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A214979 A179180 - A214977.

%I A214979 #11 Feb 05 2021 18:18:26
%S A214979 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,
%T A214979 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,
%U A214979 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
%N A214979 A179180 - A214977.
%C A214979 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.
%H A214979 Clark Kimberling, <a href="/A214979/b214979.txt">Table of n, a(n) for n = 1..10000</a>
%H A214979 Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL23/Kimberling/kimber12.html">Lucas Representations of Positive Integers</a>, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
%e A214979 (See A214977 and A179180.)
%t A214979 z = 200;
%t A214979 s = Reverse[Sort[Table[LucasL[n - 1], {n, 1, 70}]]];
%t A214979 t1 = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, s]][[2,1]], # > 0 &]] &, Range[z]];
%t A214979 u[n_] := Sum[t1[[k]], {k, 1, n}]
%t A214979 u1 = Table[u[n], {n, 1, z}] (* A214977 *)
%t A214979 s = Reverse[Table[Fibonacci[n + 1], {n, 1, 70}]];
%t A214979 t2 = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, s]][[2,1]], # > 0 &]] &, Range[z]];
%t A214979 v[n_] := Sum[t2[[k]], {k, 1, n}]
%t A214979 v1 = Table[v[n], {n, 1, z}]  (* A179180 *)
%t A214979 w=v1-u1 (* A214979 *)
%t A214979 Flatten[Position[w, 0]]  (* A214980 *)
%t A214979 (* _Peter J. C. Moses_, Oct 18 2012 *)
%Y A214979 Cf. A214977, A179180, A214980, A214981, A000032, A000045.
%K A214979 nonn
%O A214979 1,12
%A A214979 _Clark Kimberling_, Oct 22 2012