A335279 -id:A335279 - 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)

A334968 Number of possible sums of subsequences (not necessarily contiguous) of the n-th composition in standard order (A066099).

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, 6, 6, 4, 6, 6, 6, 6, 6, 6, 6, 2, 4, 4, 6, 3, 7, 7, 7, 4, 7, 4, 7, 7, 7, 7, 7, 4, 6, 7, 7, 7, 7, 7, 7, 6, 7, 7, 7, 7, 7, 7, 7, 2, 4, 4, 6, 4, 8, 8, 8, 4, 6, 6, 8, 6, 8, 8, 8, 4, 8, 6, 8, 6, 8, 8

Offset: 0

Gus Wiseman, Jun 02 2020

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 139th composition is (4,2,1,1), with possible sums of subsequences {0,1,2,3,4,5,6,7,8}, so a(139) = 9.
Triangle begins:
  1
  2
  2 3
  2 4 4 4
  2 4 3 5 4 5 5 5
  2 4 4 6 4 6 6 6 4 6 6 6 6 6 6 6
  2 4 4 6 3 7 7 7 4 7 4 7 7 7 7 7 4 6 7 7 7 7 7 7 6 7 7 7 7 7 7 7

Row lengths are A011782.
Dominated by A124771 (number of contiguous subsequences).
Dominates A333257 (the contiguous case).
Dominated by A334299 (number of subsequences).
Golomb rulers are counted by A169942 and ranked by A333222.
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 ranked by A333223.
Contiguous subsequence-sums are counted by A333224 and ranked by A333257.
Knapsack compositions are counted by A334268 and ranked by A334967.
Cf. A000120, A029931, A048793, A066099, A070939, A108917, A325769, A325770, A325778, A334300, A335279.

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

a(n) = A299701(A333219(n)).

cp's OEIS Frontend

A124771 Number of distinct subsequences for compositions in standard order.

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