A003714 -id:A003714 - cp's OEIS frontend

A227349 Product of lengths of runs of 1-bits in binary representation of n.

1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2, 4, 3, 3, 4, 5, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 2, 2, 2, 4, 2, 2, 4, 6, 3, 3, 3, 6, 4, 4, 5, 6, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2, 4, 3, 3, 4, 5, 2, 2, 2, 4, 2, 2, 4, 6, 2, 2, 2, 4, 4, 4, 6, 8, 3, 3, 3, 6, 3, 3, 6, 9, 4

Offset: 0

Antti Karttunen, Jul 08 2013

This is the Run Length Transform of S(n) = {0, 1, 2, 3, 4, 5, 6, ...}. The Run Length Transform of a sequence {S(n), n >= 0} is defined to be the sequence {T(n), n >= 0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g., 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0) = 1 (the empty product). - N. J. A. Sloane, Sep 05 2014

Like all run length transforms also this sequence satisfies for all i, j: A278222(i) = A278222(j) => a(i) = a(j). - Antti Karttunen, Apr 14 2017

			a(0) = 1, as zero has no runs of 1's, and an empty product is 1.
a(1) = 1, as 1 is "1" in binary, and the length of that only 1-run is 1.
a(2) = 1, as 2 is "10" in binary, and again there is only one run of 1-bits, of length 1.
a(3) = 2, as 3 is "11" in binary, and there is one run of two 1-bits.
a(55) = 6, as 55 is "110111" in binary, and 2 * 3 = 6.
a(119) = 9, as 119 is "1110111" in binary, and 3 * 3 = 9.
From _Omar E. Pol_, Feb 10 2015: (Start)
Written as an irregular triangle in which row lengths is A011782:
  1;
  1;
  1,2;
  1,1,2,3;
  1,1,1,2,2,2,3,4;
  1,1,1,2,1,1,2,3,2,2,2,4,3,3,4,5;
  1,1,1,2,1,1,2,3,1,1,1,2,2,2,3,4,2,2,2,4,2,2,4,6,3,3,3,6,4,4,5,6;
  ...
Right border gives A028310: 1 together with the positive integers. (End)
From _Omar E. Pol_, Mar 19 2015: (Start)
Also, the sequence can be written as an irregular tetrahedron T(s, r, k) as shown below:
  1;
  ..
  1;
  ..
  1;
  2;
  ....
  1,1;
  2;
  3;
  ........
  1,1,1,2;
  2,2;
  3;
  4;
  ................
  1,1,1,2,1,1,2,3;
  2,2,2,4;
  3,3;
  4;
  5;
  ................................
  1,1,1,2,1,1,2,3,1,1,1,2,2,2,3,4;
  2,2,2,4,2,2,4,6;
  3,3,3,6;
  4,4;
  5;
  6;
  ...
Apart from the initial 1, we have that T(s, r, k) = T(s+1, r, k). (End)

Antti Karttunen, Table of n, a(n) for n = 0..8192
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Index entries for sequences computed with run length transform

Cf. A003714 (positions of ones), A005361, A005940.
Cf. A000120 (sum of lengths of runs of 1-bits), A167489, A227350, A227193, A278222, A245562, A284562, A284569, A283972, A284582, A284583.
Run Length Transforms of other sequences: A246588, A246595, A246596, A246660, A246661, A246674.
Differs from similar A284580 for the first time at n=119, where a(119) = 9, while A284580(119) = 5.

a:= proc(n) local i, m, r; m, r:= n, 1;
      while m>0 do
        while irem(m, 2, 'h')=0 do m:=h od;
        for i from 0 while irem(m, 2, 'h')=1 do m:=h od;
        r:= r*i
      od; r
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Jul 11 2013
ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
   if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
   elif out1 = 0 and t1[i] = 1 then c:=c+1;
   elif out1 = 1 and t1[i] = 0 then c:=c;
   elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
   fi;
   if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
                   od:
a:=mul(i, i in lis);
ans:=[op(ans), a];
od:
ans;  # N. J. A. Sloane, Sep 05 2014

onBitRunLenProd[n_] := Times @@ Length /@ Select[Split @ IntegerDigits[n, 2], #[[1]] == 1 & ]; Array[onBitRunLenProd, 100, 0] (* Jean-François Alcover, Mar 02 2016 *)

from operator import mul
from functools import reduce
from re import split
def A227349(n):
    return reduce(mul, (len(d) for d in split('0+',bin(n)[2:]) if d)) if n > 0 else 1 # Chai Wah Wu, Sep 07 2014

