A029931 -id:A029931 - cp's OEIS frontend

A124771 Number of distinct subsequences for compositions in standard order.

1, 2, 2, 3, 2, 4, 4, 4, 2, 4, 3, 6, 4, 6, 6, 5, 2, 4, 4, 6, 4, 6, 6, 8, 4, 6, 6, 9, 6, 9, 8, 6, 2, 4, 4, 6, 3, 7, 7, 8, 4, 7, 4, 9, 7, 8, 9, 10, 4, 6, 7, 9, 7, 9, 8, 12, 6, 9, 9, 12, 8, 12, 10, 7, 2, 4, 4, 6, 4, 7, 7, 8, 4, 6, 6, 10, 6, 10, 10, 10, 4, 7, 6, 10, 6, 8, 9, 12, 7, 10, 9, 12, 10, 12, 12, 12, 4

Offset: 0

Franklin T. Adams-Watters, Nov 06 2006

The standard order of compositions is given by A066099.
From Vladimir Shevelev, Dec 18 2013: (Start)
Every number in binary is a concatenation of parts of the form 10...0 with k>=0 zeros. For example, 5=(10)(1), 11=(10)(1)(1), 7=(1)(1)(1). We call d>0 a c-divisor of m, if d consists of some consecutive parts of m taking from the left to the right. Note that, to d=0 corresponds an empty set of parts. So it is natural to consider 0 as a c-divisor of every m. For example, 5=(10)(1) is a divisor of 23=(10)(1)(1)(1). Analogously, 1,2,3,7,11,23 are c-divisors of 23. But 6=(1)(10) is not a c-divisor of 23.
One can prove a one-to-one correspondence between distinct subsequences for  composition no. n in standard order and c-divisors of n. So, the sequence lists also numbers of c-divisors of nonnegative integers.
(End)
These are contiguous subsequences, or restrictions to a subinterval. The case for all subsequences is A334299. - Gus Wiseman, Jun 02 2020

			Composition number 11 is 2,1,1; the subsequences are (empty); 1; 2; 1,1; 2,1; 2,1,1; so a(11) = 6.
The table starts:
1
2
1 2
1 3 3 3
Let n=11=(10)(1)(1). We have the following c-divisors of 11: 0,1,2,3,5,11. Thus a(11)=6. Note, that 3=(1)(1) is not a c-divisor of 13=(1)(10)(1) since, although it contains parts of 3=(1)(1), but in non-consecutive order. The c-divisors of 13 are 0,1,2,5,6,13. So, a(13)=6.
From _Gus Wiseman_, Jun 01 2020: (Start)
The c-divisors of n are given in column n below:
  0  0  0  0  0  0  0  0  0  0  0   0   0   0   0   0   0   0   0
     1  2  1  4  1  1  1  8  1  2   1   1   1   1   1   16  1   2
           3     2  2  3     4  10  2   4   2   2   3       8   4
                 5  6  7     9      3   12  5   3   7       17  18
                                    5       6   6   15
                                    11      13  14
(End)

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

Cf. A000005, A011782 (row lengths), A066099, A114994, A233249, A233312.
Not allowing empty subsequences gives A124770.
Dominates A333257.
The case for not just contiguous subsequences is A334299.
Positions of first appearances are A335279.
Compositions where every subinterval has a different sum are A333222.
Knapsack compositions are A333223.
Cf. A000120, A003022, A029931, A070939, A108917, A124767, A325680, A325770, A333224, A334967, A334968.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Length[Union[ReplaceList[stc[n],{_,s___,_}:>{s}]]],{n,0,100}] (* Gus Wiseman, Jun 01 2020 *)

a(n) = A124770(n) + 1.
From Vladimir Shevelev, Dec 18 2013: (Start)
a(2^n) = 2. Note that in concatenation representations of integers in binary, numbers {2^k}, k>=0, play the role of primes. So the formula is an analog of A000005(prime(n))=2.
a(2^n-1) = n+1; for n>=2, a(2^n+1) = 4.
For c-equivalent numbers n_1 and n_2 (i.e., differed only by order of parts) we have a(n_1) = a(n_2). For example, a(24)=a(17)=4. If the canonical representation of n is n=(1)^k_1[*](10)^k_2[*](100)^k_3[*]... , where [*] denotes operation of concatenation (cf. A233569), then a(n)<=(k_1+1)*(k_2+1)*...
(End)

A230877 If n = Sum_{i=0..m} c(i)2^i, c(i) = 0 or 1, then a(n) = Sum_{i=0..m} (m+1-i)c(i).

