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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A333489 Numbers k such that the k-th composition in standard order is an anti-run (no adjacent equal parts).

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 16, 17, 18, 20, 22, 24, 25, 32, 33, 34, 37, 38, 40, 41, 44, 45, 48, 49, 50, 52, 54, 64, 65, 66, 68, 69, 70, 72, 76, 77, 80, 81, 82, 88, 89, 96, 97, 98, 101, 102, 104, 105, 108, 109, 128, 129, 130, 132, 133, 134, 137, 140, 141
Offset: 1

Views

Author

Gus Wiseman, Mar 28 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 sequence together with the corresponding compositions begins:
    0: ()          33: (5,1)         70: (4,1,2)
    1: (1)         34: (4,2)         72: (3,4)
    2: (2)         37: (3,2,1)       76: (3,1,3)
    4: (3)         38: (3,1,2)       77: (3,1,2,1)
    5: (2,1)       40: (2,4)         80: (2,5)
    6: (1,2)       41: (2,3,1)       81: (2,4,1)
    8: (4)         44: (2,1,3)       82: (2,3,2)
    9: (3,1)       45: (2,1,2,1)     88: (2,1,4)
   12: (1,3)       48: (1,5)         89: (2,1,3,1)
   13: (1,2,1)     49: (1,4,1)       96: (1,6)
   16: (5)         50: (1,3,2)       97: (1,5,1)
   17: (4,1)       52: (1,2,3)       98: (1,4,2)
   18: (3,2)       54: (1,2,1,2)    101: (1,3,2,1)
   20: (2,3)       64: (7)          102: (1,3,1,2)
   22: (2,1,2)     65: (6,1)        104: (1,2,4)
   24: (1,4)       66: (5,2)        105: (1,2,3,1)
   25: (1,3,1)     68: (4,3)        108: (1,2,1,3)
   32: (6)         69: (4,2,1)      109: (1,2,1,2,1)
		

Crossrefs

Anti-runs summing to n are counted by A003242(n).
A triangle counting maximal anti-runs of compositions is A106356.
A triangle counting maximal runs of compositions is A238279 or A238130.
Partitions whose first differences are an anti-run are A238424.
All of the following pertain to compositions in standard order (A066099):
- Adjacent equal pairs are counted by A124762.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769.
- Strict compositions are ranked by A233564.
- Constant compositions are ranked by A272919.
- Normal compositions are ranked by A333217.
- Anti-runs are counted by A333381.
- Adjacent unequal pairs are counted by A333382.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,100],!MatchQ[stc[#],{_,x_,x_,_}]&]

A124767 Number of level runs for compositions in standard order.

Original entry on oeis.org

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

Views

Author

Keywords

Comments

The standard order of compositions is given by A066099.
For n > 0, a(n) is one more than the number of adjacent unequal terms in the n-th composition in standard order. Also the number of runs in the same composition. - Gus Wiseman, Apr 08 2020

Examples

			Composition number 11 is 2,1,1; the level runs are 2; 1,1; so a(11) = 2.
The table starts:
  0
  1
  1 1
  1 2 2 1
  1 2 1 2 2 3 2 1
  1 2 2 2 2 2 3 2 2 3 2 3 2 3 2 1
  1 2 2 2 1 3 3 2 2 3 1 2 3 4 3 2 2 3 3 3 3 3 4 3 2 3 2 3 2 3 2 1
The 1234567th composition in standard order is (3,2,1,2,2,1,2,5,1,1,1) with runs ((3),(2),(1),(2,2),(1),(2),(5),(1,1,1)), so a(1234567) = 8. - _Gus Wiseman_, Apr 08 2020
		

Crossrefs

Row-lengths are A011782.
Compositions counted by number of runs are A238279 or A333755.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Weakly decreasing compositions are A114994.
- Adjacent equal pairs are counted by A124762.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767 (this sequence).
- Weakly increasing compositions are A225620.
- Strict compositions A233564.
- Constant compositions are A272919.
- Anti-runs are counted by A333381.
- Adjacent unequal pairs are counted by A333382.
- Anti-run compositions are A333489.
- Runs-resistance is A333628.
- Run-lengths are A333769 (triangle).

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Length[Split[stc[n]]],{n,0,100}] (* Gus Wiseman, Apr 17 2020 *)

Formula

a(0) = 0, a(n) = 1 + Sum_{1<=i=1 0.
For n > 0, a(n) = A333382(n) + 1. - Gus Wiseman, Apr 08 2020

A238343 Triangle T(n,k) read by rows: T(n,k) is the number of compositions of n with k descents, n>=0, 0<=k<=n.

