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