Original entry on oeis.org

0, 1, 1, 3, 1, 4, 3, 6, 1, 5, 4, 8, 3, 7, 6, 10, 1, 6, 5, 10, 4, 9, 8, 13, 3, 8, 7, 12, 6, 11, 10, 15, 1, 7, 6, 12, 5, 11, 10, 16, 4, 10, 9, 15, 8, 14, 13, 19, 3, 9, 8, 14, 7, 13, 12, 18, 6, 12, 11, 17, 10, 16, 15, 21, 1, 8, 7, 14, 6, 13, 12, 19, 5, 12, 11, 18, 10, 17, 16, 23, 4, 11, 10, 17, 9, 16, 15, 22, 8, 15, 14

Offset: 0

N. J. A. Sloane, Nov 08 2013

Suggested by Jon Perry's A231204, only now the leading power of 2 gets weight 1.

			For n=13 we have 1101, so we add 1+2+4, getting a(13)=7.

Rémy Sigrist, Table of n, a(n) for n = 0..8192

Cf. A029931, A231204, A329752.

f:=proc(n) local t1,m,i;
t1:=convert(n,base,2);
m:=nops(t1)-1;
add((m+1-i)*t1[i+1], i=0..m);
end;

Array[Total[Position[IntegerDigits[#, 2], 1], 2] &, 100, 0] (* Paolo Xausa, Mar 18 2024 *)

a(n) = { my (b=binary(n)); sum(k=1, #b, b[k]*k) } \\ Rémy Sigrist, Jun 25 2021

def A230877(n): return sum(i for i, j in enumerate(bin(n)[2:],1) if j=='1') # Chai Wah Wu, Jan 09 2023

A333257 Number of distinct consecutive subsequence-sums of the k-th composition in standard order.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Mar 20 2020

nonn

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.

			The ninth composition in standard order is (3,1), which has consecutive subsequences (), (1), (3), (3,1), with sums 0, 1, 3, 4, so a(9) = 4.

Dominated by A124771.
Compositions where every subinterval has a different sum are counted by A169942 and A325677 and ranked by A333222, while 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 ranked by A299702.
Knapsack compositions are counted by A325676 and A325687 and ranked by A333223.
The version for Heinz numbers of partitions is A325770.
Not allowing empty subsequences gives A333224.
Cf. A000120, A029931, A048793, A059519, A066099, A070939, A114994, A124765, A124767, A233564, A272919, A325778, 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) = A333224(n) + 1.

A358194 Irregular triangle read by rows where T(n,k) is the number of integer partitions of n with partial sums summing to k, where k ranges from n to n(n+1)/2.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Dec 31 2022

The partial sums of a sequence (a, b, c, ...) are (a, a+b, a+b+c, ...).

			Triangle begins:
  1
  1
  1 1
  1 0 1 1
  1 0 1 1 0 1 1
  1 0 0 1 1 0 1 1 0 1 1
  1 0 0 1 1 1 1 0 1 1 0 1 1 0 1 1
  1 0 0 0 1 1 1 0 1 1 1 1 1 0 1 1 0 1 1 0 1 1
  1 0 0 0 1 1 1 1 0 1 1 1 2 1 0 1 1 1 1 1 0 1 1 0 1 1 0 1 1
For example, the T(15,59) = 5 partitions are: (8,2,2,2,1), (7,3,3,1,1), (6,5,2,1,1), (4,3,2,2,2,2), (3,3,3,3,2,1).