# uses[RLT from A246660]
A227349_list = lambda len: RLT(lambda n: n, len)
A227349_list(88) # Peter Luschny, Sep 07 2014

(define (A227349 n) (apply * (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
(define (bisect lista parity) (let loop ((lista lista) (i 0) (z (list))) (cond ((null? lista) (reverse! z)) ((eq? i parity) (loop (cdr lista) (modulo (1+ i) 2) (cons (car lista) z))) (else (loop (cdr lista) (modulo (1+ i) 2) z)))))
(define (binexp->runcount1list n) (if (zero? n) (list) (let loop ((n n) (rc (list)) (count 0) (prev-bit (modulo n 2))) (if (zero? n) (cons count rc) (if (eq? (modulo n 2) prev-bit) (loop (floor->exact (/ n 2)) rc (1+ count) (modulo n 2)) (loop (floor->exact (/ n 2)) (cons count rc) 1 (modulo n 2)))))))

A167489(n) = a(n) * A227350(n).
A227193(n) = a(n) - A227350(n).
a(n) = Product_{i in row n of table A245562} i. - N. J. A. Sloane, Aug 10 2014
From Antti Karttunen, Apr 14 2017: (Start)
a(n) = A005361(A005940(1+n)).
a(n) = A284562(n) * A284569(n).
A283972(n) = n - a(n).
(End)
a(4n+1) = a(2n) = a(n). If n is odd, then a(4n+3) = 2*a(2n+1)-a(n). If n is even, then a(4n+3) = 2*a(2n+1) = 2*a(n/2). - Chai Wah Wu, Jul 17 2025

Data section extended up to term a(120) by Antti Karttunen, Apr 14 2017

A003726 Numbers with no 3 adjacent 1's in binary expansion.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 16, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 32, 33, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 45, 48, 49, 50, 51, 52, 53, 54, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 77, 80, 81, 82

Offset: 1

N. J. A. Sloane

Positions of zeros in A014082. Could be called "tribbinary numbers" by analogy with A003714. - John Keith, Mar 07 2022

The sequence of Tribbinary numbers can be constructed by writing out the Tribonacci representations of nonnegative integers and then evaluating the result in binary. These numbers are similar to Fibbinary numbers A003714, Fibternary numbers A003726, and Tribternary numbers A356823. The number of Tribbinary numbers less than any power of two is a Tribonacci number. We can generate Tribbinary numbers recursively: Start by adding 0 and 1 to the sequence. Then, if x is a number in the sequence add 2x, 4x+1, and 8x+3 to the sequence. The n-th Tribbinary number is even if the n-th term of the Tribonacci word is a. Respectively, the n-th Tribbinary number is of the form 4x+1 if the n-th term of the Tribonacci word is b, and the n-th Tribbinary number is of the form 8x+3 if the n-th term of the Tribonacci word is c. Every nonnegative integer can be written as the sum of two Tribbinary numbers. Every number has a Tribbinary multiple. - Tanya Khovanova and PRIMES STEP Senior, Aug 30 2022

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
Index entries for 2-automatic sequences.

Cf. A278038 (binary), A063037, A000073, A014082 (number of 111).
Cf. A004781 (complement).
Cf. A007088; A003796 (no 000), A004745 (no 001), A004746 (no 010), A004744 (no 011), A003754 (no 100), A004742 (no 101), A004743 (no 110).

a003726 n = a003726_list !! (n - 1)
a003726_list = filter f [0..] where
   f x = x < 7 || (x `mod` 8) < 7 && f (x `div` 2)
-- Reinhard Zumkeller, Jun 03 2012

Select[Range[0, 82], SequenceCount[IntegerDigits[#, 2], {1, 1, 1}] == 0 &] (* Michael De Vlieger, Dec 23 2019 *)

is(n)=!bitand(bitand(n, n<<1), n<<2) \\ Charles R Greathouse IV, Feb 11 2017

There are A000073(n+3) terms of this sequence with at most n bits. In particular, a(A000073(n+3)+1) = 2^n. - Charles R Greathouse IV, Oct 22 2021

Sum_{n>=2} 1/a(n) = 9.516857810319139410424631558212354346868048230248717360943194590798113163384... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 13 2022

A087808 a(0) = 0; a(2n) = 2a(n), a(2n+1) = a(n) + 1.

