A004685 -id:A004685 - cp's OEIS frontend

A222295 Conjectured number of Fibonacci numbers with exactly n bits set in their binary representation.

1, 4, 4, 2, 3, 3, 4, 1, 4, 2, 1, 5, 5, 2, 2, 2, 5, 4, 3, 2, 2, 2, 3, 5, 3, 3, 2, 4, 2, 1, 4, 3, 2, 3, 3, 1, 6, 3, 2, 3, 3, 4, 4, 5, 0, 0, 3, 3, 2, 2, 5, 4, 3, 1, 5, 2, 2, 2, 5, 7, 3, 0, 0, 1, 2, 7, 3, 3, 2, 4, 3, 1, 2, 4, 4, 2, 0, 3, 1, 3, 7, 3, 4, 1, 3, 4, 3

Offset: 0

T. D. Noe, Feb 22 2013

			We set a(1) = 4 because Fib(1) = 1, Fib(2) = 1, Fib(3) = 2, and Fib(6) = 8.

T. D. Noe, Table of n, a(n) for n = 0..1000

Cf. A004685 (Fibonacci numbers in binary), A221158 (two bits set), A222296.
Cf. A011373 (number of bits set in each Fibonacci number).
Cf. A222601, A222602, A222757, A222758.

f = Fibonacci[Range[0,500]]; Table[Length[Select[f, Total[IntegerDigits[#, 2]] == n &]], {n, 0, 87}]

A264663 Catalan numbers written in base 2.

Original entry on oeis.org

1, 1, 10, 101, 1110, 101010, 10000100, 110101101, 10110010110, 1001011111110, 100000110011100, 1110010110100010, 110010110010001100, 10110101010111110100, 1010001100111100001000, 100100111110111001111101, 10000110111000001111100110, 111101110100011100011110110, 11100011110000011000000101100

Offset: 0

Ilya Gutkovskiy, Nov 20 2015

Eric Weisstein's World of Mathematics, Catalan Number
Eric Weisstein's World of Mathematics, Binary

Cf. A000108, A004676, A004685.

[Seqint(Intseq(Catalan(n), 2)): n in [0..20]]; // Vincenzo Librandi, Nov 21 2015

Table[FromDigits[IntegerDigits[CatalanNumber[n], 2]], {n, 0, 18}]

vector(30, n, n--; subst(Pol(binary(binomial(2*n,n)/(n+1))), 'x, 10)) \\ Altug Alkan, Nov 20 2015

a(n) = A007088(A000108(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

A214852 Indices of Fibonacci numbers with the same number of 1's and 0's in their binary representation.

Original entry on oeis.org

3, 36, 42, 59, 116, 156, 168, 211, 237, 246, 280, 335, 355, 399, 404, 416, 433, 442, 569, 580, 652, 698, 761, 770, 865, 897, 940, 989, 1041, 1049, 1101, 1144, 1214, 1286, 1335, 1352, 1369, 1395, 1698, 1726, 1810, 1928, 1940, 1951, 2055, 2159, 2326, 2332

Offset: 1

Alex Ratushnyak, Mar 08 2013

Conjecture: the sequence is infinite.

The sequence of Fibonacci numbers with the same number of 1's and 0's in their binary representation begins: 2, 14930352, 267914296, ... = A259407.

			Fibonacci(36) = 14930352 = 111000111101000110110000_2, twelve 1's and twelve 0's, therefore 36 is in the sequence.

David Radcliffe, Table of n, a(n) for n = 1..10000 (terms 1..400 from T. D. Noe).

Cf. A000045, A004685, A259407.

fQ[n_] := Module[{f = IntegerDigits[Fibonacci[n], 2]}, Count[f, 0] == Count[f, 1]]; Select[Range[3000], fQ] (* T. D. Noe, Mar 08 2013 *)
Position[Fibonacci[Range[2500]],?(DigitCount[#,2,1]==DigitCount[#,2,0]&)]//Flatten (* _Harvey P. Dale, Aug 17 2025 *)

def count10(x):
    c0, c1, m = 0, 0, 1
    while m<=x:
      if x&m:
        c1+=1
      else:
        c0+=1
      m+=m
    return c0-c1
prpr, prev = 0,1
TOP = 3000
for i in range(1,TOP):
    if count10(prev)==0:
        print(i, end=", ")
    prpr, prev = prev, prpr+prev

from sympy import fibonacci
print([n for n in range(3000) if (f := bin(fibonacci(n))[2:]).count('0') == f.count('1')]) # David Radcliffe, May 31 2025

A221158 Fibonacci numbers with two 1's in the binary representation.

