A328594 -id:A328594 - cp's OEIS frontend

A121016 Numbers whose binary expansion is properly periodic.

3, 7, 10, 15, 31, 36, 42, 45, 54, 63, 127, 136, 153, 170, 187, 204, 221, 238, 255, 292, 365, 438, 511, 528, 561, 594, 627, 660, 682, 693, 726, 759, 792, 825, 858, 891, 924, 957, 990, 1023, 2047, 2080, 2145, 2184, 2210, 2275, 2340, 2405, 2457, 2470, 2535

Offset: 1

Jacob A. Siehler, Sep 08 2006

A finite sequence is aperiodic if its cyclic rotations are all different. - Gus Wiseman, Oct 31 2019

			For example, 204=(1100 1100)_2 and 292=(100 100 100)_2 belong to the sequence, but 30=(11110)_2 cannot be split into repeating periods.
From _Gus Wiseman_, Oct 31 2019: (Start)
The sequence of terms together with their binary expansions and binary indices begins:
   3:         11 ~ {1,2}
   7:        111 ~ {1,2,3}
   10:      1010 ~ {2,4}
   15:      1111 ~ {1,2,3,4}
   31:     11111 ~ {1,2,3,4,5}
   36:    100100 ~ {3,6}
   42:    101010 ~ {2,4,6}
   45:    101101 ~ {1,3,4,6}
   54:    110110 ~ {2,3,5,6}
   63:    111111 ~ {1,2,3,4,5,6}
  127:   1111111 ~ {1,2,3,4,5,6,7}
  136:  10001000 ~ {4,8}
  153:  10011001 ~ {1,4,5,8}
  170:  10101010 ~ {2,4,6,8}
  187:  10111011 ~ {1,2,4,5,6,8}
  204:  11001100 ~ {3,4,7,8}
  221:  11011101 ~ {1,3,4,5,7,8}
  238:  11101110 ~ {2,3,4,6,7,8}
  255:  11111111 ~ {1,2,3,4,5,6,7,8}
  292: 100100100 ~ {3,6,9}
(End)

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

A020330 is a subsequence.
Numbers whose binary expansion is aperiodic are A328594.
Numbers whose reversed binary expansion is Lyndon are A328596.
Numbers whose binary indices have equal run-lengths are A164707.
Cf. A000120, A003714, A014081, A065609, A069010, A275692, A328595.

PeriodicQ[n_, base_] := Block[{l = IntegerDigits[n, base]}, MemberQ[ RotateLeft[l, # ] & /@ Most@ Divisors@ Length@l, l]]; Select[ Range@2599, PeriodicQ[ #, 2] &]

is(n)=n=binary(n);fordiv(#n,d,for(i=1,#n/d-1, for(j=1,d, if(n[j]!=n[j+i*d], next(3)))); return(d<#n)) \\ Charles R Greathouse IV, Dec 10 2013

A326774 For any number m, let m* be the bi-infinite string obtained by repetition of the binary representation of m; this sequence lists the numbers n such that for any k < n, n* does not equal k* up to a shift.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 11, 16, 17, 18, 19, 21, 23, 32, 33, 34, 35, 37, 38, 39, 43, 47, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 77, 78, 79, 85, 87, 91, 95, 128, 129, 130, 131, 132, 133, 134, 135, 137, 138, 139, 140, 141, 142, 143, 146, 147, 149, 150, 151, 154

