A069010 -id:A069010 - cp's OEIS frontend

A037800 Number of occurrences of 01 in the binary expansion of n.

0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 1, 2, 2, 2, 1, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1

Offset: 0

Clark Kimberling

Number of i such that d(i)>d(i-1), where Sum{d(i)*2^i: i=0,1,...,m} is base 2 representation of n.

This is the base-2 up-variation sequence; see A297330. - Clark Kimberling, Jan 18 2017

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Jean-Paul Allouche and Jeffrey Shallit, Sums of digits and the Hurwitz zeta function, in: K. Nagasaka and E. Fouvry (eds.), Analytic Number Theory, Lecture Notes in Mathematics, Vol. 1434, Springer, Berlin, Heidelberg, 1990, pp. 19-30.
Ralf Stephan, Some divide-and-conquer sequences with (relatively) simple ordinary generating functions, 2004.
Ralf Stephan, Table of generating functions.
Eric Weisstein's World of Mathematics, Digit Block.
Index entries for sequences related to binary expansion of n

Cf. A014081, A014082, A033264, A056974, A056975, A056976, A056977, A056978, A056979, A056980.

Cf. A030308, A231902.

a037800 = f 0 . a030308_row where
   f c [_]          = c
   f c (1 : 0 : bs) = f (c + 1) bs
   f c (_ : bs)     = f c bs
-- Reinhard Zumkeller, Feb 20 2014

Table[SequenceCount[IntegerDigits[n,2],{0,1}],{n,0,120}] (* Harvey P. Dale, Aug 10 2023 *)

a(n) = { if(n == 0, 0, -1 + hammingweight(bitnegimply(n, n>>1))) };  \\ Gheorghe Coserea, Aug 31 2015

a(2n) = a(n), a(2n+1) = a(n) + [n is even]. - Ralf Stephan, Aug 21 2003
G.f.: 1/(1-x) * Sum_{k>=0} t^5/(1+t)/(1+t^2) where t=x^2^k. - Ralf Stephan, Sep 10 2003
a(n) = A069010(n) - 1, n>0. - Ralf Stephan, Sep 10 2003
Sum_{n>=1} a(n)/(n*(n+1)) = log(2)/2 + Pi/4 - 1 = A231902 - 1 (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021

A200649 Number of 1's in the Stolarsky representation of n.

Original entry on oeis.org

0, 1, 2, 1, 3, 2, 2, 4, 1, 3, 3, 3, 5, 2, 2, 4, 2, 4, 4, 4, 6, 1, 3, 3, 3, 5, 3, 3, 5, 3, 5, 5, 5, 7, 2, 2, 4, 2, 4, 4, 4, 6, 2, 4, 4, 4, 6, 4, 4, 6, 4, 6, 6, 6, 8, 1, 3, 3, 3, 5, 3, 3, 5, 3, 5, 5, 5, 7, 3, 3, 5, 3, 5, 5, 5, 7, 3, 5, 5, 5, 7, 5, 5, 7, 5, 7, 7

Offset: 1

Casey Mongoven, Nov 19 2011

For the Stolarsky representation of n, see the C. Mongoven link.

Conjecture: a(n) is the length of row n-1 of A385886. To obtain it, first take maximal anti-run lengths of binary indices of each nonnegative integer (giving A384877), then remove all duplicate rows (giving A385886), and finally take the length of each remaining row. For sum instead of length we appear to have A200648. For runs minus 1 instead of anti-runs see A200650. - Gus Wiseman, Jul 21 2025

			The Stolarsky representation of 19 is 11101. This has 4 1's. So a(19) = 4.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Casey Mongoven, Description of Stolarsky Representations.

Cf. A072649, A130312, A135818, A200651.
For length instead of number of 1's we have A200648.
For 0's instead of 1's we have A200650.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion.
A384877 lists anti-run lengths of binary indices, duplicates removed A385886.
A384890 counts maximal anti-runs of binary indices, ranked by A385816.
Cf. A023758, A044813, A048793, A069010, A164707, A245562, A245563, A328592, A384893.

stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
a[n_] := Count[stol[n], 1]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)

stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
a(n) = vecsum(stol(n)); \\ Amiram Eldar, Jul 07 2023

a(n) = a(n - A130312(n-1)) + (A072649(n-1) - A072649(n - A130312(n-1) - 1)) mod 2 for n > 2 with a(1) = 0, a(2) = 1. - Mikhail Kurkov, Oct 19 2021 [verification needed]

a(n) = A200648(n) - A200650(n). - Amiram Eldar, Jul 07 2023

More terms from Amiram Eldar, Jul 07 2023

A243815 Number of length n words on alphabet {0,1} such that the length of every maximal block of 0's (runs) is the same.

