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 A052005 Number of Fibonacci numbers (A000045) with length n in base 2. 5
 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 (list; graph; refs; listen; history; text; internal format)
 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 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. A052006, A000045, A050815, A036284, A037093, A022927, A022934. Sequence in context: A078614 A026607 A319981 * A138702 A279620 A278109 Adjacent sequences:  A052002 A052003 A052004 * A052006 A052007 A052008 KEYWORD nonn,base AUTHOR Antti Karttunen and Patrick De Geest, Nov 15 1999. STATUS approved

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Last modified September 20 22:20 EDT 2019. Contains 327252 sequences. (Running on oeis4.)