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

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, 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, 4, 5, 6, 6, 5, 4, 5, 4, 2, 3, 4, 4, 3, 4, 5, 4, 2, 3, 4

Offset: 1

Author

Casey Mongoven, Nov 19 2011

Keywords

• nonn

Examples

			The Wythoff representation of 29 is '10110'. This has 4 equal bit-runs: '1', '0', '11' and '0'. So a(29) = 4.
		

References

• 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.]

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Aviezri S. Fraenkel, From Enmity to Amity, American Mathematical Monthly, Vol. 117, No. 7 (2010) 646-648.
• Clark Kimberling, The Zeckendorf array equals the Wythoff array, Fibonacci Quarterly 33 (1995) 3-8.

Crossrefs

Cf. A135817, A317208.

Programs

Mathematica

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 *)

Extensions

More terms from Amiram Eldar, Jul 01 2023