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

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, 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, 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

Offset

1,1

Comments

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.

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

Links

T. D. Noe, Table of n, a(n) for n = 1..1000

Formula

Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log(2)/log(phi) = A104287. - Amiram Eldar, Nov 21 2021

Example

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.

Mathematica

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 *)
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 *)
Most[Transpose[Tally[Table[Length[IntegerDigits[Fibonacci[n], 2]], {n, 140}]]][[2]]] (* T. D. Noe, Apr 16 2013 *)

Crossrefs

Cf. A000079, A001622, A052006, A000045, A050815, A036284, A037093, A022927, A022934, A104287.

Sequence in context: A078614 A026607 A319981 * A346707 A138702 A344339

Adjacent sequences: A052002 A052003 A052004 * A052006 A052007 A052008

Keyword

nonn,base

Author

Antti Karttunen and Patrick De Geest, Nov 15 1999

Status

approved