Original entry on oeis.org

0, 1, 2, 2, 4, 3, 4, 3, 8, 5, 6, 4, 8, 5, 6, 4, 16, 9, 10, 6, 12, 7, 8, 5, 16, 9, 10, 6, 12, 7, 8, 5, 32, 17, 18, 10, 20, 11, 12, 7, 24, 13, 14, 8, 16, 9, 10, 6, 32, 17, 18, 10, 20, 11, 12, 7, 24, 13, 14, 8, 16, 9, 10, 6, 64, 33, 34, 18, 36, 19, 20, 11, 40, 21, 22, 12

Offset: 0

Ralf Stephan, Oct 14 2003

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
N. J. A. Sloane and J. A. Sellers, On non-squashing partitions, Discrete Math., 294 (2005), 259-274.

This is Guy Steele's sequence GS(5, 4) (see A135416); compare GS(4, 5): A135529.
A048678(k) is where k appears first in the sequence.
Cf. A000120, A003714, A004718, A005940, A048675, A048679, A080100, A090639.
A left inverse of A277020.
Cf. also A277006.

import Data.List (transpose)
a087808 n = a087808_list !! n
a087808_list = 0 : concat
   (transpose [map (+ 1) a087808_list, map (* 2) $ tail a087808_list])
-- Reinhard Zumkeller, Mar 18 2015

S := 2; f := proc(n) global S; option remember; if n=0 then RETURN(0); fi; if n mod 2 = 0 then RETURN(S*f(n/2)); else f((n-1)/2)+1; fi; end;

a[0]=0; a[n_] := a[n] = If[EvenQ[n], 2*a[n/2], a[(n-1)/2]+1]; Array[a,76,0] (* Jean-François Alcover, Aug 12 2017 *)

a(n)=if(n<1,0,if(n%2==0,2*a(n/2),a((n-1)/2)+1))

from functools import lru_cache
@lru_cache(maxsize=None)
def A087808(n): return 0 if n == 0 else A087808(n//2) + (1 if n % 2 else A087808(n//2)) # Chai Wah Wu, Mar 08 2022

(define (A087808 n) (cond ((zero? n) n) ((even? n) (* 2 (A087808 (/ n 2)))) (else (+ 1 (A087808 (/ (- n 1) 2)))))) ;; Antti Karttunen, Oct 07 2016

a(n) = A135533(n)+1-2^(A000523(n)+1-A000120(n)). - Don Knuth, Mar 01 2008
From Antti Karttunen, Oct 07 2016: (Start)
a(n) = A048675(A005940(n+1)).
For all n >= 0, a(A003714(n)) = A048679(n).
For all n >= 0, a(A277020(n)) = n.
(End)

A372433 Binary weight (number of ones in binary expansion) of the n-th squarefree number.

Original entry on oeis.org

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

Offset: 1

Gus Wiseman, May 04 2024

Harvey P. Dale, Table of n, a(n) for n = 1..1000
MathOverflow, Are there primes of every Hamming weight?
Wikipedia, Hamming weight.

Restriction of A000120 to A005117.
For prime instead of squarefree we have A014499, zeros A035103.
Counting zeros instead of ones gives A372472, cf. A023416, A372473.
For binary length instead of weight we have A372475.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A145037 counts ones minus zeros in binary expansion, cf. A031443, A031444, A031448, A097110.
A371571 lists positions of zeros in binary expansion, sum A359359.
A371572 lists positions of ones in binary expansion, sum A230877.
A372515 lists positions of zeros in reversed binary expansion, sum A359400.
A372516 counts ones minus zeros in binary expansion of primes, cf. A177718, A177796, A372538, A372539.
Cf. A039004, A049093, A049094, A059015, A069010, A070939, A073642, A211997, A368494, A372474.

DigitCount[Select[Range[100],SquareFreeQ],2,1]
Total[IntegerDigits[#,2]]&/@Select[Range[200],SquareFreeQ] (* Harvey P. Dale, Feb 14 2025 *)

from math import isqrt
from sympy import mobius
def A372433(n):
    def f(x): return n+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return int(m).bit_count() # Chai Wah Wu, Aug 02 2024

a(n) = A000120(A005117(n)).

a(n) + A372472(n) = A372475(n) = A070939(A005117(n)).

