A052006 -id:A052006 - cp's OEIS frontend

A051392 First differences of A052006.

13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 10, 13, 13, 13, 10, 13, 13, 10

Offset: 1

Antti Karttunen and Patrick De Geest, Nov 15 1999

Does this sequence only contain 10's and 13's?

Yes. Since all blocks of terms of A052005 are of the form 1(12)n, a(n) must be congruent to 1 modulo 3. [1, 2] blocks give an average growth rate of 3/2 powers of phi for every power of two, but since phi^3 > 4, singleton 1's are required to slow growth when errors get too large. Since singleton 1's reduce the growth rate by 1/2 power of phi per power of two, they should occur roughly once every (1/2) / log_2(phi^1.5 / 2) ~ 12.088 powers of phi. Therefore, a(n) will be 13 most of the time, with 10 occurring when needed to maintain this ratio. - Charlie Neder, Oct 24 2018

Vincenzo Librandi, Table of n, a(n) for n = 1..1650

With[{F = Fibonacci}, Reap[For[n = 0, n < 1000, n++, If[F[n - 1] < 2^Floor[Log[2, F[n]]] && F[n + 1] >= 2^(Floor[Log[2, F[n]]] + 1) && F[n + 2] >= 2^(Floor[Log[2, F[n]]] + 2), Sow[n]]]][[2, 1]]] // Differences (* Jean-François Alcover, Feb 27 2016 *)

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

Original entry on oeis.org

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

Antti Karttunen and Patrick De Geest, Nov 15 1999

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

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

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

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 *)

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

cp's OEIS Frontend

A051392 First differences of A052006.

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