A295235 -id:A295235 - cp's OEIS frontend

A048793 List giving all subsets of natural numbers arranged in standard statistical (or Yates) order.

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

Offset: 0

N. J. A. Sloane

For n>0: first occurrence of n in row 2^(n-1), and when the table is seen as a flattened list at position n*2^(n-1)+1, cf. A005183. - Reinhard Zumkeller, Nov 16 2013

Row n lists the positions of 1's in the reversed binary expansion of n. Compare to triangles A112798 and A213925. - Gus Wiseman, Jul 22 2019

			From _Gus Wiseman_, Jul 22 2019: (Start)
Triangle begins:
  {}
  1
  2
  1  2
  3
  1  3
  2  3
  1  2  3
  4
  1  4
  2  4
  1  2  4
  3  4
  1  3  4
  2  3  4
  1  2  3  4
  5
  1  5
  2  5
  1  2  5
  3  5
(End)

S. Hedayat, N. J. A. Sloane and J. Stufken, Orthogonal Arrays, Springer-Verlag, NY, 1999, p. 249.

Reinhard Zumkeller, Rows n = 0..1000 of triangle, flattened

Cf. A048794.
Row lengths are A000120.
First column is A001511.
Heinz numbers of rows are A019565.
Row sums are A029931.
Reversing rows gives A272020.
Subtracting 1 from each term gives A133457; subtracting 1 and reversing rows gives A272011.
Indices of relatively prime rows are A291166 (see also A326674); arithmetic progressions are A295235; rows with integer average are A326669 (see also A326699/A326700); pairwise coprime rows are A326675.
Cf. A035327, A070939.

#include 
#include 
#define USAGE "Usage: 'A048793 num' where num is the largest number to use creating sets.\n"
#define MAX_NUM 10
#define MAX_ROW 1024
int main(int argc, char *argv[]) {
  unsigned short a[MAX_ROW][MAX_NUM]; signed short old_row, new_row, i, j, end;
  if (argc < 2) { fprintf(stderr, USAGE); return EXIT_FAILURE; }
  end = atoi(argv[1]); end = (end > MAX_NUM) ? MAX_NUM: end;
  for (i = 0; i < MAX_ROW; i++) for ( j = 0; j < MAX_NUM; j++) a[i][j] = 0;
  a[1][0] = 1; new_row = 2;
  for (i = 2; i <= end; i++) {
    a[new_row++ ][0] = i;
    for (old_row = 1; a[old_row][0] != i; old_row++) {
      for (j = 0; a[old_row][j] != 0; j++) { a[new_row][j] = a[old_row][j]; }
      a[new_row++ ][j] = i;
    }
  }
  fprintf(stdout, "Values: 0");
  for (i = 1; a[i][0] != 0; i++) for (j = 0; a[i][j] != 0; j++) fprintf(stdout, ",%d", a[i][j]);
  fprintf(stdout, "\n"); return EXIT_SUCCESS
}

a048793 n k = a048793_tabf !! n !! k
a048793_row n = a048793_tabf !! n
a048793_tabf = [0] : [1] : f [[1]] where
   f xss = yss ++ f (xss ++ yss) where
     yss = [y] : map (++ [y]) xss
     y = last (last xss) + 1
-- Reinhard Zumkeller, Nov 16 2013

T:= proc(n) local i, l, m; l:= NULL; m:= n;
      if n=0 then return 0 fi; for i while m>0 do
      if irem(m, 2, 'm')=1 then l:=l, i fi od; l
    end:
seq(T(n), n=0..50);  # Alois P. Heinz, Sep 06 2014

s[0] = {{}}; s[n_] := s[n] = Join[s[n - 1], Append[#, n]& /@ s[n - 1]]; Join[{0}, Flatten[s[6]]] (* Jean-François Alcover, May 24 2012 *)
Table[Join@@Position[Reverse[IntegerDigits[n,2]],1],{n,30}] (* Gus Wiseman, Jul 22 2019 *)

Constructed recursively: subsets that include n are obtained by appending n to all earlier subsets.

More terms from Larry Reeves (larryr(AT)acm.org), Apr 11 2000

A029931 If 2n = Sum 2^e_i, a(n) = Sum e_i.

Original entry on oeis.org

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, 6, 7, 8, 9, 9, 10, 11, 12, 10, 11, 12, 13, 13, 14, 15, 16, 11, 12, 13, 14, 14, 15, 16, 17, 15, 16, 17, 18, 18, 19, 20, 21, 7, 8, 9, 10, 10, 11, 12, 13, 11, 12, 13, 14, 14, 15, 16

Offset: 0

N. J. A. Sloane

Write n in base 2, n = sum b(i)*2^(i-1), then a(n) = sum b(i)*i. - Benoit Cloitre, Jun 09 2002
May be regarded as a triangular array read by rows, giving weighted sum of compositions in standard order. The standard order of compositions is given by A066099. - Franklin T. Adams-Watters, Nov 06 2006
Sum of all positive integer roots m_i of polynomial {m,k} - see link [Shevelev]; see also A264613. - Vladimir Shevelev, Dec 13 2015
Also the sum of binary indices of n, where a binary index of n (A048793) is any position of a 1 in its reversed binary expansion. For example, the binary indices of 11 are {1,2,4}, so a(11) = 7. - Gus Wiseman, May 22 2024

			14 = 8+4+2 so a(7) = 3+2+1 = 6.
Composition number 11 is 2,1,1; 1*2+2*1+3*1 = 7, so a(11) = 7.
The triangle starts:
  0
  1
  2 3
  3 4 5 6
The reversed binary expansion of 18 is (0,1,0,0,1) with 1's at positions {2,5}, so a(18) = 2 + 5 = 7. - _Gus Wiseman_, Jul 22 2019

T. D. Noe, Table of n, a(n) for n = 0..1023
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197, ex. 10. See also DOI.
Vladimir Shevelev, The number of permutations with prescribed up-down structure as a function of two variables, INTEGERS, 12 (2012), #A1. (See Section 3, Theorem 21 and Section 8, Theorem 50)