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219

Offset: 1

Olivier Gérard

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Fortran
```
c See link in A139351.
```

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)

for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002

That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A359359 Sum of positions of zeros in the binary expansion of n, where positions are read starting with 1 from the left (big-endian).

Original entry on oeis.org

1, 0, 2, 0, 5, 2, 3, 0, 9, 5, 6, 2, 7, 3, 4, 0, 14, 9, 10, 5, 11, 6, 7, 2, 12, 7, 8, 3, 9, 4, 5, 0, 20, 14, 15, 9, 16, 10, 11, 5, 17, 11, 12, 6, 13, 7, 8, 2, 18, 12, 13, 7, 14, 8, 9, 3, 15, 9, 10, 4, 11, 5, 6, 0, 27, 20, 21, 14, 22, 15, 16, 9, 23, 16, 17, 10

Offset: 0

Gus Wiseman, Jan 03 2023

			The binary expansion of 100 is (1,1,0,0,1,0,0), with zeros at positions {3,4,6,7}, so a(100) = 20.

The number of zeros is A023416, partial sums A059015.
For positions of 1's we have A230877, reversed A029931.
The reversed version is A359400.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion.
A039004 lists the positions of zeros in A345927.
Cf. A000120, A048793, A065359, A069010, A070939, A073642, A083652, A328594, A328595, A359402, A359495.

Table[Total[Join@@Position[IntegerDigits[n,2],0]],{n,0,100}]

a(n>0) = binomial(A029837(n)+1,2) - A230877(n).

A060142 Ordered set S defined by these rules: 0 is in S and if x is in S then 2x+1 and 4x are in S.

Original entry on oeis.org

0, 1, 3, 4, 7, 9, 12, 15, 16, 19, 25, 28, 31, 33, 36, 39, 48, 51, 57, 60, 63, 64, 67, 73, 76, 79, 97, 100, 103, 112, 115, 121, 124, 127, 129, 132, 135, 144, 147, 153, 156, 159, 192, 195, 201, 204, 207, 225, 228, 231, 240, 243, 249, 252, 255, 256, 259, 265, 268, 271

Offset: 0

Clark Kimberling, Mar 05 2001

nonn

After expelling 0 and 1, the numbers 4x occupy same positions in S that 1 occupies in the infinite Fibonacci word (A003849).
a(A026351(n)) = A219608(n); a(A004957(n)) = 4 * a(n). - Reinhard Zumkeller, Nov 26 2012
Apart from the initial term, this lists the indices of the 1's in A086747. - N. J. A. Sloane, Dec 05 2019
From Gus Wiseman, Jun 10 2020: (Start)
Numbers k such that the k-th composition in standard order has all odd parts, or numbers k such that A124758(k) is odd. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. For example, the sequence of all compositions into odd parts begins:
()              57: (1,1,3,1)          135: (5,1,1,1)
(1)             60: (1,1,1,3)          144: (3,5)
(1,1)           63: (1,1,1,1,1,1)      147: (3,3,1,1)
(3)             64: (7)                153: (3,1,3,1)
(1,1,1)         67: (5,1,1)            156: (3,1,1,3)
(3,1)           73: (3,3,1)            159: (3,1,1,1,1,1)
(1,3)           76: (3,1,3)            192: (1,7)
(1,1,1,1)       79: (3,1,1,1,1)        195: (1,5,1,1)
(5)             97: (1,5,1)            201: (1,3,3,1)
(3,1,1)        100: (1,3,3)            204: (1,3,1,3)
(1,3,1)        103: (1,3,1,1,1)        207: (1,3,1,1,1,1)
(1,1,3)        112: (1,1,5)            225: (1,1,5,1)
(1,1,1,1,1)    115: (1,1,3,1,1)        228: (1,1,3,3)
(5,1)          121: (1,1,1,3,1)        231: (1,1,3,1,1,1)
(3,3)          124: (1,1,1,1,3)        240: (1,1,1,5)
(3,1,1,1)      127: (1,1,1,1,1,1,1)    243: (1,1,1,3,1,1)
(1,5)          129: (7,1)              249: (1,1,1,1,3,1)
(1,3,1,1)      132: (5,3)              252: (1,1,1,1,1,3)
(End)
Numbers whose binary representation has the property that every run of consecutive 0's has even length. - Harry Richman, Jan 31 2024

