A022934 -id:A022934 - cp's OEIS frontend

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

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

A166986 a(n) = 2*floor((n+2)/log(2)) - 4.

Original entry on oeis.org

4, 6, 10, 12, 16, 18, 20, 24, 26, 30, 32, 36, 38, 42, 44, 46, 50, 52, 56, 58, 62, 64, 68, 70, 72, 76, 78, 82, 84, 88, 90, 94, 96, 98, 102, 104, 108, 110, 114, 116, 120, 122, 124, 128, 130, 134, 136, 140, 142, 146, 148, 150, 154, 156, 160, 162, 166, 168, 172, 174, 176

Offset: 1

Stephen Crowley, Oct 26 2009

With a different offset, partial sums of A022934, cf. formula.

The first terms appear to satisfy a linear recurrence relation of order 10 (or higher if more terms are included), but this can be proved to be impossible, cf. R. Israel's post to the SeqFan list. - M. F. Hasler, Apr 11 2019

G. C. Greubel, Table of n, a(n) for n = 1..1000
R. Israel, Re: Differences between linear recurrence, program(s) and b-file, SeqFan list, Apr 11 2019

Cf. A022934.

[2*Floor((n+2)/Log(2)) - 4: n in [1..80]]; // Vincenzo Librandi, Jul 06 2015

seq(2*floor((n+2)/log(2))-4, n=1..100); # Robert Israel, Apr 11 2019

Table[2 Floor[(n + 2)/Log[2]] - 4, {n, 1, 70}] (* Vincenzo Librandi, Jul 06 2015 *)

vector( 80, n, 2*floor((n+2)/log(2)) - 4) \\ Michel Marcus, Jul 06 2015

A166986(n)=(n+2)\log(2)*2-4 \\ M. F. Hasler, Apr 11 2019, corrected by Charles R Greathouse IV, Oct 19 2022

a(n) = 2*floor((n+2)/log(2)) - 4.

a(n) = 2*Sum_{k=2,..,n+1} A022934(k).

A078614 First differences of A072633.

Original entry on oeis.org

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, 2

Offset: 1

Jon Perry, Dec 10 2002

nonn

Differs from A022934 and A081129. - Georg Fischer, May 02 2025

			a(2)=2, as A072633(2)=4 and A072633(1)=2.

Georg Fischer, Table of n, a(n) for n = 1..1000

Cf. A072633.

pu(m,n)=local(s); s=0; for (i=1,m,s=s+i^n); s
ox=1; for (k=1,60,x=1; while (pu(x,k)<(x+1)^k,x++); print1((x-ox), ", "); ox=x)

a(1) corrected by Georg Fischer, May 02 2025

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

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A166986 a(n) = 2*floor((n+2)/log(2)) - 4.

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A078614 First differences of A072633.

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