Other sequences that are built by replacing 2^k in the binary representation with other numbers: A022290 (Fibonacci), A059590 (factorials), A073642, A089625 (primes), A116549, A326031.
Cf. A001793 (row sums), A011782 (row lengths), A059867, A066099, A124757.
Row sums of A048793 and A272020.
Cf. A035327, A056239, A291166, A295235, A326669, A326674, A326675, A326699/A326700.
Contains exactly A000009(n) copies of n.
For length instead of sum we have A000120, complement A023416.
For minimum instead of sum we have A001511, opposite A000012.
For maximum instead of sum we have A029837 or A070939, opposite A070940.
For product instead of sum we have A096111.
The reverse version is A230877, row sums of A371572.
The reverse complement is A359359, row sums of A371571.
The complement is A359400, row sums of A368494.
Numbers k such that a(k) is prime are A372689.
A014499 lists binary indices of prime numbers.
A019565 gives Heinz number of binary indices, inverse A048675.
A372471 lists binary indices of primes, row-sums A372429.
Cf. A000009, A030101, A359401, A359402.

a029931 = sum . zipWith (*) [1..] . a030308_row
-- Reinhard Zumkeller, Feb 28 2014

HammingWeight := n -> add(i, i = convert(n, base, 2)):
a := proc(n) option remember; `if`(n = 0, 0,
ifelse(n::even, a(n/2) + HammingWeight(n/2), a(n-1) + 1)) end:
seq(a(n), n = 0..78); # Peter Luschny, Oct 30 2021

a[n_] := (b = IntegerDigits[n, 2]).Reverse @ Range[Length @ b]; Array[a,78,0] (* Jean-François Alcover, Apr 28 2011, after B. Cloitre *)

for(n=0,100,l=length(binary(n)); print1(sum(i=1,l, component(binary(n),i)*(l-i+1)),","))

a(n) = my(b=binary(n)); b*-[-#b..-1]~; \\ Ruud H.G. van Tol, Oct 17 2023

def A029931(n): return sum(i if j == '1' else 0 for i, j in enumerate(bin(n)[:1:-1],1)) # Chai Wah Wu, Dec 20 2022
(C#)
ulong A029931(ulong n) {
    ulong result = 0, counter = 1;
    while(n > 0) {
        if (n % 2 == 1)
          result += counter;
        counter++;
        n /= 2;
    }
    return result;
} // Frank Hollstein, Jan 07 2023

a(n) = a(n - 2^L(n)) + L(n) + 1 [where L(n) = floor(log_2(n)) = A000523(n)] = sum of digits of A048794 [at least for n < 512]. - Henry Bottomley, Mar 09 2001
a(0) = 0, a(2n) = a(n) + e1(n), a(2n+1) = a(2n) + 1, where e1(n) = A000120(n). a(n) = log_2(A029930(n)). - Ralf Stephan, Jun 19 2003
G.f.: (1/(1-x)) * Sum_{k>=0} (k+1)*x^2^k/(1+x^2^k). - Ralf Stephan, Jun 23 2003
a(n) = Sum_{k>=0} A030308(n,k)*A000027(k+1). - Philippe Deléham, Oct 15 2011
a(n) = sum of n-th row of the triangle in A213629. - Reinhard Zumkeller, Jun 17 2012
From Reinhard Zumkeller, Feb 28 2014: (Start)
a(A089633(n)) = n and a(m) != n for m < A089633(n).
a(n) = Sum_{k=1..A070939(n)} k*A030308(n,k-1). (End)
a(n) = A073642(n) + A000120(n). - Peter Kagey, Apr 04 2016

More terms from Erich Friedman

A000975 a(2n) = 2a(2n-1), a(2n+1) = 2a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).

Original entry on oeis.org

0, 1, 2, 5, 10, 21, 42, 85, 170, 341, 682, 1365, 2730, 5461, 10922, 21845, 43690, 87381, 174762, 349525, 699050, 1398101, 2796202, 5592405, 11184810, 22369621, 44739242, 89478485, 178956970, 357913941, 715827882, 1431655765, 2863311530, 5726623061, 11453246122