Offset: 0

Rémy Sigrist, Jul 27 2019

This sequence contains every power of 2.
No term belongs to A121016.
Every terms belongs to A004761.
For any k > 0, there are A001037(k) terms with binary length k.
From Gus Wiseman, Apr 19 2020: (Start)
Also numbers k such that the k-th composition in standard order (row k of A066099) is a co-Lyndon word (regular Lyndon words being A275692). For example, the sequence of all co-Lyndon words begins:
()            37: (3,2,1)         79: (3,1,1,1,1)
(1)           38: (3,1,2)         85: (2,2,2,1)
(2)           39: (3,1,1,1)       87: (2,2,1,1,1)
(3)           43: (2,2,1,1)       91: (2,1,2,1,1)
(2,1)         47: (2,1,1,1,1)     95: (2,1,1,1,1,1)
(4)           64: (7)            128: (8)
(3,1)         65: (6,1)          129: (7,1)
(2,1,1)       66: (5,2)          130: (6,2)
(5)           67: (5,1,1)        131: (6,1,1)
(4,1)         68: (4,3)          132: (5,3)
(3,2)         69: (4,2,1)        133: (5,2,1)
(3,1,1)       70: (4,1,2)        134: (5,1,2)
(2,2,1)       71: (4,1,1,1)      135: (5,1,1,1)
(2,1,1,1)     73: (3,3,1)        137: (4,3,1)
(6)           74: (3,2,2)        138: (4,2,2)
(5,1)         75: (3,2,1,1)      139: (4,2,1,1)
(4,2)         77: (3,1,2,1)      140: (4,1,3)
(4,1,1)       78: (3,1,1,2)      141: (4,1,2,1)
(End)

			3* = ...11... equals 1* = ...1..., so 3 is not a term.
6* = ...110... equals up to a shift 5* = ...101..., so 6 is not a term.
11* = ...1011... only equals up to a shift 13* = ...1101... and 14* = ...1110..., so 11 is a term.

Rémy Sigrist, Logarithmic scatterplot of the first differences of the terms < 2^24
Rémy Sigrist, PARI program for A326774

Cf. A001037, A004761, A065609, A121016.
Necklace compositions are counted by A008965.
Lyndon compositions are counted by A059966.
Length of Lyndon factorization of binary expansion is A211100.
Numbers whose reversed binary expansion is a necklace are A328595.
Length of co-Lyndon factorization of binary expansion is A329312.
Length of Lyndon factorization of reversed binary expansion is A329313.
Length of co-Lyndon factorization of reversed binary expansion is A329326.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Necklaces are A065609.
- Sum is A070939.
- Runs are counted by A124767.
- Rotational symmetries are counted by A138904.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon compositions are A275692.
- Co-Lyndon compositions are A326774 (this sequence).
- Aperiodic compositions are A328594.
- Reversed co-necklaces are A328595.
- Rotational period is A333632.
- Co-necklaces are A333764.
- Co-Lyndon factorizations are counted by A333765.
- Lyndon factorizations are counted by A333940.
- Reversed necklaces are A333943.
- Length of co-Lyndon factorization is A334029.
Cf. A000740, A034691, A060223, A269134, A292884, A328596.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
colynQ[q_]:=Length[q]==0||Array[Union[{RotateRight[q,#],q}]=={RotateRight[q,#],q}&,Length[q]-1,1,And];
Select[Range[0,100],colynQ[stc[#]]&] (* Gus Wiseman, Apr 19 2020 *)

PARI
```
See Links section.
```

A333764 Numbers k such that the k-th composition in standard order is a co-necklace.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 15, 16, 17, 18, 19, 21, 23, 31, 32, 33, 34, 35, 36, 37, 38, 39, 42, 43, 45, 47, 63, 64, 65, 66, 67, 68, 69, 70, 71, 73, 74, 75, 77, 78, 79, 85, 87, 91, 95, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140

Offset: 1

Gus Wiseman, Apr 12 2020

nonn

A co-necklace is a finite sequence that is lexicographically greater than or equal to any cyclic rotation.

A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of 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. This gives a bijective correspondence between nonnegative integers and integer compositions

			The sequence together with the corresponding co-necklaces begins:
    1: (1)             32: (6)               69: (4,2,1)
    2: (2)             33: (5,1)             70: (4,1,2)
    3: (1,1)           34: (4,2)             71: (4,1,1,1)
    4: (3)             35: (4,1,1)           73: (3,3,1)
    5: (2,1)           36: (3,3)             74: (3,2,2)
    7: (1,1,1)         37: (3,2,1)           75: (3,2,1,1)
    8: (4)             38: (3,1,2)           77: (3,1,2,1)
    9: (3,1)           39: (3,1,1,1)         78: (3,1,1,2)
   10: (2,2)           42: (2,2,2)           79: (3,1,1,1,1)
   11: (2,1,1)         43: (2,2,1,1)         85: (2,2,2,1)
   15: (1,1,1,1)       45: (2,1,2,1)         87: (2,2,1,1,1)
   16: (5)             47: (2,1,1,1,1)       91: (2,1,2,1,1)
   17: (4,1)           63: (1,1,1,1,1,1)     95: (2,1,1,1,1,1)
   18: (3,2)           64: (7)              127: (1,1,1,1,1,1,1)
   19: (3,1,1)         65: (6,1)            128: (8)
   21: (2,2,1)         66: (5,2)            129: (7,1)
   23: (2,1,1,1)       67: (5,1,1)          130: (6,2)
   31: (1,1,1,1,1)     68: (4,3)            131: (6,1,1)