Original entry on oeis.org

1, 2, 4, 8, 14, 24, 39, 62, 97, 151, 233, 360, 557, 864, 1344, 2099, 3290, 5176, 8169, 12931, 20524, 32654, 52060, 83149, 133012, 213069, 341718, 548614, 881572, 1417722, 2281517, 3673830, 5918958, 9540577, 15384490, 24817031, 40045768, 64637963, 104358789

Offset: 0

Geoffrey Critzer, Jun 11 2014

nonn

Number of terms of A164710 with exactly n+1 binary digits. - Robert Israel, Nov 09 2015
From Gus Wiseman, Jun 23 2025: (Start)
This is the number of subsets of {1..n} with all equal lengths of runs of consecutive elements increasing by 1. For example, the runs of S = {1,2,5,6,8,9} are ((1,2),(5,6),(8,9)), with lengths (2,2,2), so S is counted under a(9). The a(0) = 1 through a(4) = 14 subsets are:
  {}  {}   {}     {}       {}
      {1}  {1}    {1}      {1}
           {2}    {2}      {2}
           {1,2}  {3}      {3}
                  {1,2}    {4}
                  {1,3}    {1,2}
                  {2,3}    {1,3}
                  {1,2,3}  {1,4}
                           {2,3}
                           {2,4}
                           {3,4}
                           {1,2,3}
                           {2,3,4}
                           {1,2,3,4}
(End)

			0110 is a "good" word because the length of both its runs of 0's is 1.
Words of the form 11...1 are good words because the condition is vacuously satisfied.
a(5) = 24 because there are 32 length 5 binary words but we do not count: 00010, 00101, 00110, 01000, 01001, 01100, 10010, 10100.

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A164710.
These subsets are ranked by A164707, complement A164708.
For distinct instead of equal lengths we have A384175, complement A384176.
For anti-runs instead of runs we have A384889, for partitions A384888.
For permutations instead of subsets we have A384892, distinct instead of equal A384891.
For partitions instead of subsets we have A384904, strict A384886.
The complement is counted by A385214.
A034839 counts subsets by number of maximal runs, for strict partitions A116674.
A049988 counts partitions with equal run-lengths, distinct A325325.
A329738 counts compositions with equal run-lengths, distinct A329739.
A384887 counts partitions with equal lengths of gapless runs, distinct A384884.
Cf. A010027, A044813, A069010, A268193, A383013, A384177, A384893.

a:= n-> 1 + add(add((d-> binomial(d+j, d))(n-(i*j-1))
          , j=1..iquo(n+1, i)), i=2..n+1):
seq(a(n), n=0..50);  # Alois P. Heinz, Jun 11 2014

nn=30;Prepend[Map[Total,Transpose[Table[Drop[CoefficientList[Series[ (1+x^k)/(1-x-x^(k+1))-1/(1-x),{x,0,nn}],x],1],{k,1,nn}]]],0]+1
Table[Length[Select[Subsets[Range[n]],SameQ@@Length/@Split[#,#2==#1+1&]&]],{n,0,10}] (* Gus Wiseman, Jun 23 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).

A124757 Zero-based weighted sum of compositions in standard order.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 2, 3, 0, 1, 2, 3, 3, 4, 5, 6, 0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14, 15, 0, 1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14

Offset: 0

Franklin T. Adams-Watters, Nov 06 2006

The standard order of compositions is given by A066099.

Sum of all positions of 1's except the last in the reversed binary expansion of n. For example, the reversed binary expansion of 14 is (0,1,1,1), so a(14) = 2 + 3 = 5. Keeping the last position gives A029931. - Gus Wiseman, Jan 17 2023

			Composition number 11 is 2,1,1; 0*2+1*1+2*1 = 3, so a(11) = 3.
The table starts:
  0
  0
  0 1
  0 1 2 3

Alois P. Heinz, Rows n = 0..14, flattened

Cf. A066099, A070939, A029931, A011782 (row lengths), A001788 (row sums).
Row sums of A048793 if we delete the last part of every row.
For prime indices instead of standard comps we have A359674, rev A359677.
Positions of first appearances are A359756.
A003714 lists numbers with no successive binary indices.
A030190 gives binary expansion, reverse A030308.
A230877 adds up positions of 1's in binary expansion, length A000120.
A359359 adds up positions of 0's in binary expansion, length A023416.
Cf. A059015, A065359, A069010, A073642, A083652, A359400, A359402, A359678.