			From _Harry Richman_, Jan 31 2024: (Start)
In the following, dots are used for zeros in the binary representation:
   n  binary(a(n))  a(n)
   0:    .......     0
   1:    ......1     1
   2:    .....11     3
   3:    ....1..     4
   4:    ....111     7
   5:    ...1..1     9
   6:    ...11..    12
   7:    ...1111    15
   8:    ..1....    16
   9:    ..1..11    19
  10:    ..11..1    25
  11:    ..111..    28
  12:    ..11111    31
  13:    .1....1    33
  14:    .1..1..    36
  15:    .1..111    39
  16:    .11....    48
  17:    .11..11    51
  18:    .111..1    57
  19:    .1111..    60
  20:    .111111    63
  21:    1......    64
  22:    1....11    67
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Clark Kimberling, A Self-Generating Set and the Golden Mean, J. Integer Sequences, 3 (2000), #00.2.8.
Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52 (2014), 195-202.
Lukasz Merta, Composition inverses of the variations of the Baum-Sweet sequence, arXiv:1803.00292 [math.NT], 2018. See l(n) p. 5.
Gus Wiseman, Statistics, classes, and transformations of standard compositions

Cf. A219608 (odd terms), A060138, A060139, A060140, A060141, A086747.
Cf. A003714 (no consecutive 1's in binary expansion).
Odd partitions are counted by A000009.
Numbers with an odd number of 1's in binary expansion are A000069.
Numbers whose binary expansion has odd length are A053738.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Compositions without odd parts are A062880.
- Sum is A070939.
- Product is A124758.
- Strict compositions are A233564.
- Heinz number is A333219.
- Number of distinct parts is A334028.

import Data.Set (singleton, deleteFindMin, insert)
a060142 n = a060142_list !! n
a060142_list = 0 : f (singleton 1) where
   f s = x : f (insert (4 * x) $ insert (2 * x + 1) s') where
       (x, s') = deleteFindMin s
-- Reinhard Zumkeller, Nov 26 2012

Take[Nest[Union[Flatten[# /. {{i_Integer -> i}, {i_Integer -> 2 i + 1}, {i_Integer -> 4 i}}]] &, {1}, 5], 32]  (* Or *)
Select[Range[124], FreeQ[Length /@ Select[Split[IntegerDigits[#, 2]], First[#] == 0 &], ?OddQ] &] (* _Birkas Gyorgy, May 29 2012 *)

is(n)=if(n<3, n<2, if(n%2,is(n\2),n%4==0 && is(n/4))) \\ Charles R Greathouse IV, Oct 21 2013

Corrected by T. D. Noe, Nov 01 2006

Definition simplified by Charles R Greathouse IV, Oct 21 2013

A359400 Sum of positions of zeros in the reversed binary expansion of n, where positions in a sequence are read starting with 1 from the left.

Original entry on oeis.org

1, 0, 1, 0, 3, 2, 1, 0, 6, 5, 4, 3, 3, 2, 1, 0, 10, 9, 8, 7, 7, 6, 5, 4, 6, 5, 4, 3, 3, 2, 1, 0, 15, 14, 13, 12, 12, 11, 10, 9, 11, 10, 9, 8, 8, 7, 6, 5, 10, 9, 8, 7, 7, 6, 5, 4, 6, 5, 4, 3, 3, 2, 1, 0, 21, 20, 19, 18, 18, 17, 16, 15, 17, 16, 15, 14, 14, 13

Offset: 0

Gus Wiseman, Jan 05 2023

			The reversed binary expansion of 100 is (0,0,1,0,0,1,1), with zeros at positions {1,2,4,5}, so a(100) = 12.