Offset: 0

Mira Bernstein, N. J. A. Sloane, Robert G. Wilson v, Sep 13 1996

Might be called the "Lichtenberg sequence" after Georg Christoph Lichtenberg, who discussed it in 1769 in connection with the Chinese Rings puzzle (baguenaudier). - Andreas M. Hinz, Feb 15 2017
Number of steps to change from a binary string of n 0's to n 1's using a Gray code. - Jon Stadler (jstadler(AT)coastal.edu)
Popular puzzles such as Spin-Out and The Brain Puzzler are based on the Gray binary system and require a(n) steps to complete for some number n.
Conjecture: {a(n)} also gives all j for which A048702(j) = A000217(j); i.e., if we take the a(n)-th triangular number (a(n)^2 + a(n))/2 and multiply it by 3, we get a(n)-th even-length binary palindrome A048701(a(n)) concatenated from a(n) and its reverse. E.g., a(4) = 10, which is 1010 in binary; the tenth triangular number is 55, and 55*3 = 165 = 10100101 in binary. - Antti Karttunen, circa 1999. (This has been now proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016
Number of ways to tie a tie of n or fewer half turns, excluding mirror images. Also number of walks of length n or less on a triangular lattice with the following restrictions; given l, r and c as the lattice axes. 1. All steps are taken in the positive axis direction. 2. No two consecutive steps are taken on the same axis. 3. All walks begin with l. 4. All walks end with rlc or lrc. - Bill Blewett, Dec 21 2000
a(n) is the minimal number of vertices to be selected in a vertex-cover of the balanced tree B_n. - Sen-peng Eu, Jun 15 2002
A087117(a(n)) = A038374(a(n)) = 1 for n > 1; see also A090050. - Reinhard Zumkeller, Nov 20 2003
Intersection of A003754 and A003714; complement of A107907. - Reinhard Zumkeller, May 28 2005
Equivalently, numbers m whose binary representation is effectively, for some number k, both the lazy Fibonacci and the Zeckendorf representation of k (in which case k = A022290(m)). - Peter Munn, Oct 06 2022
a(n+1) gives row sums of Riordan array (1/(1-x), x(1+2x)). - Paul Barry, Jul 18 2005
Total number of initial 01's in all binary words of length n+1. Example: a(3) = 5 because the binary words of length 4 that start with 01 are (01)00, (01)(01), (01)10 and (01)11 and the total number of initial 01's is 5 (shown between parentheses). a(n) = Sum_{k >= 0} k*A119440(n+1, k). - Emeric Deutsch, May 19 2006
In Norway we call the 10-ring puzzle "strikketoy" or "knitwear" (see link). It takes 682 moves to free the two parts. - Hans Isdahl, Jan 07 2008
Equals A002450 and A020988 interlaced. - Zak Seidov, Feb 10 2008
For n > 1, let B_n = the complete binary tree with vertex set V where |V| = 2^n - 1. If VC is a minimum-size vertex cover of B_n, Sen-Peng Eu points out that a(n) = |VC|. It also follows that if IS = V\VC, a(n+1) = |IS|. - K.V.Iyer, Apr 13 2009
Starting with 1 and convolved with [1, 2, 2, 2, ...] = A000295. - Gary W. Adamson, Jun 02 2009
a(n) written in base 2 is sequence A056830(n). - Jaroslav Krizek, Aug 05 2009
This is the sequence A(0, 1; 1, 2; 1) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
From Vladimir Shevelev, Jan 30 2012, Feb 13 2012: (Start)
1) Denote by {n, k} the number of permutations of 1, ..., n with the up-down index k (for definition, see comment in A203827). Then max_k{n, k} = {n, a(n)} = A000111(n).
2) a(n) is the minimal number > a(n-1) with the Hamming distance d_H(a(n-1), a(n)) = n. Thus this sequence is the Hamming analog of triangular numbers 0, 1, 3, 6, 10, ... (End)
From Hieronymus Fischer, Nov 22 2012: (Start)
Represented in binary form each term after the second one contains every previous term as a substring.
The terms a(2) = 2 and a(3) = 5 are the only primes. Proof: For even n we get a(n) = 2*(2^(2*n) - 1)/3, which shows that a(n) is even, too, and obviously a(n) > 2 for all even n > 2. For odd n we have a(n) = (2^(n+1) - 1)/3 = (2^((n+1)/2) - 1) * (2^((n+1)/2) + 1)/3. Evidently, at least one of these factors is divisible by 3, both are greater than 6, provided n > 3. Hence it follows that a(n) is composite for all odd n > 3.
Represented as a binary number, a(n+1) has exactly n prime substrings. Proof: Evidently, a(1) = 1_2 has zero and a(2) = 10_2 has 1 prime substring. Let n > 1. Written in binary, a(n+1) is 101010101...01 (if n + 1 is odd) and is 101010101...10 (if n + 1 is even) with n + 1 digits. Only the 2- and 3-digits substrings 10_2 (=2) and 101_2 (=5) are prime substrings. All the other substrings are nonprime since every substring is a previous term and all terms unequal to 2 and 5 are nonprime. For even n + 1, the number of prime substrings equal to 2 = 10_2 is (n+1)/2, and the number of prime substrings equal to 5 = 101_2 is (n-1)/2, makes a sum of n. For odd n + 1 we get n/2 for both, the number of 2's and 5's prime substrings, in any case, the sum is n. (End)
Number of different 3-colorings for the vertices of all triangulated planar polygons on a base with n+2 vertices if the colors of the two base vertices are fixed. - Patrick Labarque, Feb 09 2013
A090079(a(n)) = a(n) and A090079(m) <> a(n) for m < a(n). - Reinhard Zumkeller, Feb 16 2013
a(n) is the number of length n binary words containing at least one 1 and ending in an even number (possibly zero) of 0's. a(3) = 5 because we have: 001, 011, 100, 101, 111. - Geoffrey Critzer, Dec 15 2013
a(n) is the number of permutations of length n+1 having exactly one descent such that the first element of the permutation is an even number. - Ran Pan, Apr 18 2015
a(n) is the sequence of the last row of the Hadamard matrix H(2^n) obtained via Sylvester's construction: H(2) = [1,1;1,-1], H(2^n) = H(2^(n-1))*H(2), where * is the Kronecker product. - William P. Orrick, Jun 28 2015
Conjectured record values of A264784: a(n) = A264784(A155051(n-1)). - Reinhard Zumkeller, Dec 04 2015. (This is proved by Paul K. Stockmeyer in his arXiv:1608.08245 paper.) - Antti Karttunen, Aug 31 2016
Decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 131", based on the 5-celled von Neumann neighborhood. See A279053 for references and links. - Robert Price, Dec 05 2016
For n > 4, a(n-2) is the second-largest number in row n of A127824. - Dmitry Kamenetsky, Feb 11 2017
Conjecture: a(n+1) is the number of compositions of n with two kinds of parts, n and n', where the order of the 1 and 1' does not matter. For n=2, a(3) = 5 compositions, enumerated as follows:  2; 2'; 1,1; 1',1 = 1',1; 1',1'. - Gregory L. Simay, Sep 02 2017
Conjecture proved by recognizing the appropriate g.f. is x/(1 - x)(1 - x)(1 - 2*x^2 - 2x^3 - ...) =  x/(1 - 2*x - x^2 + 2x^3). - Gregory L. Simay, Sep 10 2017
a(n) = 2^(n-1) + 2^(n-3) + 2^(n-5) + ... a(2*k -1) = A002450(k) is the sum of the powers of 4. a(2*k) = 2*A002450(k). - Gregory L. Simay, Sep 27 2017
a(2*n) = n times the string [10] in binary representation, a(2*n+1) = n times the string [10] followed with [1] in binary representation. Example: a(7) = 85 = (1010101) in binary, a(8) = 170 = (10101010) in binary. - Ctibor O. Zizka, Nov 06 2018
Except for 0, these are the 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. Cuts-resistance 2 is A329862. - Gus Wiseman, Nov 27 2019
From Markus Sigg, Sep 14 2020: (Start)
Let s(k) be the length of the Collatz orbit of k, e.g. s(1) = 1, s(2) = 2, s(3) = 5. Then s(a(n)) = n+3 for n >= 3. Proof by induction: s(a(3)) = s(5) = 6 = 3+3. For odd n >= 5 we have s(a(n)) = s(4*a(n-2)+1) = s(12*a(n-2)+4)+1 = s(3*a(n-2)+1)+3 = s(a(n-2))+2 = (n-2)+3+2 = n+3, and for even n >= 4 this gives s(a(n)) = s(2*a(n-1)) = s(a(n-1))+1 = (n-1)+3+1 = n+3.
Conjecture: For n >= 3, a(n) is the second largest natural number whose Collatz orbit has length n+3. (End)
From Gary W. Adamson, May 14 2021: (Start)
With offset 1 the sequence equals the numbers of 1's from n = 1 to 3, 3 to 7, 7 to 15, ...; of A035263; as shown below:
..1     3           7                      15...
..1  0  1  1  1  0  1  0  1  0  1  1  1  0  1...
..1.....2...........5......................10...; a(n) = Sum_(k=1..2n-1)A035263(k)
.....1...........2.......................5...; as to zeros.
..1's in the Tower of Hanoi game represent CW moves On disks in the pattern:
..0,  1,  2,  0,  1,  2, ... whereas even numbered disks move in the pattern:
..0,  2,  1,  0,  2,  1, ... (End)
Except for 0, numbers that are repunits in Gray-code representation (A014550). - Amiram Eldar, May 21 2021
From Gus Wiseman, Apr 20 2023: (Start)
Also the number of nonempty subsets of {1..n} with integer median, where the median of a multiset is the middle part in the odd-length case, and the average of the two middle parts in the even-length case. For example, the a(1) = 1 through a(4) = 10 subsets are:
  {1}  {1}  {1}      {1}
       {2}  {2}      {2}
            {3}      {3}
            {1,3}    {4}
            {1,2,3}  {1,3}
                     {2,4}
                     {1,2,3}
                     {1,2,4}
                     {1,3,4}
                     {2,3,4}
The complement is counted by A005578.
For mean instead of median we have A051293, counting empty sets A327475.
For normal multisets we have A056450, strongly normal A361202.
For partitions we have A325347, strict A359907, complement A307683.
(End)
In binary form, because the sequence 10101_2... keeps repeating, one can represent a(n) = floor(0.(10)2 * 2^n). Because 0.(10)_2 = 0.(10)_2 * 4 - 2, then 0.(10)_2 = 2/3. Substituting this value in the first expression, we obtain a(n) = floor(2^(n+1)/3). This also agrees with the formula of Paul Barry, Oct 08 2005. - _Giovanni Ciriani, Mar 31 2025