Original entry on oeis.org

1, 1, 0, 2, 0, 0, 3, 1, 0, 0, 5, 3, 0, 0, 0, 7, 9, 0, 0, 0, 0, 11, 19, 2, 0, 0, 0, 0, 15, 41, 8, 0, 0, 0, 0, 0, 22, 77, 29, 0, 0, 0, 0, 0, 0, 30, 142, 81, 3, 0, 0, 0, 0, 0, 0, 42, 247, 205, 18, 0, 0, 0, 0, 0, 0, 0, 56, 421, 469, 78, 0, 0, 0, 0, 0, 0, 0, 0, 77, 689, 1013, 264, 5, 0, 0, 0, 0, 0, 0, 0, 0, 101, 1113, 2059, 786, 37, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Joerg Arndt and Alois P. Heinz, Feb 25 2014

Keywords

Comments

Counting ascents gives the same triangle.
For n > 0, also the number of compositions of n with k + 1 maximal weakly increasing runs. - Gus Wiseman, Mar 23 2020

Examples

			Triangle starts:
00:    1;
01:    1,    0;
02:    2,    0,    0;
03:    3,    1,    0,    0;
04:    5,    3,    0,    0,   0;
05:    7,    9,    0,    0,   0, 0;
06:   11,   19,    2,    0,   0, 0, 0;
07:   15,   41,    8,    0,   0, 0, 0, 0;
08:   22,   77,   29,    0,   0, 0, 0, 0, 0;
09:   30,  142,   81,    3,   0, 0, 0, 0, 0, 0;
10:   42,  247,  205,   18,   0, 0, 0, 0, 0, 0, 0;
11:   56,  421,  469,   78,   0, 0, 0, 0, 0, 0, 0, 0;
12:   77,  689, 1013,  264,   5, 0, 0, 0, 0, 0, 0, 0, 0;
13:  101, 1113, 2059,  786,  37, 0, 0, 0, 0, 0, 0, 0, 0, 0;
14:  135, 1750, 4021, 2097, 189, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
15:  176, 2712, 7558, 5179, 751, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
...
From _Gus Wiseman_, Mar 23 2020: (Start)
Row n = 5 counts the following compositions:
  (5)          (3,2)
  (1,4)        (4,1)
  (2,3)        (1,3,1)
  (1,1,3)      (2,1,2)
  (1,2,2)      (2,2,1)
  (1,1,1,2)    (3,1,1)
  (1,1,1,1,1)  (1,1,2,1)
               (1,2,1,1)
               (2,1,1,1)
(End)
		

Crossrefs

T(3n,n) gives A000045(n+1).
T(3n+1,n) = A136376(n+1).
Row sums are A011782.
Compositions by length are A007318.
The version for co-runs or levels is A106356.
The case of partitions (instead of compositions) is A133121.
The version for runs is A238279.
The version without zeros is A238344.
The version for weak ascents is A333213.

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1, expand(
           add(b(n-j, j)*`if`(j (p-> seq(coeff(p, x, i), i=0..n))(b(n, 0)):
    seq(T(n), n=0..20);
  • Mathematica
    b[n_, i_] := b[n, i] = If[n == 0, 1, Sum[b[n-j, j]*If[jJean-François Alcover, Jan 08 2015, translated from Maple *)
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],n==0||Length[Split[#,LessEqual]]==k+1&]],{n,0,9},{k,0,n}] (* Gus Wiseman, Mar 23 2020 *)

Formula

Sum_{k=0..n} k * T(n,k) = A045883(n-2) for n>=2.

A333381 Number of maximal anti-runs of the n-th composition in standard order.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Mar 24 2020

Keywords

Comments

Anti-runs are sequences without any adjacent equal terms.
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.
For n > 0, also one plus the number of adjacent equal pairs in the n-th composition in standard order.

Examples

			The 46th composition in standard order is (2,1,1,2), with maximal anti-runs ((2,1),(1,2)), so a(46) = 2.
		

Crossrefs

Anti-runs summing to n are counted by A003242(n).
A triangle counting maximal anti-runs of compositions is A106356.
A triangle counting maximal runs of compositions is A238279.
Partitions whose first differences are an anti-run are A238424.
All of the following pertain to compositions in standard order (A066099):
- Adjacent equal pairs are counted by A124762.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769.
- Strict compositions are ranked by A233564.
- Constant compositions are ranked by A272919.
- Normal compositions are ranked by A333217.
- Adjacent unequal pairs are counted by A333382.
- Anti-runs are ranked by A333489.

Programs

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

Formula

For n > 0, a(n) = A124762(n) + 1.