Table[Total[Most[Join@@Position[Reverse[IntegerDigits[n,2]],1]]],{n,30}]

For a composition b(1),...,b(k), a(n) = Sum_{i=1..k} (i-1)*b(i).

For n>0, a(n) = A029931(n) - A070939(n).

A278219 Filter-sequence related to base-2 run-length encoding: a(n) = A046523(A243353(n)).

Original entry on oeis.org

1, 2, 4, 2, 4, 8, 6, 2, 4, 12, 16, 8, 6, 12, 6, 2, 4, 12, 36, 12, 16, 32, 24, 8, 6, 30, 24, 12, 6, 12, 6, 2, 4, 12, 36, 12, 36, 72, 60, 12, 16, 48, 64, 32, 24, 72, 24, 8, 6, 30, 60, 30, 24, 48, 60, 12, 6, 30, 24, 12, 6, 12, 6, 2, 4, 12, 36, 12, 36, 72, 60, 12, 36, 180, 144, 72, 60, 180, 60, 12, 16, 48, 144, 48, 64, 128, 96, 32, 24, 120, 216, 72, 24, 72

Offset: 0

Antti Karttunen, Nov 16 2016

nonn

Antti Karttunen, Table of n, a(n) for n = 0..65537

Cf. A003188, A046523, A075157, A243353, A278220.
Other base-2 related filter sequences: A278217, A278222.
Sequences that (seem to) partition N into same or coarser equivalence classes are at least these: A005811, A136004, A033264, A037800, A069010, A087116, A090079 and many others like A105500, A106826, A166242, A246960, A277561, A037834, A225081 although these have not been fully checked yet.

f[n_, i_, x_] := Which[n == 0, x, EvenQ@ n, f[n/2, i + 1, x], True, f[(n - 1)/2, i, x Prime@ i]]; g[n_] := If[n == 1, 1, Times @@ MapIndexed[ Prime[First@ #2]^#1 &, Sort[FactorInteger[n][[All, -1]], Greater]]];
Table[g@ f[BitXor[n, Floor[n/2]], 1, 1], {n, 0, 93}] (* Michael De Vlieger, May 09 2017 *)

from sympy import prime, factorint
import math
def A(n): return n - 2**int(math.floor(math.log(n, 2)))
def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
def a005940(n): return b(n - 1)
def P(n):
    f = factorint(n)
    return sorted([f[i] for i in f])
def a046523(n):
    x=1
    while True:
        if P(n) == P(x): return x
        else: x+=1
def a003188(n): return n^int(n/2)
def a243353(n): return a005940(1 + a003188(n))
def a(n): return a046523(a243353(n)) # Indranil Ghosh, May 07 2017

(define (A278219 n) (A046523 (A243353 n)))

a(n) = A046523(A243353(n)).
a(n) = A278222(A003188(n)).
a(n) = A278220(1+A075157(n)).

A372540 Least k such that the k-th squarefree number has binary expansion of length n. Index of the smallest squarefree number >= 2^n.

Original entry on oeis.org

1, 2, 4, 7, 12, 21, 40, 79, 158, 315, 625, 1246, 2492, 4983, 9963, 19921, 39845, 79689, 159361, 318726, 637462, 1274919, 2549835, 5099651, 10199302, 20398665, 40797328, 81594627, 163189198, 326378285, 652756723, 1305513584, 2611027095, 5222054082, 10444108052

Offset: 0

Gus Wiseman, May 10 2024

			The squarefree numbers A005117(a(n)) together with their binary expansions and binary indices begin:
       1:                  1 ~ {1}
       2:                 10 ~ {2}
       5:                101 ~ {1,3}
      10:               1010 ~ {2,4}
      17:              10001 ~ {1,5}
      33:             100001 ~ {1,6}
      65:            1000001 ~ {1,7}
     129:           10000001 ~ {1,8}
     257:          100000001 ~ {1,9}
     514:         1000000010 ~ {2,10}
    1027:        10000000011 ~ {1,2,11}
    2049:       100000000001 ~ {1,12}
    4097:      1000000000001 ~ {1,13}
    8193:     10000000000001 ~ {1,14}
   16385:    100000000000001 ~ {1,15}
   32770:   1000000000000010 ~ {2,16}
   65537:  10000000000000001 ~ {1,17}
  131073: 100000000000000001 ~ {1,18}

