cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A124771 Number of distinct subsequences for compositions in standard order.

Original entry on oeis.org

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

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Author

Keywords

Comments

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

Examples

			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)
		

Crossrefs

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.

Programs

  • Mathematica
    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 *)

Formula

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)

A374740 Irregular triangle read by rows where row n lists the leaders of weakly decreasing runs in the n-th composition in standard order.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Jul 24 2024

Keywords

Comments

The leaders of weakly decreasing runs in a sequence are obtained by splitting it into maximal weakly decreasing subsequences 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.

Examples

			The maximal weakly decreasing subsequences of the 1234567th composition in standard order are ((3,2,1),(2,2,1),(2),(5,1,1,1)), so row 1234567 is (3,2,2,5).
The nonnegative integers, corresponding compositions, and leaders of weakly decreasing runs begin:
    0: () -> ()           15: (1,1,1,1) -> (1)
    1: (1) -> (1)         16: (5) -> (5)
    2: (2) -> (2)         17: (4,1) -> (4)
    3: (1,1) -> (1)       18: (3,2) -> (3)
    4: (3) -> (3)         19: (3,1,1) -> (3)
    5: (2,1) -> (2)       20: (2,3) -> (2,3)
    6: (1,2) -> (1,2)     21: (2,2,1) -> (2)
    7: (1,1,1) -> (1)     22: (2,1,2) -> (2,2)
    8: (4) -> (4)         23: (2,1,1,1) -> (2)
    9: (3,1) -> (3)       24: (1,4) -> (1,4)
   10: (2,2) -> (2)       25: (1,3,1) -> (1,3)
   11: (2,1,1) -> (2)     26: (1,2,2) -> (1,2)
   12: (1,3) -> (1,3)     27: (1,2,1,1) -> (1,2)
   13: (1,2,1) -> (1,2)   28: (1,1,3) -> (1,3)
   14: (1,1,2) -> (1,2)   29: (1,1,2,1) -> (1,2)
		

Crossrefs

Row-leaders are A065120.
Row-lengths are A124765.
Other types of runs are A374251, A374515, A374683, A374757.
The opposite is A374629.
Positions of distinct (strict) rows are A374701, counted by A374743.
Row-sums are A374741, opposite A374630.
Positions of identical rows are A374744, counted by A374742.
All of the following pertain to compositions in standard order:
- Length is A000120.
- Sum is A029837(n+1) (or sometimes A070939).
- Parts are listed by A066099.
- Number of adjacent equal pairs is A124762, unequal A333382.
- Number of max runs: A124765, A124766, A124767, A124768, A124769, A333381.
- Ranks of anti-run compositions are A333489, counted by A003242.
- Run-length transform is A333627, sum A070939.
- Run-compression transform is A373948, sum A373953, excess A373954.
- Ranks of contiguous compositions are A374249, counted by A274174.
- Ranks of non-contiguous compositions are A374253, counted by A335548.

Programs

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

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

Views

Author

Gus Wiseman, Mar 20 2020

Keywords

Comments

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.

Examples

			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.
		

Crossrefs

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.

Programs

  • Mathematica
    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}]

Formula

a(n) = A333224(n) + 1.

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

Views

Author

Gus Wiseman, Apr 05 2020

Keywords

Comments

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.

Examples

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

Crossrefs

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.

Programs

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

Formula

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

A373951 Triangle read by rows where T(n,k) is the number of integer compositions of n such that replacing each run of repeated parts with a single part (run-compression) yields a composition of n - k.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 3, 0, 1, 0, 4, 2, 1, 1, 0, 7, 4, 4, 0, 1, 0, 14, 5, 6, 5, 1, 1, 0, 23, 14, 10, 10, 6, 0, 1, 0, 39, 26, 29, 12, 14, 6, 1, 1, 0, 71, 46, 54, 40, 19, 16, 9, 0, 1, 0, 124, 92, 96, 82, 64, 22, 22, 8, 1, 1, 0, 214, 176, 204, 144, 137, 82, 30, 26, 10, 0, 1, 0
Offset: 0

Views

Author

Gus Wiseman, Jun 28 2024

Keywords

Examples

			Triangle begins:
    1
    1   0
    1   1   0
    3   0   1   0
    4   2   1   1   0
    7   4   4   0   1   0
   14   5   6   5   1   1   0
   23  14  10  10   6   0   1   0
   39  26  29  12  14   6   1   1   0
   71  46  54  40  19  16   9   0   1   0
  124  92  96  82  64  22  22   8   1   1   0
Row n = 6 counts the following compositions:
  (6)     (411)   (3111)   (33)     (222)  (111111)  .
  (51)    (114)   (1113)   (2211)
  (15)    (1311)  (1221)   (1122)
  (42)    (1131)  (12111)  (21111)
  (24)    (2112)  (11211)  (11112)
  (141)           (11121)
  (321)
  (312)
  (231)
  (213)
  (132)
  (123)
  (2121)
  (1212)
For example, the composition (1,2,2,1) with compression (1,2,1) is counted under T(6,2).
		

