A023416 -id:A023416 - cp's OEIS frontend

A200650 Number of 0's in Stolarsky representation of n.

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

Offset: 1

Casey Mongoven, Nov 19 2011

For the Stolarsky representation of n, see the C. Mongoven link.
a(n+1), n >= 1, gives the size of the n-th generation of each of the "[male-female] pair of Fibonacci rabbits" in the Fibonacci rabbits tree read right-to-left by row, the first pair (the root) being the 0th generation. (Cf. OEIS Wiki link below.) - Daniel Forgues, May 07 2015
From Daniel Forgues, May 07 2015: (Start)
Concatenation of:
  0:                      1,
  1:                      0,
  2:                      0,
  3:                   1, 0,
  4:                1, 1, 0,
  5:          2, 1, 1, 1, 0,
  6: 2, 2, 1, 2, 1, 1, 1, 0,
(...),
where row n, n >= 3, is row n-1 prepended by incremented row n-2. (End)
For n >= 3, this algorithm yields the next F_n terms of the sequence, where F_n is the n-th Fibonacci number (A000045). Since it is asymptotic to (phi^n)/sqrt(5), the number of terms thus obtained grows exponentially at each step! - Daniel Forgues, May 22 2015
Conjecture: a(n) is one less than the length of row n-1 of A385817. To obtain it, first take maximal run lengths of binary indices of each nonnegative integer (giving A245563), then remove all duplicate rows (giving A385817), and finally take the length of each remaining row and subtract 1. For sum instead of length we appear to have A200648. For anti-runs instead of runs we appear to have A341259 = A200649-1. - Gus Wiseman, Jul 21 2025
How is this related to A117479? - R. J. Mathar, Aug 10 2025

			The Stolarsky representation of 19 is 11101. This has one 0. So a(19) = 1.

Kenny Lau, Table of n, a(n) for n = 1..20000
Casey Mongoven, Description of Stolarsky Representations.
OEIS Wiki, Fibonacci rabbits per generation.

For length instead of number of 0's we have A200648.
For sum (or number of 1's) instead of number of 0's we have A200649.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion, 0's A023416.
A069010 counts maximal runs of binary indices, ranked by A385889.
A245563 lists maximal run lengths of binary indices, duplicates removed A385817.
Cf. A000045, A023758, A135818, A200651, A245562, A328592, A385886.

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, Count[stol[n], 0]]; 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,  my(s = stol(n)); #s - vecsum(s)); \\ Amiram Eldar, Jul 07 2023

a(n) = A200648(n) - A200649(n). - Amiram Eldar, Jul 07 2023

Corrected and extended by Kenny Lau, Jul 04 2016

A268289 a(0)=0; thereafter a(n) = a(n-1) - A037861(n).

Original entry on oeis.org

0, 1, 1, 3, 2, 3, 4, 7, 5, 5, 5, 7, 7, 9, 11, 15, 12, 11, 10, 11, 10, 11, 12, 15, 14, 15, 16, 19, 20, 23, 26, 31, 27, 25, 23, 23, 21, 21, 21, 23, 21, 21, 21, 23, 23, 25, 27, 31, 29, 29, 29, 31, 31, 33, 35, 39, 39, 41, 43, 47, 49

Offset: 0

Mark Moore, Jan 30 2016

The graph of this sequence is related to the Takagi (blancmange) curve: see Lagarias (2012), Section 9, especially Theorem 9.1. [Corrected by Laura Monroe, Oct 21 2020]
Theorem: a(n) is the cardinality of the set { 1<= m <= n, ((n-m) mod 2^floor(log_2(m)+1)) < 2^floor(log_2(m)) }. See links.
From Laura Monroe, Jun 11 2020: (Start)
Consider a full balanced binary tree with n unlabeled leaves such that for each internal node, the number of leaf descendants of the two children differs by at most 1. Call a tree with this even distribution of leaves "pairwise".
Apply labels to the internal nodes, where an internal node is labeled S if its two children have the same number of leaf descendants, and D if its two children have a different number of leaf descendants, and call this an SD-tree. (For a pairwise tree, this is equivalent to saying that a node is an S-node iff it has an even number of leaf descendants.)
a(n) is then the number of S-nodes on a pairwise SD-tree with n+1 leaves.
This is proved in Props. 17 and 18 of the Monroe et al. article in the links.
One example of such a tree is the summation tree generated by a pairwise summation on n+1 summands (see example below). Another example is the tree representing a neutral single-elimination tournament on n+1 teams, as in A096351.
(End)
From Laura Monroe, Oct 23 2020: (Start)
Subtracting a(n) from n gives a sequence of dilations of increasing length on the dyadic rational points of the Takagi function. The number of points in each dilation is 2^k and the scale of each dilation in both the x and y directions is 2^k, where k = floor(log_2(n+1)).
2^(a(n)) is the number of tree automorphisms on the pairwise (i.e., divide-and-conquer) tree with n+1 leaves.
(End)