Chai Wah Wu, Table of n, a(n) for n = 0..73

Counting zeros instead of length gives A372473, firsts of A372472.
For prime instead of squarefree we have:
- zeros A372474, firsts of A035103
- ones A372517, firsts of A014499
- bits A372684, firsts of A035100
Positions of first appearances in A372475, run-lengths A077643.
For weight instead of length we have A372541, firsts of A372433.
Indices of the squarefree numbers listed by A372683.
A000120 counts ones in binary expansion (binary weight), zeros A080791.
A005117 lists squarefree numbers.
A030190 gives binary expansion, reversed A030308.
A070939 counts bits, binary length, or length of binary expansion.
Cf. A029931, A048793, A049093, A049094, A059015, A069010, A143658, A211997, A230877.

nn=1000;
ssnm[y_]:=Max@@NestWhile[Most,y,Union[#]!=Range[Max@@#]&];
dcs=IntegerLength[Select[Range[nn],SquareFreeQ],2];
Table[Position[dcs,i][[1,1]],{i,ssnm[dcs]}]

from itertools import count
from math import isqrt
from sympy import mobius, factorint
def A372540(n): return next(sum(mobius(a)*(k//a**2) for a in range(1, isqrt(k)+1)) for k in count(1<Chai Wah Wu, May 12 2024

A005117(a(n)) = A372683(n).

a(n) = A143658(n)+1 for n > 1. - Chai Wah Wu, Aug 26 2024

a(24)-a(34) from Chai Wah Wu, May 12 2024

A200648 Length of Stolarsky representation of n.

Original entry on oeis.org

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

Offset: 1

Casey Mongoven, Nov 19 2011

For the Stolarsky representation of n, see the C. Mongoven link.

Conjecture: a(n) is the sum of row n-1 of A385886. To obtain it, first take maximal anti-run lengths of binary indices of each nonnegative integer (giving A384877), then remove all duplicate rows (giving A385886), and finally take the sum of each remaining row. For length instead of sum we appear to have A200649. - Gus Wiseman, Jul 21 2025

			The Stolarsky representation of 19 is 11101. This is of length 5. So a(19) = 5.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Casey Mongoven, Description of Stolarsky Representations.

Cf. A135817, A200651.
Counting just ones gives A200649.
Counting just zeros gives A200650.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion.
A384890 counts maximal anti-runs of binary indices, ranks A385816.
A385886 lists maximal anti-run lengths of binary indices.
Cf. A023758, A048793, A052499, A069010, A072649, A083368, A135818, A245562, A245563, A348366, A384877, A384893.

stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
a[n_] := If[n == 1, 1, Length[stol[n]]]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)

stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
a(n) = if(n == 1, 1, #stol(n)); \\ Amiram Eldar, Jul 07 2023

a(n) = A200649(n) + A200650(n). - Michel Marcus, Mar 14 2023

More terms from Amiram Eldar, Jul 07 2023

A372473 Least k such that the k-th squarefree number has exactly n zeros in its binary expansion.

Original entry on oeis.org

1, 2, 7, 12, 21, 40, 79, 158, 315, 1247, 1246, 2492, 4983, 9963, 19921, 39845, 79689, 159361, 318726, 637462, 1274919, 2549835, 5099651, 10199302, 20398665, 40797328, 81594627, 163189198, 326378285, 652756723, 1305513584, 2611027095, 5222054082, 10444108052

Offset: 0

Gus Wiseman, May 09 2024

Note that the data is not strictly increasing.

			The squarefree numbers A005117(a(n)) together with their binary expansions and binary indices begin:
     1:              1 ~ {1}
     2:             10 ~ {2}
    10:           1010 ~ {2,4}
    17:          10001 ~ {1,5}
    33:         100001 ~ {1,6}
    65:        1000001 ~ {1,7}
   129:       10000001 ~ {1,8}
   257:      100000001 ~ {1,9}
   514:     1000000010 ~ {2,10}
  2051:   100000000011 ~ {1,2,12}
  2049:   100000000001 ~ {1,12}
  4097:  1000000000001 ~ {1,13}
  8193: 10000000000001 ~ {1,14}

Chai Wah Wu, Table of n, a(n) for n = 0..57

Positions of first appearances in A372472.
For prime instead of squarefree we have A372474, A035103, A372517, A014499.
Counting bits (length) gives A372540, firsts of A372475, runs A077643.
Counting 1's (weight) instead of 0's gives A372541, firsts of A372433.
A000120 counts ones in binary expansion (binary weight), zeros A080791.
A005117 lists squarefree numbers.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A070939 gives length of binary expansion (number of bits).
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. A035100, A039004, A049093, A049094, A059015, A069010, A145037, A211997.