Row sums are A000041.
The version for compositions is A053632.
Row lengths are A152947.
The version for reversed partitions is A264034.
A048793 = partial sums of reversed standard compositions, sum A029931.
A358134 = partial sums of standard compositions, sum A359042.
A358136 = partial sums of prime indices, sum A318283.
A359361 = partial sums of reversed prime indices, sum A304818.
Cf. A000009, A325362, A358137, A359397.

Table[Length[Select[IntegerPartitions[n],Total[Accumulate[#]]==k&]],{n,0,8},{k,n,n*(n+1)/2}]

A089633 Numbers having no more than one 0 in their binary representation.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 11, 13, 14, 15, 23, 27, 29, 30, 31, 47, 55, 59, 61, 62, 63, 95, 111, 119, 123, 125, 126, 127, 191, 223, 239, 247, 251, 253, 254, 255, 383, 447, 479, 495, 503, 507, 509, 510, 511, 767, 895, 959, 991, 1007, 1015, 1019, 1021, 1022, 1023

Offset: 0

Reinhard Zumkeller, Jan 01 2004

Complement of A158582. - Reinhard Zumkeller, Apr 16 2009
Also union of A168604 and A030130. - Douglas Latimer, Jul 19 2012
Numbers of the form 2^t - 2^k - 1, 0 <= k < t.
n is in the sequence if and only if 2*n+1 is in the sequence. - Robert Israel, Dec 14 2018
Also the least binary rank of a strict integer partition of n, where the binary rank of a partition y is given by Sum_i 2^(y_i-1). - Gus Wiseman, May 24 2024

			From _Tilman Piesk_, May 09 2012: (Start)
This may also be viewed as a triangle:             In binary:
                  0                                         0
               1     2                                 01       10
             3    5    6                          011      101      110
           7   11   13   14                  0111     1011     1101     1110
        15   23   27   29   30          01111    10111    11011    11101    11110
      31  47   55   59   61   62
   63   95  111  119  123  125  126
Left three diagonals are A000225,  A055010, A086224. Right diagonal is A000918. Central column is A129868. Numbers in row n (counted from 0) have n binary 1s. (End)
From _Gus Wiseman_, May 24 2024: (Start)
The terms together with their binary expansions and binary indices begin:
   0:      0 ~ {}
   1:      1 ~ {1}
   2:     10 ~ {2}
   3:     11 ~ {1,2}
   5:    101 ~ {1,3}
   6:    110 ~ {2,3}
   7:    111 ~ {1,2,3}
  11:   1011 ~ {1,2,4}
  13:   1101 ~ {1,3,4}
  14:   1110 ~ {2,3,4}
  15:   1111 ~ {1,2,3,4}
  23:  10111 ~ {1,2,3,5}
  27:  11011 ~ {1,2,4,5}
  29:  11101 ~ {1,3,4,5}
  30:  11110 ~ {2,3,4,5}
  31:  11111 ~ {1,2,3,4,5}
  47: 101111 ~ {1,2,3,4,6}
  55: 110111 ~ {1,2,3,5,6}
  59: 111011 ~ {1,2,4,5,6}
  61: 111101 ~ {1,3,4,5,6}
  62: 111110 ~ {2,3,4,5,6}
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Vladimir Shevelev, On the Basis Polynomials in the Theory of Permutations with Prescribed Up-Down Structure, arXiv:0801.0072 [math.CO], 2007-201. See Section 14.
Vladimir Shevelev, Binomial Coefficient Predictors, Journal of Integer Sequences, Vol. 14 (2011), Article 11.2.8.
Vladimir Shevelev, The number of permutations with prescribed up-down structure as a function of two variables, INTEGERS, 12 (2012), #A1.
Index entries for sequences related to binary expansion of n.