A374629 Irregular triangle listing the leaders of maximal weakly increasing runs in the n-th composition in standard order.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Jul 20 2024

Keywords

Comments

The leaders of maximal weakly increasing runs in a sequence are obtained by splitting it into maximal weakly increasing 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 58654th composition in standard order is (1,1,3,2,4,1,1,1,2), with maximal weakly increasing runs ((1,1,3),(2,4),(1,1,1,2)), so row 58654 is (1,2,1).
The nonnegative integers, corresponding compositions, and leaders of maximal weakly increasing runs begin:
    0:      () -> ()      15: (1,1,1,1) -> (1)
    1:     (1) -> (1)     16:       (5) -> (5)
    2:     (2) -> (2)     17:     (4,1) -> (4,1)
    3:   (1,1) -> (1)     18:     (3,2) -> (3,2)
    4:     (3) -> (3)     19:   (3,1,1) -> (3,1)
    5:   (2,1) -> (2,1)   20:     (2,3) -> (2)
    6:   (1,2) -> (1)     21:   (2,2,1) -> (2,1)
    7: (1,1,1) -> (1)     22:   (2,1,2) -> (2,1)
    8:     (4) -> (4)     23: (2,1,1,1) -> (2,1)
    9:   (3,1) -> (3,1)   24:     (1,4) -> (1)
   10:   (2,2) -> (2)     25:   (1,3,1) -> (1,1)
   11: (2,1,1) -> (2,1)   26:   (1,2,2) -> (1)
   12:   (1,3) -> (1)     27: (1,2,1,1) -> (1,1)
   13: (1,2,1) -> (1,1)   28:   (1,1,3) -> (1)
   14: (1,1,2) -> (1)     29: (1,1,2,1) -> (1,1)
		

Crossrefs

Row-leaders are A065120.
Row-lengths are A124766.
Row-sums are A374630.
Positions of constant rows are A374633, counted by A374631.
Positions of strict rows are A374768, counted by A374632.
For other types of runs we have A374251, A374515, A374683, A374740, A374757.
Positions of non-weakly decreasing rows are A375137.
A011782 counts compositions.
A238130, A238279, A333755 count compositions by number of runs.
A335456 counts patterns matched by compositions.
All of the following pertain to compositions in standard order:
- Length is A000120.
- Sum is A029837(n+1).
- Leader is A065120.
- Parts are listed by A066099, reverse A228351.
- 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, length A124767, 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],LessEqual],{n,0,100}]

A124762 Number of levels for compositions in standard order.

Original entry on oeis.org

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

Views

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. This gives a bijective correspondence between nonnegative integers and integer compositions. This sequence gives the number of adjacent equal terms in the n-th composition in standard order. Alternatively, a(n) is one fewer than the number of maximal anti-runs in the same composition, where anti-runs are sequences without any adjacent equal terms. For example, the 1234567th composition in standard order is (3,2,1,2,2,1,2,5,1,1,1) with anti-runs ((3,2,1,2),(2,1,2,5,1),(1),(1)), so a(1234567) = 4 - 1 = 3. - Gus Wiseman, Apr 08 2020

Examples

			Composition number 11 is 2,1,1; 2>1=1, so a(11) = 1.
The table starts:
  0
  0
  0 1
  0 0 0 2
  0 0 1 1 0 0 1 3
  0 0 0 1 0 1 0 2 0 0 1 1 1 1 2 4
  0 0 0 1 1 0 0 2 0 0 2 2 0 0 1 3 0 0 0 1 0 1 0 2 1 1 2 2 2 2 3 5
		

Crossrefs

Cf. A066099, A124760, A124761, A124763, A124764, A011782 (row lengths), A059570 (row sums).
Anti-runs summing to n are counted by A003242(n).
A triangle counting maximal anti-runs of compositions is A106356.
A triangle counting maximal runs of compositions is A238279.
Partitions whose first differences are an anti-run are A238424.
All of the following pertain to compositions in standard order (A066099):
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Normal compositions are A333217.
- Adjacent unequal pairs are counted by A333382.
- Anti-runs are A333489.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Length[Select[Partition[stc[n],2,1],SameQ@@#&]],{n,0,100}] (* Gus Wiseman, Apr 08 2020 *)

Formula

For a composition b(1),...,b(k), a(n) = Sum_{1<=i=1
For n > 0, a(n) = A333381(n) - 1. - Gus Wiseman, Apr 08 2020

A225620 Indices of partitions in the table of compositions of A228351.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 8, 10, 12, 14, 15, 16, 20, 24, 26, 28, 30, 31, 32, 36, 40, 42, 48, 52, 56, 58, 60, 62, 63, 64, 72, 80, 84, 96, 100, 104, 106, 112, 116, 120, 122, 124, 126, 127, 128, 136, 144, 160, 164, 168, 170, 192, 200, 208, 212, 224, 228, 232, 234, 240, 244, 248, 250, 252, 254, 255
Offset: 1