The non-"co" version is A065609.
The reversed version is A328595.
Binary necklaces are A000031.
Necklace compositions are A008965.
Necklaces covering an initial interval are A019536.
Numbers whose prime signature is a necklace are A329138.
Length of co-Lyndon factorization of binary expansion is A329312.
Length of Lyndon factorization of reversed binary expansion is A329313.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Runs are counted by A124767.
- Rotational symmetries are counted by A138904.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon compositions are A275692.
- Co-Lyndon compositions are A326774.
- Aperiodic compositions are A328594.
- Length of Lyndon factorization is A329312.
- Rotational period is A333632.
- Reversed necklaces are A333943.
Cf. A000740, A001037, A027375, A059966, A211100, A302291, A328596, A329142, A333765, A333939, A333941.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
coneckQ[q_]:=Array[OrderedQ[{RotateRight[q,#],q}]&,Length[q]-1,1,And];
Select[Range[100],coneckQ[stc[#]]&]

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

A057335 a(0) = 1, and for n > 0, a(n) = A000040(A000120(n)) * a(floor(n/2)); essentially sequence A055932 generated using A000120, hence sorted by number of factors.

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 18, 30, 16, 24, 36, 60, 54, 90, 150, 210, 32, 48, 72, 120, 108, 180, 300, 420, 162, 270, 450, 630, 750, 1050, 1470, 2310, 64, 96, 144, 240, 216, 360, 600, 840, 324, 540, 900, 1260, 1500, 2100, 2940, 4620, 486, 810, 1350, 1890, 2250, 3150, 4410

Offset: 0

Alford Arnold, Aug 27 2000

Note that for n>0 the prime divisors of a(n) are consecutive primes starting with 2. All of the least prime signatures (A025487) are included; with the other values forming A056808.
Using the formula, terms of b(n)= a(n)/A057334(n) are: 1, 1, 2, 2, 4, 4, 6, 6, 8, ..., indeed a(n) repeated. - Michel Marcus, Feb 09 2014
a(n) is the unique normal number whose unsorted prime signature is the k-th composition in standard order (graded reverse-lexicographic). This composition (row k of 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. A number is normal if its prime indices cover an initial interval of positive integers. Unsorted prime signature is the sequence of exponents in a number's prime factorization. - Gus Wiseman, Apr 19 2020

			From _Gus Wiseman_, Apr 19 2020: (Start)
The sequence of terms together with their prime indices begins:
      1: {}
      2: {1}
      4: {1,1}
      6: {1,2}
      8: {1,1,1}
     12: {1,1,2}
     18: {1,2,2}
     30: {1,2,3}
     16: {1,1,1,1}
     24: {1,1,1,2}
     36: {1,1,2,2}
     60: {1,1,2,3}
     54: {1,2,2,2}
     90: {1,2,2,3}
    150: {1,2,3,3}
    210: {1,2,3,4}
     32: {1,1,1,1,1}
     48: {1,1,1,1,2}
For example, the 27th composition in standard order is (1,2,1,1), and the normal number with prime signature (1,2,1,1) is 630 = 2*3*3*5*7, so a(27) = 630.
(End)

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Cf. A000120, A057334, A055932 and A056808.
Cf. A324939.
Unsorted prime signature is A124010.
Numbers whose prime signature is aperiodic are A329139.
The reversed version is A334031.
A partial inverse is A334032.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Aperiodic compositions are A328594.
- Normal compositions are A333217.
- Permutations are A333218.
- Heinz number is A333219.
Cf. A005867, A029931, A048793, A052409, A056239, A066099, A112798, A124767, A228351, A233249, A333220, A345974.
Related to A019565 via A122111 and to A000079 via A336321.