Cf. A000120, A007088, A011371, A014132, A023416, A023532, A029931.
Cf. A181741 (primes), union of A081118 and A000918, apart from initial -1.
Cf. A065442, A160502, A265705.
For least binary index (instead of rank) we have A001511.
Applying A019565 (Heinz number of binary indices) gives A077011.
For greatest binary index we have A029837 or A070939, opposite A070940.
Row minima of A118462 (binary ranks of strict partitions).
For sum instead of minimum we have A372888, non-strict A372890.
A000009 counts strict partitions, ranks A005117.
A048675 gives binary rank of prime indices, distinct A087207.
A048793 lists binary indices, product A096111, reverse A272020.
A277905 groups all positive integers by binary rank of prime indices.
Cf. A000041, A005940, A018819, A147655, A225620, A246868.

a089633 n = a089633_list !! (n-1)
a089633_list = [2 ^ t - 2 ^ k - 1 | t <- [1..], k <- [t-1,t-2..0]]
-- Reinhard Zumkeller, Feb 23 2012

seq(seq(2^a-1-2^b,b=a-1..0,-1),a=1..11); # Robert Israel, Dec 14 2018

fQ[n_] := DigitCount[n, 2, 0] < 2; Select[ Range[0, 2^10], fQ] (* Robert G. Wilson v, Aug 02 2012 *)

{insq(n) = local(dd, hf, v); v=binary(n);hf=length(v);dd=sum(i=1,hf,v[i]);if(dd<=hf-2,-1,1)}
{for(w=0,1536,if(insq(w)>=0,print1(w,", ")))}
\\ Douglas Latimer, May 07 2013

isoka(n) = #select(x->(x==0), binary(n)) <= 1; \\ Michel Marcus, Dec 14 2018

from itertools import count, islice
def A089633_gen(): # generator of terms
    return ((1<A089633_list = list(islice(A089633_gen(),30)) # Chai Wah Wu, Feb 10 2023

from math import isqrt, comb
def A089633(n): return (1<<(a:=(isqrt((n<<3)+1)-1>>1)+1))-(1<Chai Wah Wu, Dec 19 2024

A023416(a(n)) <= 1; A023416(a(n)) = A023532(n-2) for n>1;
A000120(a(u)) <= A000120(a(v)) for uA000120(a(n)) = A003056(n).
a(0)=0, n>0: a(n+1) = Min{m>n: BinOnes(a(n))<=BinOnes(m)} with BinOnes=A000120.
If m = floor((sqrt(8*n+1) - 1) / 2), then a(n) = 2^(m+1) - 2^(m*(m+3)/2 - n) - 1. - Carl R. White, Feb 10 2009
A029931(a(n)) = n and A029931(m) != n for m < a(n). - Reinhard Zumkeller, Feb 28 2014
A265705(a(n),k) = A265705(a(n),a(n)-k), k = 0 .. a(n). - Reinhard Zumkeller, Dec 15 2015
a(A014132(n)-1) = 2*a(n-1)+1 for n >= 1. - Robert Israel, Dec 14 2018
Sum_{n>=1} 1/a(n) = A065442 + A160502 = 3.069285887459... . - Amiram Eldar, Jan 09 2024
A019565(a(n)) = A077011(n). - Gus Wiseman, May 24 2024

A326753 Number of connected components of the set-system with BII-number n.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jul 23 2019

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793. We define the set-system with BII-number n to be obtained by taking the binary indices of each binary index of n. Every finite set of finite nonempty sets has a different BII-number. For example, 18 has reversed binary expansion (0,1,0,0,1), and since the binary indices of 2 and 5 are {2} and {1,3} respectively, the BII-number of {{2},{1,3}} is 18. Elements of a set-system are sometimes called edges.

			The set-system {{1,2},{1,4},{3}} with BII-number 268 has two connected components, so a(268) = 2.

John Tyler Rascoe, Table of n, a(n) for n = 0..10000
John Tyler Rascoe, Log scatterplot of a(n), n=0..32906.

Positions of 0's and 1's are A326749.
Cf. A000120, A001187, A029931, A048143, A048793, A070939, A072639, A304716, A305078, A305079 (same for MM-numbers), A323818, A326031, A326702.
Ranking sequences using BII-numbers: A309314 (hyperforests), A326701 (set partitions), A326703 (chains), A326704 (antichains), A326750 (clutters), A326751 (blobs), A326752 (hypertrees), A326754 (covers).

