A116543 Number of terms in greedy representation of n in terms of the Lucas numbers.

0, 1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 3, 3, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 4, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 4, 2, 3, 3, 3, 3, 4, 4, 3, 4, 4, 4, 1, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 4, 2, 3, 3, 3, 3, 4, 4, 3, 4, 4, 4, 2

Offset: 0

Author

James E Davis, Mar 28 2006, Jun 07 2006

Keywords

• nonn
• base

Comments

I have been studying A007895 and similar sequences and created this sequence as an analog of A007895 for the Lucas sequence (A000032).

Examples

			a(12)=2 because 12=11+1.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..10000
• Ron Knott, Using the Fibonacci numbers to represent whole numbers.

Crossrefs

Cf. A000032, A007895, A007953, A130310, A131343.

Programs

Mathematica

s = Reverse[Sort[Table[LucasL[n - 1], {n, 1, 22}]]];
t = Map[Length[Select[Reap[FoldList[(Sow[Quotient[#1, #2]]; Mod[#1, #2]) &, #, s]][[2,1]], # > 0 &]] &, Range[1000]] (* Peter J. C. Moses, Oct 18 2012 *)

Formula

Let L(n) = max(Lucas numbers < n). Then a(0) = 0, a(n) = 1 + a(n-L(n)).

a(n) = A007953(A130310(n)). - Amiram Eldar, Feb 17 2022

Extensions

Edited by N. J. A. Sloane, Aug 10 2007

a(0) added by Amiram Eldar, Feb 17 2022