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A133775 Number of 0's in the minimal "phinary" (A130600) representation of n.

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%I A133775 #18 Apr 21 2023 07:58:11
%S A133775 0,2,3,2,5,5,7,6,5,5,4,8,8,8,7,8,8,11,10,9,9,8,8,8,9,8,7,7,6,11,11,11,
%T A133775 10,11,11,12,11,10,10,9,11,11,11,10,11,11,15,14,13,13,12,12,12,13,12,
%U A133775 11,11,10,11,11,11,10,11,11,13,12,11,11,10,10,10,11,10,9,9,8,14,14,14,13,14,14
%N A133775 Number of 0's in the minimal "phinary" (A130600) representation of n.
%D A133775 Zeckendorf, E., Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège 41, 179-182, 1972.
%H A133775 Casey Mongoven and T. D. Noe, <a href="/A133775/b133775.txt">Table of n, a(n) for n = 1..1000</a>
%H A133775 Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/phigits.html">Using Powers of Phi to represent Integers</a>.
%F A133775 For n > 1, a(n) <= A190796(n) - 2. - _Charles R Greathouse IV_, Apr 21 2023
%e A133775 A130600(5)=10001001, which has five 0's. So a(5)=5.
%t A133775 nn = 100; len = 2*Ceiling[Log[GoldenRatio, nn]]; Table[d = RealDigits[n, GoldenRatio, len]; last1 = Position[d[[1]], 1][[-1, 1]]; Count[Take[d[[1]], last1], 0], {n, 1, nn}] (* _T. D. Noe_, May 20 2011 *)
%Y A133775 Cf. A133776, A130600.
%K A133775 nonn
%O A133775 1,2
%A A133775 _Casey Mongoven_, Sep 23 2007