			a(4)=10 since 0001, 0011, 0010, 0110, 0111, 0101, 0100, 1100, 1101, 1111 are the 10 binary strings switching 0000 to 1111.
a(3) = 1 because "lrc" is the only way to tie a tie with 3 half turns, namely, pass the business end of the tie behind the standing part to the left, bring across the front to the right, then behind to the center. The final motion of tucking the loose end down the front behind the "lr" piece is not considered a "step".
a(4) = 2 because "lrlc" is the only way to tie a tie with 4 half turns. Note that since the number of moves is even, the first step is to go to the left in front of the tie, not behind it. This knot is the standard "four in hand", the most commonly known men's tie knot. By contrast, the second most well known tie knot, the Windsor, is represented by "lcrlcrlc".
a(n) = (2^0 - 1) XOR (2^1 - 1) XOR (2^2 - 1) XOR (2^3 - 1) XOR ... XOR (2^n - 1). - _Paul D. Hanna_, Nov 05 2011
G.f. = x + 2*x^2 + 5*x^3 + 10*x^4 + 21*x^5 + 42*x^6 + 85*x^7 + 170*x^8 + ...
a(9) = 341 = 2^8 + 2^6 + 2^4 + 2^2 + 2^0 = 4^4 + 4^3 + 4^2 + 4^1 + 4^0 = A002450(5). a(10) = 682 = 2*a(9) = 2*A002450(5). - _Gregory L. Simay_, Sep 27 2017

Thomas Fink and Yong Mao, The 85 Ways to Tie a Tie, Broadway Books, New York (1999), p. 138.
Clifford A. Pickover, The Math Book, From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics, Sterling Publ., NY, 2009.

