A214974 Numbers k for which A116543(k) = A007895(k); i.e., the Lucas and Zeckendorf representations of k have the same length.
1, 2, 3, 6, 9, 10, 14, 15, 17, 22, 27, 28, 36, 38, 41, 43, 44, 46, 52, 58, 59, 61, 62, 66, 69, 74, 75, 81, 84, 94, 95, 96, 98, 107, 112, 114, 117, 119, 120, 122, 128, 131, 136, 139, 148, 152, 153, 154, 155, 159, 161, 164, 173, 175, 176, 181, 182, 184, 185
Offset: 1
Keywords
Examples
k...Lucas.....Zeckendorf....counter 1...1.........1.............a(1)= 1 2...2.........2.............a(2)= 2 3...3.........3.............a(3)= 3 4...4.........3+1 5...4+1.......5 6...4+2.......5+1...........a(4)= 6 7...7.........5+2 8...7+1.......8 9...7+2.......8+1...........a(5)= 9
Links
- Clark Kimberling, Table of n, a(n) for n = 1..8000
Programs
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Mathematica
u = Reverse[Sort[Table[LucasL[n - 1], {n, 1, 50}]]]; u1 = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, u]][[2,1]], # > 0 &]] &, Range[1000]]; v = Reverse[Table[Fibonacci[n + 1], {n, 1, 50}]]; v1 = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, v]][[2,1]], # > 0 &]] &, Range[1000]]; s[n_] := If[u1[[n]] == v1[[n]], 1, 0]; s1 = Table[s[n], {n, 1, 200}]; f1 = Flatten[Position[s1, 1]] (* A214974 *) s[n_] := If[u1[[n]] < v1[[n]], 1, 0]; s2 = Table[s[n], {n, 1, 200}]; f2 = Flatten[Position[s2, 1]] (* A214975 *) s[n_] := If[u1[[n]] > v1[[n]], 1, 0]; s3 = Table[s[n], {n, 1, 200}]; f3 = Flatten[Position[s3, 1]] (* A214976 *) (* Peter J. C. Moses *)