A036284 -id:A036284 - 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

A136381 Sequence A136380 shown in octal base.

Original entry on oeis.org

30, 240, 27300, 275332400, 27624273321353000, 277524424264553332245513535524000, 27762724550512424245125524562733322130552452655353526564552130000, 277751305605652455261312526532424241366545132655245452272135533332224427213254552451226545102753535225125262712455250570562640000

Offset: 1

Antti Karttunen, Dec 29 2007

A. Karttunen, Table of n, a(n) for n = 1..10
A. Karttunen, C program for computing this sequence

Cf. A036285, A136383 (shifted two bits right), A136385.

a(n) = A007094(A136380(n)).

A136383 Sequence A136382 shown in octal base.

Original entry on oeis.org

6, 50, 5660, 57266500, 5745056664272600, 57725105055132666451322727325000, 5774565132122505051225325134566664426132512553272725535132426000, 57772261341352513254262525526505050275531226553251312456427326666445105642653132512245531220572727245225254562513252136134550000

Offset: 1

Antti Karttunen, Dec 29 2007

A. Karttunen, Table of n, a(n) for n = 1..11
A. Karttunen, C program for computing this sequence

Cf. A036285, A136381 (shifted two bits left), A136385.

a(n) = A007094(A136382(n)).

A136385 Sequence A136384 shown in octal base.

Original entry on oeis.org

4, 60, 6440, 65444600, 6506064447454400, 65231606066154444716234545546000, 6527131623634606061431546150644447033623163115454546626154744000, 65254336276263162310334631144606060326621433115461563064745544444706171303315623163471156340654545471431467134615463624150660000

Offset: 1

Antti Karttunen, Dec 29 2007

A. Karttunen, Table of n, a(n) for n = 1..10
A. Karttunen, C program for computing this sequence

Cf. A036285, A136381, A136383.

a(n) = A007094(A136384(n)).

A036286 Periodic vertical binary vectors of Fibonacci numbers, topmost bits being most significant.

Original entry on oeis.org

3, 6, 90, 202474, 802914372650, 124876754670311211270396330, 2261740218128437766312179308277308483058208661638110890, 7527129205899945471753233641719262207829849606092782843679109711117799287001392666047916596823438974998183293610

Offset: 0

Antti Karttunen, Nov 01 1998

			When Fibonacci numbers are written in binary (see A004685), under each other as:
0000000 (0)
0000001 (1)
0000001 (1)
0000010 (2)
0000011 (3)
0000101 (5)
0001000 (8)
0001101 (13)
0010101 (21)
0100010 (34)
0110111 (55)
1011001 (89)
it can be seen that the bits in the n-th column from right repeat after the period of A007283(n): 3, 6, 12, 24, ... (see also A001175). This sequence is formed from those bits: 011, binary for 3, thus a(0) = 3. 000110, binary for 6, thus a(1) = 6, 000001011010, binary for 90, thus a(2) = 90. Cf. A036284.

A. Karttunen, Table of n, a(n) for n = 0..10
A. Karttunen, C program for computing this sequence

See comments at A036284. a(n)/A036287(n) can be interpreted as fractions.

a(n) = Sum_{k=0..A007283(n)-1} ([A000045((A007283(n)-1)-k)/(2^n)] mod 2) * 2^k, where [] stands for floor function.

Entry revised Dec 29 2007

A052006 Numbers k for which Fibonacci(k) is the first member of a 1,1 pair (A052005).

Original entry on oeis.org

17, 30, 43, 53, 66, 79, 89, 102, 115, 125, 138, 151, 161, 174, 187, 200, 210, 223, 236, 246, 259, 272, 282, 295, 308, 321, 331, 344, 357, 367, 380, 393, 403, 416, 429, 442, 452, 465, 478, 488, 501, 514, 524, 537, 550, 560, 573, 586, 599, 609, 622, 635, 645

Offset: 0

Antti Karttunen and Patrick De Geest, Nov 15 1999

Keep adding the terms of sequence A052005 up to the first member of the next 1,1 pair to yield the terms of this sequence. - Patrick De Geest

Those k for which F(k-1) < 2^(floor(log_2(F(k)))) and F(k+1) >= 2^(floor(log_2(F(k)))+1) and F(k+2) >= 2^(floor(log_2(F(k)))+2).

Vincenzo Librandi, Table of n, a(n) for n = 0..1650

The first differences are A051392.

Cf. A052005, A000045, A050815, A036284, A037093.

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), Print[n]; Sow[n]]]][[2, 1]]] (* Jean-François Alcover, Feb 27 2016 *)

A036287 a(n) = ((2^(3*(2^n))) - 1).

Original entry on oeis.org

7, 63, 4095, 16777215, 281474976710655, 79228162514264337593543950335, 6277101735386680763835789423207666416102355444464034512895, 39402006196394479212279040100143613805079739270465446667948293404245721771497210611414266254884915640806627990306815

Offset: 0

Antti Karttunen

Also "Denominators for Fibonacci Binary Verticals viewed as Periodic Binary Fractions": The cycle of bit-n of Fibonacci numbers in binary is (3*(2^n)). Looking from top to bottom they can be viewed as non-finite periodic binary fractions, with each fraction computed as the n-th element of A036286 divided by the n-th element of A036287.

Antti Karttunen, C program related to this sequence

Cf. A036286, A000045.

2^(3 2^Range[0,10])-1 (* Harvey P. Dale, Jun 19 2011 *)

A036287(n) = ((2^(3*(2^n))) - 1); \\ Antti Karttunen, Jan 14 2024

A136387 Sequence A136386 written in base 2.

Original entry on oeis.org

100, 1000, 101100000, 111010010000, 110110001010011100100000000100000, 101010010000001101001101111001000100001000011000101100000000, 100100101011010111000001100000001000110011001010001110000001111101101000000000100010010000010000010100010010010100000010011000000

Offset: 3

Antti Karttunen, Dec 29 2007

A. Karttunen, Table of n, a(n) for n = 3..11
A. Karttunen, C program for computing this sequence

a(n) = A007088(A136386(n)).

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A136381 Sequence A136380 shown in octal base.

Views

Author

Keywords

Links

Crossrefs

Formula

A136383 Sequence A136382 shown in octal base.

Views

Author

Keywords

Links

Crossrefs

Formula

A136385 Sequence A136384 shown in octal base.

Views

Author

Keywords

Links

Crossrefs

Formula

A036286 Periodic vertical binary vectors of Fibonacci numbers, topmost bits being most significant.

Views

Author

Keywords

Examples

Links

Crossrefs

Formula

Extensions

A052006 Numbers k for which Fibonacci(k) is the first member of a 1,1 pair (A052005).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

A036287 a(n) = ((2^(3*(2^n))) - 1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

A136387 Sequence A136386 written in base 2.

Views

Author

Keywords

Links

Formula