G. C. Greubel, Table of n, a(n) for n = 0..3300 (terms 0..300 from T. D. Noe)
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019.
Sergei L. Bezrukov et al., The congestion of n-cube layout on a rectangular grid, Discrete Mathematics 213.1-3 (2000): 13-19. See Theorem 1.
F. Chapoton and S. Giraudo, Enveloping operads and bicoloured noncrossing configurations, arXiv:1310.4521 [math.CO], 2013. Is the sequence in Table 2 this sequence? - _N. J. A. Sloane_, Jan 04 2014. (Yes, it is. See Stockmeyer's arXiv:1608.08245 2016 paper for the proof.)
Ji Young Choi, Ternary Modified Collatz Sequences And Jacobsthal Numbers, Journal of Integer Sequences, Vol. 19 (2016), #16.7.5.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
Madeleine Goertz and Aaron Williams, The Quaternary Gray Code and How It Can Be Used to Solve Ziggurat and Other Ziggu Puzzles, arXiv:2411.19291 [math.CO], 2024. See pp. 1, 17, 42.
David Hayes, Kaveh Khodjasteh, Lorenza Viola and Michael J. Biercuk, Reducing sequencing complexity in dynamical quantum error suppression by Walsh modulation, arXiv:1109.6002 [quant-ph], 2011.
Clemens Heuberger and Daniel Krenn, Esthetic Numbers and Lifting Restrictions on the Analysis of Summatory Functions of Regular Sequences, arXiv:1808.00842 [math.CO], 2018. See p. 10.
Clemens Heuberger and Daniel Krenn, Asymptotic Analysis of Regular Sequences, arXiv:1810.13178 [math.CO], 2018. See p. 29.
Andreas M. Hinz, The Lichtenberg sequence, Fib. Quart., 55 (2017), 2-12.
Andreas M. Hinz, Sandi Klavžar, Uroš Milutinović, and Ciril Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 56. Book's website
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Jia Huang, Nonassociativity of the Norton Algebras of some distance regular graphs, arXiv:2001.05547 [math.CO], 2020.
Jia Huang, Norton algebras of the Hamming Graphs via linear characters, arXiv:2101.05711 [math.CO], 2021.
Jia Huang and Erkko Lehtonen, Associative-commutative spectra for some varieties of groupoids, arXiv:2401.15786 [math.CO], 2024. See p. 17.
Jia Huang, Madison Mickey, and Jianbai Xu, The Nonassociativity of the Double Minus Operation, Journal of Integer Sequences, Vol. 20 (2017), #17.10.3.
Ryota Inagaki, Tanya Khovanova, and Austin Luo, On Chip-Firing on Undirected Binary Trees, Ann. Comb. (2025). See p. 10.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 394
Hans Isdahl, "Knitwear" puzzle
D. E. Knuth and O. P. Lossers, Partitions of a circular set, Problem 11151 in Amer. Math. Monthly 114 (3) (2007) p 265, E_3.
S. Lafortune, A. Ramani, B. Grammaticos, Y. Ohta and K.M. Tamizhmani, Blending two discrete integrability criteria: singularity confinement and algebraic entropy, arXiv:nlin/0104020 [nlin.SI], 2001.
Robert L. Lamphere, A Recurrence Relation in the Spinout Puzzle, The College Mathematics Journal, Vol. 27, Nbr. 4, Page 289, Sep. 96.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
Georg Christoph Lichtenberg, Vermischte Schriften, Band 6 (1805). See chapter 6, p. 257.
Saad Mneimneh, Simple Variations on the Tower of Hanoi to Guide the Study of Recurrences and Proofs by Induction, Department of Computer Science, Hunter College, CUNY, 2019.
Richard Moot, Partial Orders, Residuation, and First-Order Linear Logic, arXiv:2008.06351 [cs.LO], 2020.
Gregg Musiker and Son Nguyen, Labeled Chip-firing on Binary Trees, arXiv:2206.02007 [math.CO], 2022.
Kival Ngaokrajang, Illustration of initial terms
Ahmet Öteleş, On the sum of Pell and Jacobsthal numbers by the determinants of Hessenberg matrices, AIP Conference Proceedings 1863, 310003 (2017).
Vladimir Shevelev, On the Basis Polynomials in the Theory of Permutations with Prescribed Up-Down Structure, arXiv:0801.0072 [math.CO], 2007-2010. See Example 3.
Chloe E. Shiff and Noah A. Rosenberg, Enumeration of rooted binary perfect phylogenies, arXiv:2410.14915 [q-bio.PE], 2024. See pp. 15, 17.
A. V. Sills and H. Wang, On the maximal Wiener index and related questions, Discrete Applied Mathematics, Volume 160, Issues 10-11, July 2012, Pages 1615-1623 - _N. J. A. Sloane_, Sep 21 2012
N. J. A. Sloane, Transforms
Elen Viviani Pereira Spreafico, Francisco Regis Vieira Alves, Paula Maria Machado Cruz Catarino, and Eudes Antonio Costa, A brief study of the one parameter Lichtenberg numbers, VI Int'l Conf. Math. Appl. Sci. Eng. (ICMASE 2025).
Paul K. Stockmeyer, An Exploration of Sequence A000975, arXiv:1608.08245 [math.CO], 2016; Fib. Quart. 55 (2017) 174.
Eric Weisstein's World of Mathematics, Baguenaudier
A. K. Whitford, Binet's Formula Generalized, Fibonacci Quarterly, Vol. 15, No. 1, 1979, pp. 21, 24, 29.
Index entries for sequences related to binary expansion of n
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Partial sums of A001045.
Row sums of triangle A013580.
Equals A026644/2.
Union of the bijections A002450 and A020988. - Robert G. Wilson v, Jun 09 2014
Column k=3 of A261139.
Cf. A000295, A005578, A015441, A043291, A053404, A059260, A077854, A119440, A127824, A130125, A135228, A155051, A179970, A264784.
Complement of A107907.
Row 3 of A300653.
Other sequences that relate to the binary representation of the terms: A003714, A003754, A007088, A022290, A056830, A104161, A107909.
Cf. A000120, A001511, A003242, A027383, A070939, A164707, A295235, A319416.
Cf. A005186, A033491, A153772.
Cf. A014550, A035263
Cf. A051293, A067659, A079309, A231147, A325347, A359893, A359907, A361801.

List([0..35],n->(2^(n+1)-2+(n mod 2))/3); # Muniru A Asiru, Nov 01 2018

a000975 n = a000975_list !! n
a000975_list = 0 : 1 : map (+ 1)
   (zipWith (+) (tail a000975_list) (map (* 2) a000975_list))
-- Reinhard Zumkeller, Mar 07 2012

[(2^(n+1) - 2 + (n mod 2))/3: n in [0..40]]; // Vincenzo Librandi, Mar 18 2015

A000975 := proc(n) option remember; if n <= 1 then n else if n mod 2 = 0 then 2*A000975(n-1) else 2*A000975(n-1)+1 fi; fi; end;
seq(iquo(2^n,3),n=1..33); # Zerinvary Lajos, Apr 20 2008
f:=n-> if n mod 2 = 0 then (2^n-1)/3 else (2^n-2)/3; fi; [seq(f(n),n=0..40)]; # N. J. A. Sloane, Mar 21 2017

Array[Ceiling[2(2^# - 1)/3] &, 41, 0]
RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == a[n - 1] + 2a[n - 2] + 1}, a, {n, 40}] (* or *)
LinearRecurrence[{2, 1, -2}, {0, 1, 2}, 40] (* Harvey P. Dale, Aug 10 2013 *)
f[n_] := Block[{exp = n - 2}, Sum[2^i, {i, exp, 0, -2}]]; Array[f, 33] (* Robert G. Wilson v, Oct 30 2015 *)
f[s_List] := Block[{a = s[[-1]]}, Append[s, If[OddQ@ Length@ s, 2a + 1, 2a]]]; Nest[f, {0}, 32] (* Robert G. Wilson v, Jul 20 2017 *)
NestList[2# + Boole[EvenQ[#]] &, 0, 39] (* Alonso del Arte, Sep 21 2018 *)