Table[Times @@ Map[If[# == 0, 1, Prime@ #] &, Accumulate@ IntegerDigits[n, 2]], {n, 0, 54}] (* Michael De Vlieger, May 23 2017 *)

mg(n) = if (n==0, 1, prime(hammingweight(n))); \\ A057334
lista(nn) = {my(v = vector(nn)); v[1] = 1; for (i=2, nn, v[i] = mg(i-1)*v[(i+1)\2];); v;} \\ Michel Marcus, Feb 09 2014

A057335(n) = if(0==n,1,prime(hammingweight(n))*A057335(n\2)); \\ Antti Karttunen, Jul 20 2020

a(n) = A057334(n) * a (repeated).
A334032(a(n)) = n; a(A334032(n)) = A071364(n). - Gus Wiseman, Apr 19 2020
a(n) = A122111(A019565(n)); A019565(n) = A122111(a(n)). - Peter Munn, Jul 18 2020
a(n) = A336321(2^n). - Peter Munn, Mar 04 2022
Sum_{n>=0} 1/a(n) = Sum_{n>=0} 1/A005867(n) = 2.648101... (A345974). - Amiram Eldar, Jun 26 2025

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 29 2003

New primary name from Antti Karttunen, Jul 20 2020

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).

A333632 Rotational period of the k-th composition in standard order; a(0) = 0.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 3, 2, 3, 3, 1, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 1, 1, 2, 2, 3, 1, 3, 3, 4, 2, 3, 1, 4, 3, 2, 4, 5, 2, 3, 3, 4, 3, 4, 2, 5, 3, 4, 4, 5, 4, 5, 5, 1, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4

Offset: 0

Gus Wiseman, Apr 12 2020

nonn

A composition of n is a finite sequence of positive integers summing to n. 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. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The a(299) = 5 rotations:
  (1,1,3,2,2)
  (1,3,2,2,1)
  (3,2,2,1,1)
  (2,2,1,1,3)
  (2,1,1,3,2)
The a(9933) = 4 rotations:
  (1,2,1,3,1,2,1,3)
  (1,3,1,2,1,3,1,2)
  (2,1,3,1,2,1,3,1)
  (3,1,2,1,3,1,2,1)

Aperiodic compositions are counted by A000740.
Aperiodic binary words are counted by A027375.
The orderless period of prime indices is A052409.
Numbers whose binary expansion is periodic are A121016.
Periodic compositions are counted by A178472.
The version for binary expansion is A302291.
Numbers whose prime signature is aperiodic are A329139.
Compositions by number of distinct rotations are A333941.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Necklaces are A065609.
- Sum is A070939.
- Equal runs are counted by A124767.
- Rotational symmetries are counted by A138904.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon compositions are A275692.
- Co-Lyndon compositions are A326774.
- Aperiodic compositions are A328594.
- Rotational period is A333632 (this sequence).
- Co-necklaces are A333764.
- Reversed necklaces are A333943.
Cf. A000031, A001037, A008965, A019536, A211100, A328595, A328596, A329312, A329313, A329326.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Length[Union[Array[RotateRight[stc[n],#]&,DigitCount[n,2,1]]]],{n,0,100}]

a(n) = A000120(n)/A138904(n) = A302291(n) - A023416(n)/A138904(n).

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.

A087117 Number of zeros in the longest string of consecutive zeros in the binary representation of n.

Original entry on oeis.org

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

Offset: 0

Reinhard Zumkeller, Aug 14 2003

The following four statements are equivalent: a(n) = 0; n = 2^k - 1 for some k > 0; A087116(n) = 0; A023416(n) = 0.

