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A214979
5
0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 2, 1, 1, 1, 0, 0, 1, 2, 3, 2, 2, 1, 0, 1, 0, 0, 0, 1, 2, 3, 4, 6, 4, 3, 3, 2, 2, 1, 2, 2, 1, 1, 1, 0, 0, 1, 2, 3, 4, 6, 6, 7, 9, 7, 6, 5, 5, 5, 4, 4, 4, 2, 1, 2, 2, 3, 2, 2, 1, 0, 1, 0, 0, 0, 1, 2, 3, 4, 6, 6, 7, 9, 9, 10, 11, 13, 15, 13, 12, 11, 9, 8, 8, 8, 8
OFFSET
1,12
COMMENTS
Let U(n) and V(n) be the number of terms in the Lucas representations and Zeckendorf (Fibonacci) representations, respectively, of all the numbers 1,2,...,n. Then a(n) = V(n) - U(n). Conjecture: a(n) >= 0 for all n, and a(n) = 0 for infinitely many n.
LINKS
Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
EXAMPLE
(See A214977 and A179180.)
MATHEMATICA
z = 200;
s = Reverse[Sort[Table[LucasL[n - 1], {n, 1, 70}]]];
t1 = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, s]][[2, 1]], # > 0 &]] &, Range[z]];
u[n_] := Sum[t1[[k]], {k, 1, n}]
u1 = Table[u[n], {n, 1, z}] (* A214977 *)
s = Reverse[Table[Fibonacci[n + 1], {n, 1, 70}]];
t2 = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, s]][[2, 1]], # > 0 &]] &, Range[z]];
v[n_] := Sum[t2[[k]], {k, 1, n}]
v1 = Table[v[n], {n, 1, z}] (* A179180 *)
w=v1-u1 (* A214979 *)
Flatten[Position[w, 0]] (* A214980 *)
(* Peter J. C. Moses, Oct 18 2012 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Oct 22 2012
STATUS
approved