{a(n) = if( n<0, 0, 2 * 2^n \ 3)}; /* Michael Somos, Sep 04 2006 */

a(n)=if(n<=0,0,bitxor(a(n-1),2^n-1)) \\ Paul D. Hanna, Nov 05 2011

concat(0, Vec(x/(1-2*x-x^2+2*x^3) + O(x^100))) \\ Altug Alkan, Oct 30 2015

{a(n) = (4*2^n - 3 - (-1)^n) / 6}; /* Michael Somos, Jul 23 2017 */

def a(n): return (2**(n+1) - 2 + (n%2))//3
print([a(n) for n in range(35)]) # Michael S. Branicky, Dec 19 2021

a(n) = ceiling(2*(2^n-1)/3).
Alternating sum transform (PSumSIGN) of {2^n - 1} (A000225).
a(n) = a(n-1) + 2*a(n-2) + 1.
a(n) = 2*2^n/3 - 1/2 - (-1)^n/6.
a(n) = Sum_{i = 0..n} A001045(i), partial sums of A001045. - Bill Blewett
a(n) = n + 2*Sum_{k = 1..n-2} a(k).
G.f.: x/((1+x)*(1-x)*(1-2*x)) = x/(1-2*x-x^2+2*x^3). - Paul Barry, Feb 11 2003
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3). - Paul Barry, Feb 11 2003
a(n) = Sum_{k = 0..floor((n-1)/2)} 2^(n-2*k-1). - Paul Barry, Nov 11 2003
a(n+1) = Sum_{k=0..floor(n/2)} 2^(n-2*k); a(n+1) = Sum_{k = 0..n} Sum_{j = 0..k} (-1)^(j+k)*2^j. - Paul Barry, Nov 12 2003
(-1)^(n+1)*a(n) = Sum_{i = 0..n} Sum_{k = 1..i} k!*k* Stirling2(i, k)*(-1)^(k-1) = (1/3)*(-2)^(n+1)-(1/6)(3*(-1)^(n+1)-1). - Mario Catalani (mario.catalani(AT)unito.it), Dec 22 2003
a(n+1) = (n!/3)*Sum_{i - (-1)^i + j = n, i = 0..n, j = 0..n} 1/(i - (-1)^i)!/j!. - Benoit Cloitre, May 24 2004
a(n) = A001045(n+1) - A059841(n). - Paul Barry, Jul 22 2004
a(n) = Sum_{k = 0..n} 2^(n-k-1)*(1-(-1)^k), row sums of A130125. - Paul Barry, Jul 28 2004
a(n) = Sum_{k = 0..n} binomial(k, n-k+1)2^(n-k); a(n) = Sum_{k = 0..floor(n/2)} binomial(n-k, k+1)2^k. - Paul Barry, Oct 07 2004
a(n) = A107909(A104161(n)); A007088(a(n)) = A056830(n). - Reinhard Zumkeller, May 28 2005
a(n) = floor(2^(n+1)/3) = ceiling(2^(n+1)/3) - 1 = A005578(n+1) - 1. - Paul Barry, Oct 08 2005
Convolution of "Number of fixed points in all 231-avoiding involutions in S_n." (A059570) with "1-n" (A024000), treating the result as if offset was 0. - Graeme McRae, Jul 12 2006
a(n) = A081254(n) - 2^n. - Philippe Deléham, Oct 15 2006
Starting (1, 2, 5, 10, 21, 42, ...), these are the row sums of triangle A135228. - Gary W. Adamson, Nov 23 2007
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0] = [A005578(n), A001045(n), a(n-1)]. - Gary W. Adamson, Dec 25 2007
2^n = 2*A005578(n-1) + 2*A001045(n) + 2*a(n-2). - Gary W. Adamson, Dec 25 2007
If we define f(m,j,x) = Sum_{k=j..m} binomial(m,k)*stirling2(k,j)*x^(m-k) then a(n-3) = (-1)^(n-1)*f(n,3,-2), (n >= 3). - Milan Janjic, Apr 26 2009
a(n) + A001045(n) = A166920(n). a(n) + A001045(n+2) = A051049(n+1). - Paul Curtz, Oct 29 2009
a(n) = floor(A051049(n+1)/3). - Gary Detlefs, Dec 19 2010
a(n) = round((2^(n+2)-3)/6) = floor((2^(n+1)-1)/3) = round((2^(n+1)-2)/3); a(n) = a(n-2) + 2^(n-1), n > 1. - Mircea Merca, Dec 27 2010
a(n) = binary XOR of 2^k-1 for k=0..n. - Paul D. Hanna, Nov 05 2011
E.g.f.: 2/3*exp(2*x) - 1/2*exp(x) - 1/6*exp(-x) = 2/3*U(0); U(k) = 1 - 3/(4*(2^k) - 4*(2^k)/(1+3*(-1)^k - 24*x*(2^k)/(8*x*(2^k)*(-1)^k - (k+1)/U(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2011
Starting with "1" = triangle A059260 * [1, 2, 2, 2, ...] as a vector. - Gary W. Adamson, Mar 06 2012
a(n) = 2*(2^n - 1)/3, for even n; a(n) = (2^(n+1) - 1)/3 = (1/3)*(2^((n+1)/2) - 1)*(2^((n+1)/2) + 1), for odd n. - Hieronymus Fischer, Nov 22 2012
a(n) + a(n+1) = 2^(n+1) - 1. - Arie Bos, Apr 03 2013
G.f.: Q(0)/(3*(1-x)), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 21 2013
floor(a(n+2)*3/5) = A077854(n), for n >= 0. - Armands Strazds, Sep 21 2014
a(n) = (2^(n+1) - 2 + (n mod 2))/3. - Paul Toms, Mar 18 2015
a(0) = 0, a(n) = 2*(a(n-1)) + (n mod 2). - Paul Toms, Mar 18 2015
Binary: a(n) = (a(n-1) shift left 1) + (a(n-1)) NOR (...11110). - Paul Toms, Mar 18 2015
Binary: for n > 1, a(n) = 2*a(n-1) OR a(n-2). - Stanislav Sykora, Nov 12 2015
a(n) = A266613(n) - 20*2^(n-5), for n > 2. - Andres Cicuttin, Mar 31 2016
From Michael Somos, Jul 23 2017: (Start)
a(n) = -(2^n)*a(-n) for even n; a(n) = -(2^(n+1))*a(-n) + 1 for odd n.
0 = +a(n)*(+2 +4*a(n) -4*a(n+1)) + a(n+1)*(-1 +a(n+1)) for all n in Z. (End)
G.f.: (x^1+x^3+x^5+x^7+...)/(1-2*x). - Gregory L. Simay, Sep 27 2017
a(n+1) = A051049(n) + A001045(n). - Yuchun Ji, Jul 12 2018
a(n) = A153772(n+3)/4. - Markus Sigg, Sep 14 2020
a(4*k+d) = 2^(d+1)*a(4*k-1) + a(d), a(n+4) = a(n) + 2^n*10, a(0..3) = [0,1,2,5]. So the last digit is always 0,1,2,5 repeated. - Yuchun Ji, May 22 2023

