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 A214979 A179180 - A214977. 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 (list; graph; refs; listen; history; text; internal format)
 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, 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. 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 Cf. A214977, A179180, A214980, A214981, A000032, A000045. Sequence in context: A316675 A111405 A089053 * A068462 A054973 A030351 Adjacent sequences:  A214976 A214977 A214978 * A214980 A214981 A214982 KEYWORD nonn AUTHOR Clark Kimberling, Oct 22 2012 STATUS approved

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Last modified May 15 17:50 EDT 2021. Contains 343920 sequences. (Running on oeis4.)