The k-th composition in standard order (row k of 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. Then a(k) is the maximum part of this composition, minus one. The maximum part is A333766(k). - Gus Wiseman, Apr 09 2020

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for sequences related to binary expansion of n

Cf. A023416, A007088, A007814.
Cf. A025480, A038374, A090046, A090047, A090048, A090049, A090050.
Positions of zeros are A000225.
Positions of terms <= 1 are A003754.
Positions of terms > 0 are A062289.
Positions of first appearances are A131577.
The version for prime indices is A252735.
The proper maximum is A333766.
The version for minimum is A333767.
Maximum prime index is A061395.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Runs are counted by A124767.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Runs-resistance is A333628.
- Weakly decreasing compositions are A114994.
- Weakly increasing compositions are A225620.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
Cf. A029931, A048793, A061395, A087117, A228351, A328594, A333217, A333218, A333219, A333767, A333768.

import Data.List (unfoldr, group)
a087117 0       = 1
a087117 n
  | null $ zs n = 0
  | otherwise   = maximum $ map length $ zs n where
  zs = filter ((== 0) . head) . group .
       unfoldr (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2)
-- Reinhard Zumkeller, May 01 2012

A087117 := proc(n)
    local d,l,zlen ;
    if n = 0 then
        return 1 ;
    end if;
    d := convert(n,base,2) ;
    for l from nops(d)-1 to 0 by -1 do
        zlen := [seq(0,i=1..l)] ;
        if verify(zlen,d,'sublist') then
            return l ;
        end if;
    end do:
    return 0 ;
end proc; # R. J. Mathar, Nov 05 2012

nz[n_]:=Max[Length/@Select[Split[IntegerDigits[n,2]],MemberQ[#,0]&]]; Array[nz,110,0]/.-\[Infinity]->0 (* Harvey P. Dale, Sep 05 2017 *)

h(n)=if(n<2, return(0)); my(k=valuation(n,2)); if(k, max(h(n>>k), k), n++; n>>=valuation(n,2); h(n-1))
a(n)=if(n,h(n),1) \\ Charles R Greathouse IV, Apr 06 2022

a(n) = max(A007814(n), a(A025480(n-1))) for n >= 2. - Robert Israel, Feb 19 2017
a(2n+1) = a(n) (n>=1); indeed, the binary form of 2n+1 consists of the binary form of n with an additional 1 at the end - Emeric Deutsch, Aug 18 2017
For n > 0, a(n) = A333766(n) - 1. - Gus Wiseman, Apr 09 2020

A329139 Numbers whose prime signature is an aperiodic word.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 11, 12, 13, 16, 17, 18, 19, 20, 23, 24, 25, 27, 28, 29, 31, 32, 37, 40, 41, 43, 44, 45, 47, 48, 49, 50, 52, 53, 54, 56, 59, 60, 61, 63, 64, 67, 68, 71, 72, 73, 75, 76, 79, 80, 81, 83, 84, 88, 89, 90, 92, 96, 97, 98, 99, 101, 103, 104

Offset: 1

Gus Wiseman, Nov 09 2019

nonn

First differs from A319161 in having 1260 = 2*2 * 3^2 * 5^1 * 7^1. First differs from A325370 in having 420 = 2^2 * 3^1 * 5^1 * 7^1.
A number's prime signature (A124010) is the sequence of positive exponents in its prime factorization.
A sequence is aperiodic if its cyclic rotations are all different.

			The sequence of terms together with their prime signatures begins:
   1: ()
   2: (1)
   3: (1)
   4: (2)
   5: (1)
   7: (1)
   8: (3)
   9: (2)
  11: (1)
  12: (2,1)
  13: (1)
  16: (4)
  17: (1)
  18: (1,2)
  19: (1)
  20: (2,1)
  23: (1)
  24: (3,1)
  25: (2)
  27: (3)

Complement of A329140.
Aperiodic compositions are A000740.
Aperiodic binary words are A027375.
Numbers whose binary expansion is aperiodic are A328594.
Numbers whose prime signature is a Lyndon word are A329131.
Numbers whose prime signature is a necklace are A329138.
Cf. A025487, A097318, A112798, A124010, A178472, A181819, A304678, A329133, A329135, A329136, A329137, A329142.

aperQ[q_]:=Array[RotateRight[q,#1]&,Length[q],1,UnsameQ];
Select[Range[100],aperQ[Last/@FactorInteger[#]]&]

cp's OEIS Frontend

A121016 Numbers whose binary expansion is properly periodic.

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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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A333632 Rotational period of the k-th composition in standard order; a(0) = 0.

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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.

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