Original entry on oeis.org

3, 5, 34, 144

Offset: 1

Alex Ratushnyak, Feb 20 2013

Fibonacci numbers of the form 2^a + 2^b, a>b.
Elkies (2014) proved that there are no other terms.
This sequence is one row of A222296. - T. D. Noe, Mar 08 2013

			144 = 128 + 16 = 2^7 + 2^4, thus it is in the sequence.

Noam D. Elkies, Fibonacci numbers with Hamming weight 2, Mathoverflow, 2014.

Cf. A000045, A004685, A222296.

Select[Fibonacci[Range[1000]], DigitCount[#, 2, 1] == 2 &] (* Alonso del Arte, Feb 21 2013 *)

from sympy import fibonacci
print([f for n in range(100) if (f := int(fibonacci(n))).bit_count() == 2]) # David Radcliffe, Jul 03 2025

full, fini keywords added by Max Alekseyev, May 13 2014

A196024 Odious Fibonacci numbers.

Original entry on oeis.org

1, 2, 8, 13, 21, 55, 233, 1597, 4181, 28657, 121393, 196418, 317811, 1346269, 2178309, 3524578, 9227465, 165580141, 1134903170, 1836311903, 2971215073, 20365011074, 32951280099, 53316291173, 225851433717, 2504730781961, 6557470319842, 17167680177565, 27777890035288

Offset: 1

Kausthub Gudipati, Sep 27 2011

Intersection of A000069 (odious numbers) and A000045 (Fibonacci numbers).
The k-th Fibonacci number is odious for k = 1, 2, 3, 6, 7, 10, 13, 17, 19, 23, 26, 27, 28, 30, 31, 32, 33, 35, 41, 45, ...
The k-th odious number is a Fibonacci number for k = 1, 2, 5, 7, 11, 28, 117, 799, 2091, ...

			8 is a Fibonacci number that equals 1000 in binary, which contains one (odd number) 1.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000045, A000120, A007088, A004685.

isA000069 := proc(n)
        type(wt(n),'odd') ;
end proc:
for n from 1 to 300 do
        F := combinat[fibonacci](n) ;
        if isA000069(F) then printf("%d,",F) ; end if;
end do: # R. J. Mathar, Oct 15 2011

Rest[Select[Fibonacci[Range[100]],OddQ[DigitCount[#,2,1]]&]] (* Harvey P. Dale, Oct 16 2011 *)

A214326 Square array read by antidiagonals in which T(n,b) gives the n-th Fibonacci number written in base b with n,b >= 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 11, 1, 1, 10, 111, 1, 1, 2, 11, 11111, 1, 1, 2, 10, 101, 11111111, 1, 1, 2, 3, 12, 1000, 1111111111111, 1, 1, 2, 3, 11, 22, 1101, 111111111111111111111, 1, 1, 2, 3, 10, 20, 111, 10101, 1111111111111111111111111111111111, 1, 1, 2, 3, 5, 13, 31, 210, 100010

Offset: 1

Alois P. Heinz, Jul 24 2012

For b > 10, some terms cannot be properly notated using only decimal characters.

			Square array A(n,b) begins:
              1,    1,   1,  1,  1,  1,  1,  1,  1,  1,  1,  1, ...
              1,    1,   1,  1,  1,  1,  1,  1,  1,  1,  1,  1, ...
             11,   10,   2,  2,  2,  2,  2,  2,  2,  2,  2,  2, ...
            111,   11,  10,  3,  3,  3,  3,  3,  3,  3,  3,  3, ...
          11111,  101,  12, 11, 10,  5,  5,  5,  5,  5,  5,  5, ...
       11111111, 1000,  22, 20, 13, 12, 11, 10,  8,  8,  8,  8, ...
  1111111111111, 1101, 111, 31, 23, 21, 16, 15, 14, 13, 12, 11, ...

Alois P. Heinz, Antidiagonals n = 1..13

Columns b=1-10, 12, 13 give: A108047, A004685, A004686, A004687, A004688, A004689, A004690, A004691, A004692, A000045, A004693, A004694.

Cf. A037842, A131293.

A:= proc(n, b) local f, l; f:= combinat[fibonacci](n);
      if b=1 then parse(cat(1$f))
    else l:= NULL;
         while f>0 do l:= irem(f, b, 'f'), l od;
         parse(cat(l))
      fi
    end:
seq(seq(A(n, 1+d-n), n=1..d), d=1..10);

A214853 Fibonacci numbers with only one 0 in the binary representation.