			From _Laura Monroe_, Jun 11 2020: (Start)
For n=2, the pairwise summation on 2+1=3 summands takes the form ((a+b)+c). The corresponding summation tree and SD-tree look like:
       +            D
      / \          / \
     +   c        S   c
    / \          / \
   a   b        a   b
and exactly 1 internal node has an even number of leaf descendants, hence is an S-node.
For n=3, the pairwise summation on 3+1=4 summands takes the form ((a+b)+(c+d)). The corresponding summation tree and SD-tree look like:
       +            S
      / \          / \
     +   +        S   S
    /|   |\      /|   |\
   a b   c d    a b   c d
and exactly 3 internal nodes have an even number of leaf descendants, hence are S-nodes.
(End)

Laura Monroe, Table of n, a(n) for n = 0..16383 (first 10000 terms from N. J. A. Sloane)
Thomas Baruchel, Flattening Karatsuba's recursion tree into a single summation, arXiv:1902.08982 [math.NT], 2019. See (5) conjecture of theorem.
Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019. See 2.4 proof of theorem.
Thomas Baruchel, Flattening Karatsuba's Recursion Tree into a Single Summation, SN Computer Science (2020) Vol. 1, Article No. 48.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint 2016.
Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
Jeffrey C. Lagarias, The Takagi function and its properties, arXiv:1112.4205 [math.CA], 2011-2012.
Jeffrey C. Lagarias, The Takagi function and its properties, In Functions in number theory and their probabilistic aspects, 153--189, RIMS Kôkyûroku Bessatsu, B34, Res. Inst. Math. Sci. (RIMS), Kyoto, 2012. MR3014845.
Laura Monroe, A Few Identities of the Takagi Function on Dyadic Rationals, arXiv:2111.05996 [math.CO], 2021.
Laura Monroe, Takagi Function Identities on Dyadic Rationals, J. Int. Seq (2024) Vol. 27, Art. 24.2.7.
Laura Monroe and Vanessa Job, Computationally Inequivalent Summations and Their Parenthetic Forms, arXiv:2005.05387 [cs.DM], 2020. See Prop. 26, the alternate proof.

Partial sums of A145037.

Cf. A000120, A000788, A023416, A037861, A096351, A181132, A269735, A233271, A296062.

int a(int n)   {
    int m=n+1;
    int result=0;
    int i=0;
    while (n) {
        int ith_bit_set = m&(1<>= 1;
    }
   return result;
}
/* Laura Monroe, Jun 17 2020 */

function A268289List(len)
    A = zeros(Int, len)
    for n in 1:len-1
        a, b, c = n, n & 1, 1
        while (a >>= 1) != 0
            b += a & 1
            c += 1
        end
        A[n+1] = A[n] + <<(b, 1) - c
    end
    A
end; println(A268289List(61)) # Peter Luschny, Jun 22 2020

a000120 := proc(n) add(i, i=convert(n, base, 2)) end:
a023416 := proc(n) if n = 0 then 1; else add(1-e, e=convert(n, base, 2)) ; end if; end proc:
a268289:=proc(n) option remember; global a000120, a023416;
if n=0 then 0 else a268289(n-1)+a000120(n)-a023416(n); fi; end;
[seq(a268289(n),n=0..132)];
# N. J. A. Sloane, Mar 07 2016
# second Maple program:
a:= proc(n) option remember; `if`(n<0, 0,
      a(n-1)+add(2*i-1, i=Bits[Split](n)))
    end:
seq(a(n), n=0..60);  # Alois P. Heinz, Jan 18 2022

