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%I #5 Dec 04 2016 19:46:31
%S 1,1,2,1,2,1,3,4,4,3,5,5,4,6,6,5,5,7,7,7,6,8,8,8,8,7,7,7,9,9,9,9,9,8,
%T 8,10,10,10,10,10,10,9,9,9,9,9,11,11,11,11,11,11,11,11,10,10,10,12,12,
%U 12,12,12,12,12,12,12,12,11,11,11,11,11,11,11,11
%N Zeckendorf distance between n and floor(n/2).
%C Zeckendorf distance is defined at A226207.
%H Clark Kimberling, <a href="/A226212/b226212.txt">Table of n, a(n) for n = 1..1000</a>
%e 11 = 8 + 3 -> 5 + 2 -> 3 + 1 -> 2, and 5 -> 3 -> 2. The total number of Zeckendorf downshifts (i.e., arrows) is 5, so that a(11) = D(11,5) = 5.
%t zeck[n_Integer] := Block[{k = Ceiling[Log[GoldenRatio, n*Sqrt[5]]], t = n, z = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[z, 1]; t = t - Fibonacci[k], AppendTo[z, 0]]; k--]; If[n > 0 && z[[1]] == 0, Rest[z], z]]; d[n1_, n2_] := Module[{z1 = zeck[n1], z2 = zeck[n2]}, Length[z1] + Length[z2] - 2 (NestWhile[# + 1 &, 1, z1[[#]] == z2[[#]] &, 1,
%t Min[{Length[z1], Length[z2]}]] - 1)]; lst = Map[d[#, Floor[#/2]] &, Range[100]] (* _Peter J. C. Moses_, May 30 2013 *)
%Y Cf. A226080, A226207, A226211.
%K nonn,easy
%O 1,3
%A _Clark Kimberling_, May 31 2013
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