Original entry on oeis.org

0, 2, 5, 13, 55

Offset: 1

Alex Ratushnyak, Mar 08 2013

Conjecture: the sequence is finite.
No more terms below 2*10^301. - Matthew House, Sep 06 2015
No more terms below 10^162809483. (This number could easily be raised. Of the Fibonacci numbers less than 2^32 -- i.e., F(0) through F(47) -- F(10)=55 is the largest that has only one 0 in its binary representation, and of those not less than 2^32, the smallest one whose 32 least significant bits include fewer than 2 zero bits is Fibonacci(779038816), which exceeds 10^162809483.) - Jon E. Schoenfield, Sep 07 2015

			55 is 110111 in binary, thus 55 is in the sequence.

Cf. A004685, A221158.

Intersection of A030130 and A000045.

Select[Fibonacci@ Range[0, 120], Last@ DigitCount[#, 2] == 1 &] (* Michael De Vlieger, Sep 07 2015 *)

def count0(x):
    c = 0
    while x:
        c+= 1 - (x&1)
        if c>1:
            return 2
        x>>=1
    return c
prpr, prev = 0,1
TOP = 1<<12
print(0, end=',')
for i in range(1,TOP):
    if count0(prpr)==1:
        print(prpr, end=',')
    if (i&4095)==0:
        print('.', end=',')
    prpr, prev = prev, prpr+prev

A287014 Bell numbers written in base 2.

Original entry on oeis.org

1, 1, 10, 101, 1111, 110100, 11001011, 1101101101, 1000000101100, 101001010011011, 11100010100000111, 10100101101010101010, 10000000100101101011101, 1101001011101001000010101, 1011011000001110010001111010, 1010010011011100100010111010001

Offset: 0

Vincenzo Librandi, May 22 2017

Robert Israel, Table of n, a(n) for n = 0..214

Cf. A000110, A004685, A007088,

[Seqint(Intseq(Bell(n), 2)): n in [0..20]];

seq(convert(combinat:-bell(n),binary),n=0..20); # Robert Israel, Aug 12 2018

Table[FromDigits[IntegerDigits[BellB[n], 2]], {n, 0, 30}]

a(n) = A007088(A000110(n)).

A381704 Fibonacci numbers having a Fibonacci number of 1's in their binary representation.

Original entry on oeis.org

0, 1, 2, 3, 5, 8, 13, 21, 34, 55, 144, 233, 987, 4181, 6765, 17711, 832040, 3524578, 1836311903, 2971215073, 225851433717, 259695496911122585, 3928413764606871165730, 26925748508234281076009, 9969216677189303386214405760200, 638817435613190341905763972389505493

Offset: 1

Ctibor O. Zizka, Mar 04 2025

			F(10) = (55)_10 = (110111)_2 has five 1's in binary, 5 is a Fibonacci number, thus 55 is a term.
F(12) = (144)_10 = (10010000)_2 has two 1's in binary, 2 is a Fibonacci number, thus 144 is a term.

Robert Israel, Table of n, a(n) for n = 1..49

Cf. A000045, A004685, A010056, A382053.

isfib:= n -> issqr(5*n^2+4) or issqr(5*n^2-4):
select(n -> isfib(convert(convert(n,base,2),`+`)), map(combinat:-fibonacci,[0,$2..1000])); # Robert Israel, Mar 13 2025

With[{f = Fibonacci[Range[0, 200]]}, DeleteDuplicates[Select[f, MemberQ[f, DigitCount[#, 2, 1]] &]]] (* Amiram Eldar, Mar 04 2025 *)

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8)); \\ A010056
lista(nn) = for (n=2, nn, my(f = fibonacci(n)); if (isfib(hammingweight(f)), print1(f, ", "));); \\ Michel Marcus, Mar 04 2025

a(1) = 0 inserted by Robert Israel, Mar 13 2025

cp's OEIS Frontend

A222295 Conjectured number of Fibonacci numbers with exactly n bits set in their binary representation.

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A264663 Catalan numbers written in base 2.

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Magma

Mathematica

PARI

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

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A214852 Indices of Fibonacci numbers with the same number of 1's and 0's in their binary representation.

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Mathematica

Python

Python

A221158 Fibonacci numbers with two 1's in the binary representation.

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Mathematica

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A196024 Odious Fibonacci numbers.

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Maple

Mathematica

A214326 Square array read by antidiagonals in which T(n,b) gives the n-th Fibonacci number written in base b with n,b >= 1.

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Maple

A214853 Fibonacci numbers with only one 0 in the binary representation.

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