Join[{0}, Table[DigitCount[n, 2, 1] - DigitCount[n, 2, 0], {n, 1, 100}] // Accumulate] (* Jean-François Alcover, Oct 24 2016 *)

a(n) = if (n==0, 0, if (n%2, 2*a((n-1)/2)+1, a(n/2) + a(n/2-1))); \\ Michel Marcus, Jun 16 2020

a(n) = my(v=binary(n+1),s=-1); for(i=1,#v, v[i]=if(v[i],s++,s--;1)); fromdigits(v,2); \\ Kevin Ryde, Jun 16 2020

def A268289(n): return (sum(i.bit_count() for i in range(1,n+1))<<1)-1-(n+1)*(m:=(n+1).bit_length())+(1<Chai Wah Wu, Mar 01 2023

def A268289(n): return sum((n+1)%m if (n+1)&(m:=1<Chai Wah Wu, Nov 11 2024

From N. J. A. Sloane, Mar 11 2016: (Start)
a(0)=0; for n > 0, a(n) = a(n-1) + A000120(n) - A023416(n) = A000788(n) - A181132(n).
a(0)=0; thereafter a(2*n) = a(n) + a(n-1), a(2*n+1) = 2*a(n) + 1.
G.f.: (1/(1-x)^2) * Sum_{k >= 0} x^(2^k)*(1-x^(2^k))/(1+x^(2^k)).
a(2^k-1) = 2^k-1, a(3*2^k-1) = 2^(k+1)-1, a(5*2^k-1) = 3*2^k-1, etc.
(End)
From Laura Monroe, Jun 11 2020: (Start)
a(n-1) = Sum_{i=0..floor(log_2(n))} (((floor(n/(2^i))+1) mod 2)*(2^i)+(-1)^((floor(n/(2^i))+1) mod 2)*(n mod (2^i))), for n>=1.
This is an explicit formula for this sequence, and is O(log(n)). This formula is proven in Prop. 18, in the Monroe et al. reference in the links. (End)
From Laura Monroe, Oct 23 2020: (Start)
a(n) = n - A296062(n).
a(n+1) = (n+1) - (2^k)*tau(x/(2^k)), where tau is the Takagi function and n+1 = (2^k)+x with x < 2^k. (End)

Simplified definition following a suggestion from Michel Marcus. Corrected start, added more terms. - N. J. A. Sloane, Mar 07 2016

A372472 Number of zeros in the binary expansion of the n-th squarefree number.

Original entry on oeis.org

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

Offset: 1

Gus Wiseman, May 09 2024

			The 12th squarefree number is 17, with binary expansion (1,0,0,0,1), so a(12) = 3.

Positions of first appearances are A372473.
Restriction of A023416 to A005117.
For prime instead of squarefree we have A035103, ones A014499, bits A035100.
Counting 1's instead of 0's (so restrict A000120 to A005117) gives A372433.
For binary length we have A372475, run-lengths A077643.
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, A039004, A049093, A049094, A059015, A069010, A070939, A073642, A145037, A211997, A368494, A372474, A372516.

A372583 := proc(n)
    A023416(A005117(n)) ;
end proc:
seq(A372583(n),n=1..200) ; # R. J. Mathar, May 20 2024

DigitCount[Select[Range[100],SquareFreeQ],2,0]

a(n) = A023416(A005117(n)).

a(n) + A372433(n) = A070939(A005117(n)) = A372475(n).

A031448 Numbers whose base-2 representation has one fewer 0's than 1's.

Original entry on oeis.org

1, 5, 6, 19, 21, 22, 25, 26, 28, 71, 75, 77, 78, 83, 85, 86, 89, 90, 92, 99, 101, 102, 105, 106, 108, 113, 114, 116, 120, 271, 279, 283, 285, 286, 295, 299, 301, 302, 307, 309, 310, 313, 314, 316, 327, 331, 333, 334, 339, 341, 342

Offset: 1

Clark Kimberling

A037861(a(n)) = -1. - Reinhard Zumkeller, Mar 31 2015

The viabin numbers of the integer partitions in which the number of parts is equal to the largest part (for the definition of viabin number see comment in A290253). For example, 99 is in the sequence because it is the viabin number of the integer partition [4,2,2,2]. - Emeric Deutsch, Aug 29 2017

			99 is in the sequence because its binary form is 1100011. - _Emeric Deutsch_, Aug 29 2017

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to binary expansion of n

Cf. A007088, A023416, A000120, A031444, subsequence of A089648.