nn=10000;
spnm[y_]:=Max@@NestWhile[Most,y,Union[#]!=Range[0,Max@@#]&];
dcs=DigitCount[Select[Range[nn],SquareFreeQ],2,0];
Table[Position[dcs,i][[1,1]],{i,0,spnm[dcs]}]

from math import isqrt
from itertools import count
from sympy import factorint, mobius
from sympy.utilities.iterables import multiset_permutations
def A372473(n):
    if n==0: return 1
    for l in count(n):
        m = 1<Chai Wah Wu, May 10 2024

a(23)-a(33) from Chai Wah Wu, May 10 2024

A372474 Least k such that the k-th prime number has exactly n zeros in its binary expansion.

Original entry on oeis.org

2, 1, 8, 7, 19, 32, 99, 55, 174, 310, 565, 1029, 1902, 3513, 6544, 6543, 23001, 43395, 82029, 155612, 295957, 564164, 1077901, 3957811, 3965052, 7605342, 14630844, 28194383, 54400029, 105097568, 393615809, 393615807, 762939128, 1480206930, 2874398838, 5586502349

Offset: 0

Gus Wiseman, May 11 2024

			The prime numbers A000040(a(n)) together with their binary expansions and binary indices begin:
         3:                          11 ~ {1,2}
         2:                          10 ~ {2}
        19:                       10011 ~ {1,2,5}
        17:                       10001 ~ {1,5}
        67:                     1000011 ~ {1,2,7}
       131:                    10000011 ~ {1,2,8}
       523:                  1000001011 ~ {1,2,4,10}
       257:                   100000001 ~ {1,9}
      1033:                 10000001001 ~ {1,4,11}
      2053:                100000000101 ~ {1,3,12}
      4099:               1000000000011 ~ {1,2,13}
      8209:              10000000010001 ~ {1,5,14}
     16417:             100000000100001 ~ {1,6,15}
     32771:            1000000000000011 ~ {1,2,16}
     65539:           10000000000000011 ~ {1,2,17}
     65537:           10000000000000001 ~ {1,17}
    262147:         1000000000000000011 ~ {1,2,19}
    524353:        10000000000001000001 ~ {1,7,20}
   1048609:       100000000000000100001 ~ {1,6,21}
   2097169:      1000000000000000010001 ~ {1,5,22}
   4194433:     10000000000000010000001 ~ {1,8,23}
   8388617:    100000000000000000001001 ~ {1,4,24}
  16777729:   1000000000000001000000001 ~ {1,10,25}
  67108913: 100000000000000000000110001 ~ {1,5,6,27}
  67239937: 100000000100000000000000001 ~ {1,18,27}

Positions of first appearances in A035103.
For squarefree instead of prime we have A372473, firsts of A372472.
Counting ones (weight) gives A372517, firsts of A014499.
Counting squarefree bits gives A372540, firsts of A372475, runs A077643.
Counting squarefree ones gives A372541, firsts of A372433.
Counting bits (length) gives A372684, firsts of A035100.
A000120 counts ones in binary expansion (binary weight), zeros A080791.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A070939 gives length of binary expansion (number of bits).
Cf. A059015, A066195, A069010, A073642, A145037, A211997, A230877, A359359, A359400, A371571, A372516.

nn=10000;
spnm[y_]:=Max@@NestWhile[Most,y,Union[#]!=Range[0,Max@@#]&];
dcs=DigitCount[Select[Range[nn],PrimeQ],2,0];
Table[Position[dcs,i][[1,1]],{i,0,spnm[dcs]}]

from itertools import count
from sympy import isprime, primepi
from sympy.utilities.iterables import multiset_permutations
def A372474(n):
    for l in count(n):
        m = 1<Chai Wah Wu, May 13 2024

a(n) = A000720(A066195(n)). - Robert Israel, May 13 2024

a(22)-a(35) from and offset corrected by Chai Wah Wu, May 13 2024

cp's OEIS Frontend

A037800 Number of occurrences of 01 in the binary expansion of n.

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A200649 Number of 1's in the Stolarsky representation of n.

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A243815 Number of length n words on alphabet {0,1} such that the length of every maximal block of 0's (runs) is the same.

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

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A124757 Zero-based weighted sum of compositions in standard order.

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A278219 Filter-sequence related to base-2 run-length encoding: a(n) = A046523(A243353(n)).

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A372540 Least k such that the k-th squarefree number has binary expansion of length n. Index of the smallest squarefree number >= 2^n.

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