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
csm[s_]:=With[{c=Select[Tuples[Range[Length[s]],2],And[OrderedQ[#],UnsameQ@@#,Length[Intersection@@s[[#]]]>0]&]},If[c=={},s,csm[Sort[Append[Delete[s,List/@c[[1]]],Union@@s[[c[[1]]]]]]]]];
Table[Length[csm[bpe/@bpe[n]]],{n,0,100}]

from sympy.utilities.iterables import connected_components
def bin_i(n): #binary indices
    return([(i+1) for i, x in enumerate(bin(n)[2:][::-1]) if x =='1'])
def A326753(n):
    E,a = [],[bin_i(k) for k in bin_i(n)]
    m = len(a)
    for i in range(m):
        for j in a[i]:
            for k in range(m):
                if j in a[k]:
                    E.append((i,k))
    return(len(connected_components((list(range(m)),E)))) # John Tyler Rascoe, Jul 16 2024

a(A072639(n)) = n. - John Tyler Rascoe, Jul 15 2024

A333766 Maximum part of the n-th composition in standard order. a(0) = 0.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Apr 05 2020

nonn

One plus the longest run of 0's in the binary expansion of n.

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 100th composition in standard order is (1,3,3), so a(100) = 3.

Positions of ones are A000225.
Positions of terms <= 2 are A003754.
The version for prime indices is A061395.
Positions of terms > 1 are A062289.
Positions of first appearances are A131577.
The minimum part is given by A333768.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Compositions without 1's are A022340.
- Sum is A070939.
- Product is A124758.
- 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, A087117, A228351, A328594, A333217, A333218, A333219, A333632, A333767.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[If[n==0,0,Max@@stc[n]],{n,0,100}]

For n > 0, a(n) = A087117(n) + 1.

A320387 Number of partitions of n into distinct parts such that the successive differences of consecutive parts are nonincreasing, and first difference <= first part.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 3, 2, 2, 4, 3, 4, 5, 3, 5, 7, 4, 7, 8, 6, 8, 11, 7, 9, 13, 9, 11, 16, 12, 15, 18, 13, 17, 20, 17, 21, 24, 19, 24, 30, 22, 28, 34, 26, 34, 38, 30, 37, 43, 37, 42, 48, 41, 50, 58, 48, 55, 64, 53, 64, 71, 59, 73, 81, 69, 79, 89, 79, 90, 101, 87, 100, 111

Offset: 0

Seiichi Manyama, Oct 12 2018

nonn

Partitions are usually written with parts in descending order, but the conditions are easier to check "visually" if written in ascending order.
Generating function of the "second integrals" of partitions: given a partition (p_1, ..., p_s) written in weakly decreasing order, write the sequence B = (b_1, b_2, ..., b_s) = (p_1, p_1 + p_2, ..., p_1 + ... + p_s).  The sequence gives the coefficients of the generating function summing q^(b_1 + ... + b_s) over all partitions of all nonnegative integers. - William J. Keith, Apr 23 2022
From Gus Wiseman, Jan 17 2023: (Start)
Equivalently, a(n) is the number of multisets (weakly increasing sequences of positive integers) with weighted sum n. For example, the Heinz numbers of the a(0) = 1 through a(15) = 7 multisets are:
  1  2  3  4  7  6   8   10  15  12  16  18  20  26  24  28
           5     11  9   17  19  14  21  22  27  41  30  32
                     13          23  29  31  33  55  39  34
                                 25      35  37      43  45
                                             49      77  47
                                                         65
                                                         121
These multisets are counted by A264034. The reverse version is A007294. The zero-based version is A359678.
(End)

			There are a(29) = 15 such partitions of 29:
  01: [29]
  02: [10, 19]
  03: [11, 18]
  04: [12, 17]
  05: [13, 16]
  06: [14, 15]
  07: [5, 10, 14]
  08: [6, 10, 13]
  09: [6, 11, 12]
  10: [7, 10, 12]
  11: [8, 10, 11]
  12: [3, 6, 9, 11]
  13: [5, 7, 8, 9]
  14: [2, 4, 6, 8, 9]
  15: [3, 5, 6, 7, 8]
There are a(30) = 18 such partitions of 30:
  01: [30]
  02: [10, 20]
  03: [11, 19]
  04: [12, 18]
  05: [13, 17]
  06: [14, 16]
  07: [5, 10, 15]
  08: [6, 10, 14]
  09: [6, 11, 13]
  10: [7, 10, 13]
  11: [7, 11, 12]
  12: [8, 10, 12]
  13: [3, 6, 9, 12]
  14: [9, 10, 11]
  15: [4, 7, 9, 10]
  16: [2, 4, 6, 8, 10]
  17: [6, 7, 8, 9]
  18: [4, 5, 6, 7, 8]