a031448 n = a031448_list !! (n-1)
a031448_list = filter ((== -1) . a037861) [1..]
-- Reinhard Zumkeller, Mar 31 2015

vitopart := proc (n) local L, i, j, N, p, t: N := 2*n; L := ListTools:-Reverse(convert(N, base, 2)): j := 0: for i to nops(L) do if L[i] = 0 then j := j+1: p[j] := numboccur(L[1 .. i], 1) end if end do: sort([seq(p[t], t = 1 .. j)], `>=`) end proc: A := {}; for m to 500 do if nops(vitopart(m)) = max(vitopart(m)) then A := `union`(A, {m}) else  end if end do: A; # program is based on my comment; the command vitopart(n) yields the integer partition having viabin number n. - Emeric Deutsch, Aug 29 2017

Select[Range[400],DigitCount[#,2,1]==DigitCount[#,2,0]+1&] (* Harvey P. Dale, May 24 2019 *)

A064194 a(2n) = 3a(n), a(2n+1) = 2a(n+1)+a(n), with a(1) = 1.

Original entry on oeis.org

1, 3, 7, 9, 17, 21, 25, 27, 43, 51, 59, 63, 71, 75, 79, 81, 113, 129, 145, 153, 169, 177, 185, 189, 205, 213, 221, 225, 233, 237, 241, 243, 307, 339, 371, 387, 419, 435, 451, 459, 491, 507, 523, 531, 547, 555, 563, 567, 599, 615, 631, 639, 655, 663, 671, 675

Offset: 1

Guillaume Hanrot and Paul Zimmermann, Sep 21 2001

Number of ring multiplications needed to multiply two degree-n polynomials using Karatsuba's algorithm.
Number of gates in the AND/OR problem (see Chang/Tsai reference).
a(n) is also the number of odd elements in the n X n symmetric Pascal matrix. - Stefano Spezia, Nov 14 2022

A. A. Karatsuba and Y. P. Ofman, Multiplication of multiplace numbers by automata. Dokl. Akad. Nauk SSSR 145, 2, 293-294 (1962).

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
K.-N. Chang and S.-C. Tsai, Exact solution of a minimal recurrence, Inform. Process. Lett. 75 (2000), 61-64.
P. J. Grabner and H.-K. Hwang, Digital sums and divide-and-conquer recurrences: Fourier expansions and absolute convergence, Constructive Approximation, Jan. 2005, Volume 21, Issue 2, pp. 149-179.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, pp. 27-28.

Cf. A023416, A267584, A047999 (Sierpinski triangle).
Cf. also A268514.
Sequences of form a(n)=r*a(ceil(n/2))+s*a(floor(n/2)), a(1)=1, for (r,s) = (1,1), (1,2), (2,1), (1,3), (2,2), (3,1), (1,4), (2,3), (3,2), (4,1): A000027, A006046, A064194, A130665, A073121, A268524, A116520, A268525, A268526, A268527.

[n le 1 select 1 else Self(Floor(n/2)) + 2*Self(Ceiling(n/2)): n in [1..60]]; // Vincenzo Librandi, Aug 30 2016

f:=proc(n) option remember; if n=1 then 1 elif n mod 2 = 0 then 3*f(n/2) else 2*f((n+1)/2)+f((n-1)/2); fi; end; [seq(f(n),n=1..60)]; # N. J. A. Sloane, Jan 17 2016

a[n_] := a[n] = If[EvenQ[n], 3 a[n/2], 2 a[# + 1] + a[#] &[(n - 1)/2]]; a[1] = 1; Array[a, 56] (* Michael De Vlieger, Oct 29 2022 *)

a(n) = sum(i=0, n-1, sum(j=0, n-1, binomial(i+j, i) % 2)); \\ Michel Marcus, Aug 25 2013

Partial sums of the sequence { b(1)=1, b(n)=2^(e0(n-1)+1) } (essentially A267584), where e0(n)=A023416(n) is the number of zeros in the binary expansion of n. [Chang/Tsai] - Ralf Stephan, Jul 29 2003
a(1) = 1, a(n) = a(floor(n/2)) + 2*a(ceiling(n/2)), n > 1.
a(n+1) = Sum_{0<=i, j<=n} (binomial(i+j, i) mod 2). - Benoit Cloitre, Mar 07 2005
In particular, a(2^k)=3^k, a(3*2^k)=7*3^k. - N. J. A. Sloane, Jan 18 2016
a(n) = 2*A268514(n-1) + 1. - N. J. A. Sloane, Feb 07 2016

Edited with clearer definition by N. J. A. Sloane, Jan 18 2016

A066195 Smallest prime containing n zeros in its binary expansion.