Additional comments from Barry E. Williams, Jan 10 2000

A333254 Lengths of maximal runs in the sequence of prime gaps (A001223).

Original entry on oeis.org

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

Offset: 1

Gus Wiseman, Mar 20 2020

nonn

Prime gaps are differences between adjacent prime numbers.

Also lengths of maximal arithmetic progressions of consecutive primes.

			The prime gaps split into the following runs: (1), (2,2), (4), (2), (4), (2), (4), (6), (2), (6), (4), (2), (4), (6,6), (2), (6), (4), ...

Robert Israel, Table of n, a(n) for n = 1..10000
Wikipedia, Arithmetic progression
Wikipedia, Longest increasing subsequence

The version for A000002 is A000002. Similarly for A001462.
The unequal version is A333216.
The weakly decreasing version is A333212.
The weakly increasing version is A333215.
The strictly decreasing version is A333252.
The strictly increasing version is A333253.
Positions of first appearances are A335406.
The first term of the first length-n arithmetic progression of consecutive primes is A006560(n), with index A089180(n).
Prime gaps are A001223.
Positions of adjacent equal prime gaps are A064113.
Positions of adjacent unequal prime gaps are A333214.
Cf. A000040, A031217, A054800, A059044, A084758, A090832, A124767, A238279, A295235.

p:= 3: t:= 1: R:= NULL: s:= 1: count:= 0:
for i from 2 while count < 100 do
  q:= nextprime(p);
  g:= q-p; p:= q;
  if g = t then s:= s+1
  else count:= count+1; R:= R, s; t:= g; s:= 1;
  fi
od:
R; # Robert Israel, Jan 06 2021

Length/@Split[Differences[Array[Prime,100]],#1==#2&]//Most

Partial sums are A333214.

A291166 Connected Haar graph numbers.

Original entry on oeis.org

1, 3, 5, 6, 7, 9, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 35, 37, 38, 39, 41, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81

Offset: 1

Eric W. Weisstein, Aug 19 2017

nonn

Complement of A291165.

These appear to be numbers whose positions of 1's in their reversed binary expansion are relatively prime. If so, this sequence lists all positions of 1's in A326674. Numbers whose positions of 1's in their reversed binary expansion are pairwise coprime (as opposed to relatively prime) are A326675. - Gus Wiseman, Jul 19 2019

Eric Weisstein's World of Mathematics, Connected Graph
Eric Weisstein's World of Mathematics, Haar Graph

Cf. A000120, A029931, A048793, A059519, A070939, A289508, A291165, A295235, A319826, A326031, A326669, A326672, A326673.

A333224 Number of distinct positive consecutive subsequence-sums of the k-th composition in standard order.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Mar 18 2020

nonn

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.

			The composition (4,3,1,2) has positive subsequence-sums 1, 2, 3, 4, 6, 7, 8, 10, so a(550) = 8.

Dominated by A124770.
Compositions where every subinterval has a different sum are counted by A169942 and A325677 and ranked by A333222. The case of partitions is counted by A325768 and ranked by A325779.
Positive subset-sums of partitions are counted by A276024 and A299701.
Knapsack partitions are counted by A108917 and A325592 and ranked by A299702.
Strict knapsack partitions are counted by A275972 and ranked by A059519 and A301899.
Knapsack compositions are counted by A325676 and A325687 and ranked by A333223. The case of partitions is counted by A325769 and ranked by A325778, for which the number of distinct consecutive subsequences is given by A325770.
Allowing empty subsequences gives A333257.
Cf. A000120, A003022, A029931, A048793, A066099, A070939, A103295, A124767, A143823, A295235, A325680, A333217.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Length[Union[ReplaceList[stc[n],{_,s__,_}:>Plus[s]]]],{n,0,100}]

a(n) = A333257(n) - 1.

A333222 Numbers k such that every restriction of the k-th composition in standard order to a subinterval has a different sum.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 9, 12, 16, 17, 18, 20, 24, 32, 33, 34, 40, 41, 48, 50, 64, 65, 66, 68, 69, 70, 72, 80, 81, 88, 96, 98, 104, 128, 129, 130, 132, 133, 134, 144, 145, 160, 161, 176, 192, 194, 196, 208, 256, 257, 258, 260, 261, 262, 264, 265, 268, 272, 274

Offset: 1

Gus Wiseman, Mar 17 2020

nonn

Also numbers whose binary indices together with 0 define a Golomb ruler.

			The list of terms together with the corresponding compositions begins:
    0: ()        41: (2,3,1)    130: (6,2)      262: (6,1,2)
    1: (1)       48: (1,5)      132: (5,3)      264: (5,4)
    2: (2)       50: (1,3,2)    133: (5,2,1)    265: (5,3,1)
    4: (3)       64: (7)        134: (5,1,2)    268: (5,1,3)
    5: (2,1)     65: (6,1)      144: (3,5)      272: (4,5)
    6: (1,2)     66: (5,2)      145: (3,4,1)    274: (4,3,2)
    8: (4)       68: (4,3)      160: (2,6)      276: (4,2,3)
    9: (3,1)     69: (4,2,1)    161: (2,5,1)    288: (3,6)
   12: (1,3)     70: (4,1,2)    176: (2,1,5)    289: (3,5,1)
   16: (5)       72: (3,4)      192: (1,7)      290: (3,4,2)
   17: (4,1)     80: (2,5)      194: (1,5,2)    296: (3,2,4)
   18: (3,2)     81: (2,4,1)    196: (1,4,3)    304: (3,1,5)
   20: (2,3)     88: (2,1,4)    208: (1,2,5)    320: (2,7)
   24: (1,4)     96: (1,6)      256: (9)        321: (2,6,1)
   32: (6)       98: (1,4,2)    257: (8,1)      324: (2,4,3)
   33: (5,1)    104: (1,2,4)    258: (7,2)      328: (2,3,4)
   34: (4,2)    128: (8)        260: (6,3)      352: (2,1,6)
   40: (2,4)    129: (7,1)      261: (6,2,1)    384: (1,8)