The number of zeros is A023416, partial sums A059015.
Row sums of A368494.
For positions of 1's we have A029931, non-reversed A230877.
The non-reversed version is A359359.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion, reverse A030308.
A039004 lists the positions of zeros in A345927.
Cf. A000120, A048793, A069010, A070939, A073642, A328594, A328595, A344618, A359402, A359495.

long A359400(long n) {
  long result = 0, counter = 1;
  do {
    if (n % 2 == 0)
      result += counter;
    counter++;
    n /= 2;
  } while (n > 0);
  return result; } // Frank Hollstein, Jan 06 2023

Table[Total[Join@@Position[Reverse[IntegerDigits[n,2]],0]],{n,0,100}]

def a(n): return sum(i for i, bi in enumerate(bin(n)[:1:-1], 1) if bi=='0')
print([a(n) for n in range(78)]) # Michael S. Branicky, Jan 09 2023

a(n) = binomial(A029837(n)+1, 2) - A029931(n), for n>0.

A107907 Numbers having consecutive zeros or consecutive ones in binary representation.

Original entry on oeis.org

3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77

Offset: 1

Reinhard Zumkeller, May 28 2005

Also positive integers whose binary expansion has cuts-resistance > 1. For the operation of shortening all runs by 1, cuts-resistance is the number of applications required to reach an empty word. - Gus Wiseman, Nov 27 2019

			From _Gus Wiseman_, Nov 27 2019: (Start)
The sequence of terms together with their binary expansions begins:
    3:      11
    4:     100
    6:     110
    7:     111
    8:    1000
    9:    1001
   11:    1011
   12:    1100
   13:    1101
   14:    1110
   15:    1111
   16:   10000
   17:   10001
   18:   10010
(End)

Amiram Eldar, Table of n, a(n) for n = 1..10000

Union of A003754 and A003714.
Complement of A000975.
Cf. A000120, A007088, A070939, A107909, A329862.

Select[Range[100],MatchQ[IntegerDigits[#,2],{_,x_,x_,_}]&] (* Gus Wiseman, Nov 27 2019 *)
Select[Range[80],SequenceCount[IntegerDigits[#,2],{x_,x_}]>0&] (* or *) Complement[Range[85],Table[FromDigits[PadRight[{},n,{1,0}],2],{n,7}]] (* Harvey P. Dale, Jul 31 2021 *)

def A107907(n): return (m:=n-2+(k:=(3*n+3).bit_length()))+(m>=(1<Chai Wah Wu, Apr 21 2025

a(A000247(n)) = A000225(n+2).
a(A000295(n+2)) = A000079(n+2).
a(A000325(n+2)) = A000051(n+2) for n>0.
a(n) = m+1 if m >= floor(2^k/3) otherwise a(n) = m where k = floor(log_2(3*(n+1))) and m = n-2+k. - Chai Wah Wu, Apr 21 2025

Offset changed to 1 by Chai Wah Wu, Apr 21 2025

A372475 Length of binary expansion (or number of bits) of the n-th squarefree number.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset: 1

Gus Wiseman, May 09 2024

			The 10th squarefree number is 14, with binary expansion (1,1,1,0), so a(10) = 4.

For prime instead of squarefree we have A035100, 1's A014499, 0's A035103.
Restriction of A070939 to A005117.
Run-lengths are A077643.
For weight instead of length we have A372433 (restrict A000120 to A005117).
For zeros instead of length we have A372472, firsts A372473.
Positions of first appearances are A372540.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A371571 lists positions of zeros in binary expansion, sum A359359.
A371572 lists positions of ones in binary expansion, sum A230877.
A372515 lists positions of zeros in reversed binary expansion, sum A359400.
Cf. A003714, A023416, A049093, A049094, A059015, A069010, A073642, A145037, A211997, A368494, A372474, A372516.

IntegerLength[Select[Range[1000],SquareFreeQ],2]

from math import isqrt
from sympy import mobius
def A372475(n):
    def f(x): return n+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return int(m).bit_length() # Chai Wah Wu, Aug 02 2024

a(n) = A070939(A005117(n)).

a(n) = A372472(n) + A372433(n).

cp's OEIS Frontend

A227349 Product of lengths of runs of 1-bits in binary representation of n.

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A003726 Numbers with no 3 adjacent 1's in binary expansion.

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A087808 a(0) = 0; a(2n) = 2a(n), a(2n+1) = a(n) + 1.

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A372433 Binary weight (number of ones in binary expansion) of the n-th squarefree number.

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A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

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A359359 Sum of positions of zeros in the binary expansion of n, where positions are read starting with 1 from the left (big-endian).

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A060142 Ordered set S defined by these rules: 0 is in S and if x is in S then 2x+1 and 4x are in S.

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