A214979 A179180 - A214977.

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

Author

Clark Kimberling, Oct 22 2012

Keywords

• nonn

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.

Examples

			(See A214977 and A179180.)
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..10000
• Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.

Crossrefs

Cf. A214977, A179180, A214980, A214981, A000032, A000045.

Programs

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 *)