Fausto A. C. Cariboni, Table of n, a(n) for n = 0..2000 (terms 0..300 from Seiichi Manyama)

Cf. A007294, A179254, A179255, A179269, A320382, A320385, A320388.
Number of appearances of n > 0 in A304818, reverse A318283.
A053632 counts compositions by weighted sum.
A358194 counts partitions by weighted sum, reverse A264034.
Weighted sum of prime indices: A359497, A359676, A359682, A359754, A359755.
Cf. A029931, A359361, A359397, A359674, A359677, A359678.

prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
ots[y_]:=Sum[i*y[[i]],{i,Length[y]}];
Table[Length[Select[Range[2^n],ots[prix[#]]==n&]],{n,10}] (* Gus Wiseman, Jan 17 2023 *)

seq(n)={Vec(sum(k=1, (sqrtint(8*n+1)+1)\2, my(t=binomial(k,2)); x^t/prod(j=1, k-1, 1 - x^(t-binomial(j,2)) + O(x^(n-t+1)))))} \\ Andrew Howroyd, Jan 22 2023

def partition(n, min, max)
  return [[]] if n == 0
  [max, n].min.downto(min).flat_map{|i| partition(n - i, min, i - 1).map{|rest| [i, *rest]}}
end
def f(n)
  return 1 if n == 0
  cnt = 0
  partition(n, 1, n).each{|ary|
    ary << 0
    ary0 = (1..ary.size - 1).map{|i| ary[i - 1] - ary[i]}
    cnt += 1 if ary0.sort == ary0
  }
  cnt
end
def A320387(n)
  (0..n).map{|i| f(i)}
end
p A320387(50)

G.f.: Sum_{k>=1} x^binomial(k,2)/Product_{j=1..k-1} (1 - x^(binomial(k,2)-binomial(j,2))). - Andrew Howroyd, Jan 22 2023

A368109 Number of ways to choose a binary index of each binary index of n.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 4, 4, 4, 4, 8, 8, 8, 8, 3, 3, 3, 3, 6, 6, 6, 6, 3, 3, 3, 3, 6, 6, 6, 6, 6, 6, 6, 6, 12, 12, 12

Offset: 0

Gus Wiseman, Dec 12 2023

nonn

First differs from A367912 at a(52) = 8, A367912(52) = 7.
A binary index of n (row n of A048793) is any position of a 1 in its reversed binary expansion. For example, 18 has reversed binary expansion (0,1,0,0,1) and binary indices {2,5}.
Run-lengths are all 4 or 8.

			The binary indices of binary indices of 20 are {{1,2},{1,3}}, with choices (1,1), (1,3), (2,1), (2,3), so a(20) = 4.
The binary indices of binary indices of 52 are {{1,2},{1,3},{2,3}}, with choices (1,1,1), (1,1,3), (1,3,2), (1,3,3), (2,1,2), (2,1,3), (2,3,2), (2,3,3), so a(52) = 8.

All entries appear to belong to A003586.
Positions of ones are A253317.
The version for prime indices is A355741, for multisets A355744.
Choosing a multiset (not sequence) gives A367912, firsts A367913.
Positions of first appearances are A368111, sorted A368112.
A048793 lists binary indices, length A000120, sum A029931.
A058891 counts set-systems, covering A003465, connected A323818.
A070939 gives length of binary expansion.
A096111 gives product of binary indices.
Cf. A072639, A309326, A326031, A326702, A326753, A355731, A355739, A367771, A367905, A367906, A367915.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n, 2]],1];
Table[Length[Tuples[bpe/@bpe[n]]], {n,0,100}]

a(n) = Product_{k in A048793(n)} A000120(k).

