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A052005 Number of Fibonacci numbers (A000045) with length n in base 2.

%I A052005 #26 Nov 21 2021 07:39:42
%S A052005 2,2,1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,2,1,1,2,1,2,1,
%T A052005 2,1,1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,1,2,1,2,1,2,1,
%U A052005 2,1,1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,1,2,1,2,1,2,1,2,1,1,2,1,2,1,2,1,2,1,1
%N A052005 Number of Fibonacci numbers (A000045) with length n in base 2.
%C A052005 There are no double 2's except at the very start because multiplying by phi^3 adds at least 2 to Fn's binary length. For a similar reason there aren't any 3's because multiplying by phi^2 increments at least by one F(n)'s binary length.
%C A052005 Also a(n) is the number of Fibonacci numbers F(k) between powers of 2 such that 2^n <= F(k) < 2^(n+1). - _Frank M Jackson_, Apr 14 2013
%H A052005 T. D. Noe, <a href="/A052005/b052005.txt">Table of n, a(n) for n = 1..1000</a>
%F A052005 Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log(2)/log(phi) = A104287. - _Amiram Eldar_, Nov 21 2021
%e A052005 F(17) = 1597{10} = 11000111101{2} the only one of length 11 and F(18) = 2584{10} = 101000011000{2} the only one of length 12 so both a(11) and a(12) equal 1.
%t A052005 nmax = 105; kmax = Floor[ k /. FindRoot[ Log[2, Fibonacci[k]] == nmax, {k, nmax, 2*nmax}]]; A052005 = Tally[ Length /@ IntegerDigits[ Fibonacci[ Range[kmax]], 2]][[All, 2]] (* _Jean-François Alcover_, May 07 2012 *)
%t A052005 termcount[n1_] := (m1=0; While[Fibonacci[m1]<2^n1, m1++]; m1); Table[termcount[n+1]-termcount[n], {n, 0, 200}] (* _Frank M Jackson_, Apr 14 2013 *)
%t A052005 Most[Transpose[Tally[Table[Length[IntegerDigits[Fibonacci[n], 2]], {n, 140}]]][[2]]] (* _T. D. Noe_, Apr 16 2013 *)
%Y A052005 Cf. A000079, A001622, A052006, A000045, A050815, A036284, A037093, A022927, A022934, A104287.
%K A052005 nonn,base
%O A052005 1,1
%A A052005 _Antti Karttunen_ and _Patrick De Geest_, Nov 15 1999