Crossrefs

Column k = 0 is A003242 (anti-runs or compressed compositions).
Row-sums are A011782.
Same as A373949 with rows reversed.
Column k = 1 is A373950.
This statistic is represented by A373954, difference A373953.
A114901 counts compositions with no isolated parts.
A116861 counts partitions by compressed sum, by compressed length A116608.
A124767 counts runs in standard compositions, anti-runs A333381.
A240085 counts compositions with no unique parts.
A333755 counts compositions by compressed length.
A373948 represents the run-compression transformation.

Programs

  • Mathematica
    Table[Length[Select[Join@@Permutations /@ IntegerPartitions[n], Total[First/@Split[#]]==n-k&]],{n,0,10},{k,0,n}]

A374251 Irregular triangle read by rows where row n is the run-compression of the n-th composition in standard order.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 1, 2, 1, 4, 3, 1, 2, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 5, 4, 1, 3, 2, 3, 1, 2, 3, 2, 1, 2, 1, 2, 2, 1, 1, 4, 1, 3, 1, 1, 2, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 6, 5, 1, 4, 2, 4, 1, 3, 3, 2, 1, 3, 1, 2, 3, 1, 2, 4, 2, 3, 1, 2, 2, 1, 2, 1, 3
Offset: 1

Views

Author

Gus Wiseman, Jul 09 2024

Keywords

Comments

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.
We define the run-compression of a sequence to be the anti-run obtained by reducing each run of repeated parts to a single part. Alternatively, run-compression removes all parts equal to the part immediately to their left. For example, (1,1,2,2,1) has run-compression (1,2,1).

Examples

			The standard compositions and their run-compressions begin:
   0: ()        --> ()
   1: (1)       --> (1)
   2: (2)       --> (2)
   3: (1,1)     --> (1)
   4: (3)       --> (3)
   5: (2,1)     --> (2,1)
   6: (1,2)     --> (1,2)
   7: (1,1,1)   --> (1)
   8: (4)       --> (4)
   9: (3,1)     --> (3,1)
  10: (2,2)     --> (2)
  11: (2,1,1)   --> (2,1)
  12: (1,3)     --> (1,3)
  13: (1,2,1)   --> (1,2,1)
  14: (1,1,2)   --> (1,2)
  15: (1,1,1,1) --> (1)
		

Crossrefs

Last column is A001511.
First column is A065120.
Row-lengths are A124767.
Using prime indices we get A304038, row-sums A066328.
Row n has A334028(n) distinct elements.
Rows are ranked by A373948 (standard order).
Row-sums are A373953.
A003242 counts run-compressed compositions, i.e., anti-runs, ranks A333489.
A007947 (squarefree kernel) represents run-compression of multisets.
A037201 run-compresses first differences of primes, halved A373947.
A066099 lists the parts of compositions in standard order.
A116861 counts partitions by sum of run-compression.
A238279 and A333755 count compositions by number of runs.
A373949 counts compositions by sum of run-compression, opposite A373951.

Programs

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

A335458 Number of normal patterns contiguously matched by the n-th composition in standard order (A066099).

Original entry on oeis.org

1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 5, 3, 5, 5, 5, 2, 3, 3, 5, 3, 5, 5, 7, 3, 5, 5, 8, 5, 8, 7, 6, 2, 3, 3, 5, 3, 4, 5, 7, 3, 5, 4, 7, 5, 7, 8, 9, 3, 5, 5, 8, 4, 8, 7, 11, 5, 8, 7, 11, 7, 11, 9, 7, 2, 3, 3, 5, 3, 4, 5, 7, 3, 5, 5, 7, 5, 7, 8, 9, 3, 5, 5, 8, 5, 7
Offset: 0

Views

Author

Gus Wiseman, Jun 21 2020

Keywords

Comments

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.
We define a (normal) pattern to be a finite sequence covering an initial interval of positive integers. Patterns are counted by A000670 and ranked by A333217. A sequence S is said to match a pattern P if there is a not necessarily contiguous subsequence of S whose parts have the same relative order as P. For example, (3,1,1,3) matches (1,1,2), (2,1,1), and (2,1,2), but avoids (1,2,1), (1,2,2), and (2,2,1).

Examples

			The a(180) = 7 patterns are: (), (1), (1,2), (2,1), (1,2,3), (2,1,2), (2,1,2,3).
		

Crossrefs

The non-contiguous version is A335454.
Summing over indices with binary length n gives A335457(n).
The nonempty version is A335474.
Patterns are counted by A000670 and ranked by A333217.
The n-th composition has A124771(n) distinct consecutive subsequences.
Knapsack compositions are counted by A325676 and ranked by A333223.
The n-th composition has A333257(n) distinct subsequence-sums.
The n-th composition has A334299(n) distinct subsequences.
Minimal avoided patterns are counted by A335465.

Programs

  • Mathematica
    stc[n_]:=Reverse[Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]];
    mstype[q_]:=q/.Table[Union[q][[i]]->i,{i,Length[Union[q]]}];
    Table[Length[Union[mstype/@ReplaceList[stc[n],{_,s___,_}:>{s}]]],{n,0,30}]

Formula

a(n) = A335474(n) + 1.

