A226214 Zeckendorf distance between n and n^2.
0, 1, 2, 5, 3, 6, 4, 6, 10, 9, 11, 12, 9, 11, 13, 12, 8, 6, 10, 15, 12, 14, 16, 16, 13, 15, 15, 11, 17, 11, 13, 18, 18, 15, 17, 17, 19, 19, 19, 16, 16, 18, 18, 18, 14, 14, 8, 12, 14, 16, 21, 21, 21, 21, 18, 18, 20, 20, 20, 22, 22, 22, 22, 22, 19, 19, 19, 21
Offset: 1
Examples
7 = 5 + 2, and 7^2 = 34 + 13 + 2 -> 21 + 8 + 1 -> 13 + 5 -> 8 + 3 -> 5 + 2. The total number of Zeckendorf downshifts (i.e., arrows) is 4, so that a(7) = D(7,49) = 4.
Links
- Clark Kimberling, Table of n, a(n) for n = 1..1000
Programs
-
Mathematica
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, Min[{Length[z1], Length[z2]}]] - 1)]; lst = Map[d[#, #^2] &, Range[100]] (* Peter J. C. Moses, May 30 2013 *)
Comments