A374515 Irregular triangle read by rows where row n lists the leaders of anti-runs in the n-th composition in standard order.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jul 31 2024

Anti-runs summing to n are counted by A003242(n).
The leaders of anti-runs in a sequence are obtained by splitting it into maximal consecutive anti-runs (sequences with no adjacent equal terms) and taking the first term of each.
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 maximal anti-runs of the 1234567th composition in standard order are ((3,2,1,2),(2,1,2,5,1),(1),(1)), so row 1234567 is (3,2,1,1).
The nonnegative integers, corresponding compositions, and leaders of anti-runs begin:
    0:      () -> ()        15: (1,1,1,1) -> (1,1,1,1)
    1:     (1) -> (1)       16:       (5) -> (5)
    2:     (2) -> (2)       17:     (4,1) -> (4)
    3:   (1,1) -> (1,1)     18:     (3,2) -> (3)
    4:     (3) -> (3)       19:   (3,1,1) -> (3,1)
    5:   (2,1) -> (2)       20:     (2,3) -> (2)
    6:   (1,2) -> (1)       21:   (2,2,1) -> (2,2)
    7: (1,1,1) -> (1,1,1)   22:   (2,1,2) -> (2)
    8:     (4) -> (4)       23: (2,1,1,1) -> (2,1,1)
    9:   (3,1) -> (3)       24:     (1,4) -> (1)
   10:   (2,2) -> (2,2)     25:   (1,3,1) -> (1)
   11: (2,1,1) -> (2,1)     26:   (1,2,2) -> (1,2)
   12:   (1,3) -> (1)       27: (1,2,1,1) -> (1,1)
   13: (1,2,1) -> (1)       28:   (1,1,3) -> (1,1)
   14: (1,1,2) -> (1,1)     29: (1,1,2,1) -> (1,1)

Gus Wiseman, Sequences counting and ranking compositions by their leaders (for six types of runs).

Row-leaders of nonempty rows are A065120.
Row-lengths are A333381.
Row-sums are A374516.
Positions of identical rows are A374519 (counted by A374517).
Positions of distinct (strict) rows are A374638 (counted by A374518).
A106356 counts compositions by number of maximal anti-runs.
A238279 counts compositions by number of maximal runs
A238424 counts partitions whose first differences are an anti-run.
All of the following pertain to compositions in standard order:
- Length is A000120.
- Sum is A029837(n+1).
- Parts are listed by A066099.
- Number of adjacent equal pairs is A124762, unequal A333382.
- Anti-runs are ranked by A333489, counted by A003242.
- Run-length transform is A333627, sum A070939.
- Run-compression is A373948 or A374251, sum A373953, excess A373954.
- Ranks of contiguous compositions are A374249, counted by A274174.
Six types of maximal runs:
- Count: A124766, A124765, A124768, A124769, A333381, A124767.
- Leaders: A374629, A374740, A374683, A374757, A374515, A374251.
- Rank: A375123, A375124, A375125, A375126, A375127, A373948.
Cf. A029931, A059893, A228351, A233564, A272919, A333213, A373949.

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[First/@Split[stc[n],UnsameQ],{n,0,100}]

cp's OEIS Frontend

A124771 Number of distinct subsequences for compositions in standard order.

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A230877 If n = Sum_{i=0..m} c(i)*2^i, c(i) = 0 or 1, then a(n) = Sum_{i=0..m} (m+1-i)*c(i).

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A333257 Number of distinct consecutive subsequence-sums of the k-th composition in standard order.

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A358194 Irregular triangle read by rows where T(n,k) is the number of integer partitions of n with partial sums summing to k, where k ranges from n to n(n+1)/2.

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A089633 Numbers having no more than one 0 in their binary representation.

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A326753 Number of connected components of the set-system with BII-number n.

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A333766 Maximum part of the n-th composition in standard order. a(0) = 0.

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A230877 If n = Sum_{i=0..m} c(i)2^i, c(i) = 0 or 1, then a(n) = Sum_{i=0..m} (m+1-i)c(i).