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

Views

Author

Gus Wiseman, Jul 31 2024

Keywords

Comments

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.

Examples

			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)
		

Crossrefs

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:

Programs

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

A353849 Number of distinct positive run-sums of the n-th composition in standard order.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 1, 3, 3, 1, 2, 3, 1, 2, 3, 2, 1, 2, 2, 2, 3, 3, 3, 2, 2, 3, 2, 3, 2, 1, 1, 3, 2, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3
Offset: 0

Views

Author

Gus Wiseman, May 30 2022

Keywords

Comments

Every sequence can be uniquely split into a sequence of non-overlapping runs. For example, the runs of (2,2,1,1,1,3,2,2) are ((2,2),(1,1,1),(3),(2,2)), with sums (4,3,3,4).
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.

Examples

			Composition 462903 in standard order is (1,1,4,7,1,2,1,1,1), with run-sums (2,4,7,1,2,3), of which a(462903) = 5 are distinct.
		

Crossrefs

Counting repeated runs also gives A124767.
Positions of first appearances are A246534.
For distinct runs instead of run-sums we have A351014 (firsts A351015).
A version for partitions is A353835, weak A353861.
Positions of 1's are A353848, counted by A353851.
The version for binary expansion is A353929 (firsts A353930).
The run-sums themselves are listed by A353932, with A353849 distinct terms.
For distinct run-lengths instead of run-sums we have A354579.
A005811 counts runs in binary expansion.
A066099 lists compositions in standard order.
A165413 counts distinct run-lengths in binary expansion.
A297770 counts distinct runs in binary expansion, firsts A350952.
A353847 represents the run-sum transformation for compositions.
A353853-A353859 pertain to composition run-sum trajectory.
Selected statistics of standard compositions:
- Length is A000120.
- Sum is A070939.
- Heinz number is A333219.
- Number of distinct parts is A334028.
Selected classes of standard compositions:
- Partitions are A114994, strict A333256.
- Multisets are A225620, strict A333255.
- Strict compositions are A233564.
- Constant compositions are A272919.

Programs

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

A124758 Product of the parts of the compositions in standard order.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 2, 1, 4, 3, 4, 2, 3, 2, 2, 1, 5, 4, 6, 3, 6, 4, 4, 2, 4, 3, 4, 2, 3, 2, 2, 1, 6, 5, 8, 4, 9, 6, 6, 3, 8, 6, 8, 4, 6, 4, 4, 2, 5, 4, 6, 3, 6, 4, 4, 2, 4, 3, 4, 2, 3, 2, 2, 1, 7, 6, 10, 5, 12, 8, 8, 4, 12, 9, 12, 6, 9, 6, 6, 3, 10, 8, 12, 6, 12, 8, 8, 4, 8, 6, 8, 4, 6, 4, 4, 2, 6, 5, 8, 4, 9, 6
Offset: 0

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Author

Keywords

Comments

The standard order of compositions is given by A066099.
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. - Gus Wiseman, Apr 03 2020

Examples

			Composition number 11 is 2,1,1; 2*1*1 = 2, so a(11) = 2.
The table starts:
  1
  1
  2 1
  3 2 2 1
  4 3 4 2 3 2 2 1
  5 4 6 3 6 4 4 2 4 3 4 2 3 2 2 1
The 146-th composition in standard order is (3,3,2), with product 18, so a(146) = 18. - _Gus Wiseman_, Apr 03 2020
		

Crossrefs

Cf. A066099, A118851, A011782 (row lengths), A001906 (row sums).
The lengths of standard compositions are given by A000120.
The version for prime indices is A003963.
The version for binary indices is A096111.
Taking the sum instead of product gives A070939.
The sum of binary indices is A029931.
The sum of prime indices is A056239.
Taking GCD instead of product gives A326674.
Positions of first appearances are A331579.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Times@@stc[n],{n,0,100}] (* Gus Wiseman, Apr 03 2020 *)

Formula

For a composition b(1),...,b(k), a(n) = Product_{i=1}^k b(i).
a(A164894(n)) = a(A246534(n)) = n!. - Gus Wiseman, Apr 03 2020
a(A233249(n)) = a(A333220(n)) = A003963(n). - Gus Wiseman, Apr 03 2020
From Mikhail Kurkov, Jul 11 2021: (Start)
Some conjectures:
a(2n+1) = a(n) for n >= 0.
a(2n) = (1 + 1/A001511(n))*a(n) = 2*a(n) + a(n - 2^f(n)) - a(2n - 2^f(n)) for n > 0 with a(0)=1 where f(n) = A007814(n).
From the 1st formula for a(2n) we get a(4n+2) = 2*a(n), a(4n) = 2*a(2n) - a(n).
Sum_{k=0..2^n - 1} a(k) = A001519(n+1) for n >= 0.
a((4^n - 1)/3) = A011782(n) for n >= 0.
a(2^m*(2^n - 1)) = m + 1 for n > 0, m >= 0. (End)
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