Wikipedia, Golomb ruler

A subset of A233564.
Also a subset of A333223.
These compositions are counted by A169942 and A325677.
The number of distinct nonzero subsequence-sums is A333224.
The number of distinct subsequence-sums is A333257.
Lengths of optimal Golomb rulers are A003022.
Inequivalent optimal Golomb rulers are counted by A036501.
Complete rulers are A103295, with perfect case A103300.
Knapsack partitions are counted by A108917, with strict case A275972.
Distinct subsequences are counted by A124770 and A124771.
Golomb subsets are counted by A143823.
Heinz numbers of knapsack partitions are A299702.
Knapsack compositions are counted by A325676.
Maximal Golomb rulers are counted by A325683.
Cf. A000120, A029931, A048793, A066099, A070939, A228351, A295235, A325678, A325680, A325768, A325779, A333217.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Select[Range[0,300],UnsameQ@@ReplaceList[stc[#],{_,s__,_}:>Plus[s]]&]

A059519 Number of partitions of n all of whose subpartitions sum to distinct values. Partition(n) = [a, b, c...] where 2n = 2^a + 2^b + 2^c + ...

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 14, 16, 17, 18, 19, 20, 21, 24, 26, 28, 32, 33, 34, 35, 36, 37, 38, 40, 41, 44, 48, 50, 52, 56, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 80, 81, 84, 88, 96, 98, 100, 104, 112, 116, 128, 129, 130, 131, 132, 133, 134, 136, 137, 138, 139, 140

Offset: 1

Marc LeBrun, Jan 19 2001

Partition encoding as in A029931. Complement of A059520.
From Gus Wiseman, Jul 22 2019: (Start)
These are numbers whose positions of 1's in their reversed binary expansion form a strict knapsack partition (A275972). The initial terms together with their corresponding partitions are:
(1)
(2)
(2,1)
(3)
(3,1)
(3,2)
(4)
(4,1)
(4,2)
(4,2,1)
(4,3)
(4,3,2)
(5)
(5,1)
(5,2)
(5,2,1)
(5,3)
(End)

			14=2+4+8 so Partition(14) = [2,3,4], whose sub-sums are 0,2,3,4,5,6,7 and 14.

Cf. A000120, A029931, A048793, A059520, A070939, A108917, A272020, A275972, A299702, A326015.

Other sequences classifying numbers by their binary indices: A291166 (relatively prime), A295235 (arithmetic progression), A326669 (integer average), A326675 (pairwise coprime).

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Select[Range[100],UnsameQ@@Total/@Subsets[bpe[#]]&] (* Gus Wiseman, Jul 22 2019 *)

A326699 Numerator of the average position of a 1 in the reversed binary expansion of n.

Original entry on oeis.org

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

Offset: 1

Gus Wiseman, Jul 20 2019

			The sequence of fractions begins: 1, 2, 3/2, 3, 2, 5/2, 2, 4, 5/2, 3, 7/3, 7/2, 8/3, 3, 5/2, 5, 3, 7/2, 8/3, 4.
For example, the reversed binary expansion of 18 is (0,1,0,0,1), and the average of {2,5} is 7/2, so a(18) = 7.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000120, A029931, A035327, A048793, A070939, A291166, A295235, A326669, A326674, A326700 (denominators).

f:= proc(n) local L;
  L:= convert(n,base,2);
  L:= select(t -> L[t]=1, [$1..nops(L)]);
  numer(convert(L,`+`)/nops(L))
end proc:
map(f, [$1..100]); # Robert Israel, Oct 07 2019

Table[Numerator[Mean[Join@@Position[Reverse[IntegerDigits[n,2]],1]]],{n,100}]

A326700 Denominator of the average position of a 1 in the reversed binary expansion of n.

Original entry on oeis.org

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

Offset: 1

Gus Wiseman, Jul 20 2019

The sequence of fractions begins: 1, 2, 3/2, 3, 2, 5/2, 2, 4, 5/2, 3, 7/3, 7/2, 8/3, 3, 5/2, 5, 3, 7/2, 8/3, 4.
For example, the reversed binary expansion of 18 is (0,1,0,0,1), and the average of {2,5} is 7/2, so a(18) = 2.
a(n) divides A000120(n). - Robert Israel, Oct 07 2019

Robert Israel, Table of n, a(n) for n = 1..10000

Positions of 1's are A326669.

Cf. A000120, A029931, A035327, A048793, A070939, A291166, A295235, A326674, A326699 (numerators).

f:= proc(n) local L;
  L:= convert(n,base,2);
  L:= select(t -> L[t]=1, [$1..nops(L)]);
  denom(convert(L,`+`)/nops(L))
end proc:
map(f, [$1..100]); # Robert Israel, Oct 07 2019

Table[Denominator[Mean[Join@@Position[Reverse[IntegerDigits[n,2]],1]]],{n,100}]

cp's OEIS Frontend

A048793 List giving all subsets of natural numbers arranged in standard statistical (or Yates) order.

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A029931 If 2n = Sum 2^e_i, a(n) = Sum e_i.

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A000975 a(2n) = 2*a(2n-1), a(2n+1) = 2*a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).

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A333254 Lengths of maximal runs in the sequence of prime gaps (A001223).

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A291166 Connected Haar graph numbers.

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A333224 Number of distinct positive consecutive subsequence-sums of the k-th composition in standard order.

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A000975 a(2n) = 2a(2n-1), a(2n+1) = 2a(2n)+1 (also a(n) is the n-th number without consecutive equal binary digits).