Original entry on oeis.org

3, 2, 19, 17, 67, 131, 523, 257, 1033, 2053, 4099, 8209, 16417, 32771, 65539, 65537, 262147, 524353, 1048609, 2097169, 4194433, 8388617, 16777729, 67108913, 67239937, 134250497, 268435459, 536903681, 1073741827, 2147483713, 8589934627, 8589934609, 17179869697

Offset: 0

Robert G. Wilson v, Dec 15 2001

A023416(a(n)) = n and A023416(m) <> n for m < A049084(a(m)).

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

Cf. A061712.

import Data.List (find)
import Data.Maybe (fromJust)
a066195 n = fromJust $ find ((== n) . a023416) a000040_list
-- Reinhard Zumkeller, Feb 19 2013

Do[ k = 1; While[ Count[ IntegerDigits[ Prime[ k ], 2 ], 0 ] != n, k++ ]; Print[ Prime[ k ] ], {n, 1, 24} ]

a(25)-a(32) from Alois P. Heinz, Jun 28 2015

A072600 Numbers which in base 2 have fewer 0's than 1's.

Original entry on oeis.org

1, 3, 5, 6, 7, 11, 13, 14, 15, 19, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 39, 43, 45, 46, 47, 51, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 71, 75, 77, 78, 79, 83, 85, 86, 87, 89, 90, 91, 92, 93, 94, 95, 99, 101, 102, 103, 105, 106, 107, 108, 109, 110, 111, 113, 114, 115

Offset: 1

Reinhard Zumkeller, Jun 23 2002

A037861(a(n)) < 0.

b_k = {a(n) | for all n s.t. a(n) contains k binary digits equal to 1} is the list of all valid win/loss round sequences in a "best of 2k-1" two player game, where 1 is a win and 0 is a loss. For example 19 = 10011b represents a game where the winner won the first two rounds, lost the next two, and won the last one. |b_k| = A001700(k). - Philippe Beaudoin, May 14 2014

			11 is present because '1011' contains 1 '0' and 3 '1's: 1<3.

T. D. Noe, Table of n, a(n) for n = 1..4733 ( numbers < 2^13)
Jason Bell, Thomas Finn Lidbetter, Jeffrey Shallit, Additive Number Theory via Approximation by Regular Languages, arXiv:1804.07996 [cs.FL], 2018.
Thomas Finn Lidbetter, Counting, Adding, and Regular Languages, Master's Thesis, University of Waterloo, Ontario, Canada, 2018.
Index entries for sequences related to binary expansion of n

Cf. A007088, A000120, A023416, A072601, A031443, A072602, A072603, A037861

a072600 n = a072600_list !! (n-1)
a072600_list = filter ((< 0) . a037861) [0..]
-- Reinhard Zumkeller, Mar 31 2015

Select[Range[130],DigitCount[#,2,0]Harvey P. Dale, Jan 12 2011 *)

is(n)=2*hammingweight(n)>exponent(n)+1 \\ Charles R Greathouse IV, Apr 18 2020

A372541 Least k such that the k-th squarefree number has exactly n ones in its binary expansion.

Original entry on oeis.org

1, 3, 6, 11, 20, 60, 78, 157, 314, 624, 1245, 3736, 4982, 9962, 19920, 39844, 79688, 239046, 318725, 956194, 1912371, 2549834, 5099650, 15298984, 20398664, 40797327, 81594626, 163189197, 326378284, 979135127, 1305513583, 2611027094, 5222054081, 10444108051

Offset: 0

Gus Wiseman, May 09 2024

			The squarefree numbers A005117(a(n)) together with their binary expansions and binary indices begin:
       1:                   1 ~ {1}
       3:                  11 ~ {1,2}
       7:                 111 ~ {1,2,3}
      15:                1111 ~ {1,2,3,4}
      31:               11111 ~ {1,2,3,4,5}
      95:             1011111 ~ {1,2,3,4,5,7}
     127:             1111111 ~ {1,2,3,4,5,6,7}
     255:            11111111 ~ {1,2,3,4,5,6,7,8}
     511:           111111111 ~ {1,2,3,4,5,6,7,8,9}
    1023:          1111111111 ~ {1,2,3,4,5,6,7,8,9,10}
    2047:         11111111111 ~ {1,2,3,4,5,6,7,8,9,10,11}
    6143:       1011111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,13}
    8191:       1111111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,12,13}
   16383:      11111111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,12,13,14}
   32767:     111111111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
   65535:    1111111111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}
  131071:   11111111111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17}

Chai Wah Wu, Table of n, a(n) for n = 0..56
Wikipedia, Hamming weight.

