 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 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 4,5,6,6,5,4,5,4,2,3,4,4,3,4,5,4,2,3,4
%N Number of equal bit-runs in Wythoff representation of n.
%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="/A200647/b200647.txt">Table of n, a(n) for n = 1..10000</a>
%H 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 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 (1995) 3-8.
%e The Wythoff representation of 29 is '10110'. This has 4 equal bit-runs: '1', '0', '11' and '0'. So a(29) = 4.
%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]; a[n_] := Length[Split[w[n]]]; Array[a, 100] (* _Amiram Eldar_, Jul 01 2023 *)
%Y Cf. A135817, A317208.
%K nonn
%O 1,3
%A _Casey Mongoven_, Nov 19 2011
%E More terms from _Amiram Eldar_, Jul 01 2023
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