A226215 Zeckendorf distance between n! and (n+1)!.
1, 2, 5, 11, 18, 20, 36, 45, 49, 53, 69, 83, 94, 105, 116, 122, 122, 146, 164, 178, 191, 204, 217, 229, 244, 253, 265, 263, 293, 309, 328, 339, 357, 372, 385, 400, 415, 430, 447, 462, 476, 490, 504, 516, 541, 536, 573, 580, 600, 618, 636, 654, 671, 686, 704
Offset: 1
Examples
4! = 21 + 3 -> 13 + 2 -> 8 + 1 -> 5 -> 3, and 5! = 89 + 21 + 8 + 2 -> 55 + 13 + 5 + 1 -> 34 + 8 + 3 -> 21 + 5 + 2 -> 13 + 3 + 1 -> 8 + 2 -> 5 + 1 -> 3. The total number of Zeckendorf downshifts (i.e., arrows) is 4 + 7, so that a(4) = D(24,120) = 11.
Links
- Clark Kimberling, Table of n, a(n) for n = 1..200
Programs
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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[#!, (#+1)!] &, Range[100]] (* Peter J. C. Moses, May 30 2013 *)
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