Positions of firsts appearances in A372433.
Counting zeros instead of ones gives A372473, firsts in A372472.
For prime instead of squarefree we have A372517, firsts of A014499.
Counting bits (length) gives A372540, firsts of A372475, runs A077643.
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.
A145037, A097110 count ones minus zeros, for primes A372516, A177796.
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. A023416, A049093, A049094, A069010, A070939, A071403, A211997, A280296, A372474.

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

from math import isqrt
from itertools import count
from sympy import factorint, mobius
from sympy.utilities.iterables import multiset_permutations
def A372541(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

A031444 Numbers whose base-2 representation has one more 0 than 1's.

Original entry on oeis.org

4, 17, 18, 20, 24, 67, 69, 70, 73, 74, 76, 81, 82, 84, 88, 97, 98, 100, 104, 112, 263, 267, 269, 270, 275, 277, 278, 281, 282, 284, 291, 293, 294, 297, 298, 300, 305, 306, 308, 312, 323, 325, 326, 329, 330, 332, 337, 338, 340, 344

Offset: 1

Clark Kimberling

If m is a term, then also 4*m+1. - Reinhard Zumkeller, Mar 31 2015

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to binary expansion of n.

Cf. A007088, A023416, A000120, A031448, A037861, A095072 (subsequence).

Subsequence of A089648.

a031444 n = a031444_list !! (n-1)
a031444_list = filter ((== 1) . a037861) [1..]
-- Reinhard Zumkeller, Mar 31 2015

Select[Range[350], (Differences@ DigitCount[#, 2])[[1]] == 1 &] (* Amiram Eldar, Aug 03 2023 *)

A037861(a(n)) = 1. - Reinhard Zumkeller, Mar 31 2015

A200714 Stolarsky representation interpreted as binary to decimal integers.

Original entry on oeis.org

0, 1, 3, 2, 7, 5, 6, 15, 4, 11, 13, 14, 31, 10, 9, 23, 12, 27, 29, 30, 63, 8, 21, 19, 22, 47, 26, 25, 55, 28, 59, 61, 62, 127, 20, 17, 43, 18, 39, 45, 46, 95, 24, 53, 51, 54, 111, 58, 57, 119, 60, 123, 125, 126, 255, 16, 41, 35, 42, 87, 37, 38, 79, 44, 91, 93

Offset: 1

Casey Mongoven, Nov 20 2011

See explanation of Stolarsky representations in the C. Mongoven link.

			The Stolarsky representation of 19 is 11101. In binary this is equal to 29. So a(19) = 29.

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

Cf. A048679, A200648, A200649, A200650, A200651.

Cf. A000120, A005811, A023416, A070939.

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_] := FromDigits[stol[n], 2]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)

a(n) = {if (n == 1, return (0)); tau = (1 + sqrt(5))/2; mn = 0; while ((m = round(mn*tau)) < n, mn++;); if (m == n, return (2*a(mn)+1)); mn = 0; while ((m = floor(mn*(1+tau)-tau/2)) < n, mn++;); if (m == n, return (2*a(mn))); error("neither A nor B !!");} \\ (cf C. Mongoven link) Michel Marcus, May 21 2013, Sep 02 2013

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) = fromdigits(stol(n), 2); \\ Amiram Eldar, Jul 07 2023

From Amiram Eldar, Jul 07 2023: (Start)
A000120(a(n)) = A200649(n).
A023416(a(n)) = A200650(n).
A070939(a(n)) = A200648(n).
A005811(a(n)) = A200651(n). (End)
Conjecture: a(n) = A367306(A358654(n-1)). - Mikhail Kurkov, Oct 17 2024

More terms from Amiram Eldar, Jul 07 2023

cp's OEIS Frontend

A200650 Number of 0's in Stolarsky representation of n.

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A268289 a(0)=0; thereafter a(n) = a(n-1) - A037861(n).

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A372472 Number of zeros in the binary expansion of the n-th squarefree number.

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A031448 Numbers whose base-2 representation has one fewer 0's than 1's.

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A064194 a(2n) = 3*a(n), a(2n+1) = 2*a(n+1)+a(n), with a(1) = 1.

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A066195 Smallest prime containing n zeros in its binary expansion.

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A064194 a(2n) = 3a(n), a(2n+1) = 2a(n+1)+a(n), with a(1) = 1.