Author

Omar E. Pol, Aug 03 2013

Keywords

Comments

Also triangle read by rows in which T(n,k) is the decimal representation of a binary number whose mirror represents the k-th partition of n according with the list of juxtaposed reverse-lexicographically ordered partitions of the positive integers (A026792).
In order to construct this sequence as a triangle we use the following rules:
- In the list of A026792 we replace each part of size j of the k-th partition of n by concatenation of j - 1 zeros and only one 1.
- Then replace this new set of parts by the concatenation of its parts.
- Then replace this string by its mirror version which is a binary number.
T(n,k) is the decimal value of this binary number, which represents the k-th partition of n (see example).
The partitions of n are represented by a subsequence with A000041(n) integers starting with 2^(n-1) and ending with 2^n - 1, n >= 1. The odd numbers of the sequence are in A000225.
First differs from A065609 at a(23).
Conjecture: this sequence is a sorted version of b(n) where b(2^k) = 2^k for k >= 0, b(n) = A080100(n)*(2*b(A053645(n)) + 1) otherwise. - Mikhail Kurkov, Oct 21 2023

Examples

			T(6,8) = 58 because 58 in base 2 is 111010 whose mirror is 010111 which is the concatenation of 01, 01, 1, 1, whose number of digits are 2, 2, 1, 1, which are also the 8th partition of 6.
Illustration of initial terms:
The sequence represents a table of partitions (see below):
--------------------------------------------------------
.            Binary                        Partitions
n  k  T(n,k) number  Mirror   Diagram       (A026792)
.                                          1 2 3 4 5 6
--------------------------------------------------------
.                             _
1  1     1       1    1        |           1,
.                             _ _
1  1     2      10    01      _  |           2,
2  2     3      11    11       | |         1,1,
.                             _ _ _
3  1     4     100    001     _ _  |           3,
3  2     6     110    011     _  | |         2,1,
3  3     7     111    111      | | |       1,1,1,
.                             _ _ _ _
4  1     8    1000    0001    _ _    |           4,
4  2    10    1010    0101    _ _|_  |         2,2,
4  3    12    1100    0011    _ _  | |         3,1,
4  4    14    1110    0111    _  | | |       2,1,1,
4  5    15    1111    1111     | | | |     1,1,1,1,
.                             _ _ _ _ _
5  1    16   10000    00001   _ _ _    |           5,
5  2    20   10100    00101   _ _ _|_  |         3,2,
5  3    24   11000    00011   _ _    | |         4,1,
5  4    26   11010    01011   _ _|_  | |       2,2,1,
5  5    28   11100    00111   _ _  | | |       3,1,1,
5  6    30   11110    01111   _  | | | |     2,1,1,1,
5  7    31   11111    11111    | | | | |   1,1,1,1,1,
.                             _ _ _ _ _ _
6  1    32  100000    000001  _ _ _      |           6
6  2    36  100100    001001  _ _ _|_    |         3,3,
6  3    40  101000    000101  _ _    |   |         4,2,
6  4    42  101010    010101  _ _|_ _|_  |       2,2,2,
6  5    48  110000    000011  _ _ _    | |         5,1,
6  6    52  110100    001011  _ _ _|_  | |       3,2,1,
6  7    56  111000    000111  _ _    | | |       4,1,1,
6  8    58  111010    010111  _ _|_  | | |     2,2,1,1,
6  9    60  111100    001111  _ _  | | | |     3,1,1,1,
6  10   62  111110    011111  _  | | | | |   2,1,1,1,1,
6  11   63  111111    111111   | | | | | | 1,1,1,1,1,1,
.
Triangle begins:
  1;
  2,   3;
  4,   6,  7;
  8,  10, 12, 14, 15;
  16, 20, 24, 26, 28, 30, 31;
  32, 36, 40, 42, 48, 52, 56, 58, 60, 62, 63;
  ...
From _Gus Wiseman_, Apr 01 2020: (Start)
Using the encoding of A066099, this sequence ranks all finite nonempty multisets, as follows.
   1: {1}
   2: {2}
   3: {1,1}
   4: {3}
   6: {1,2}
   7: {1,1,1}
   8: {4}
  10: {2,2}
  12: {1,3}
  14: {1,1,2}
  15: {1,1,1,1}
  16: {5}
  20: {2,3}
  24: {1,4}
  26: {1,2,2}
  28: {1,1,3}
  30: {1,1,1,2}
  31: {1,1,1,1,1}
(End)
		

Crossrefs

Column 1 is A000079. Row n has length A000041(n). Right border gives A000225.
The case covering an initial interval is A333379 or A333380.
All of the following pertain to compositions in the order of A066099.
- The weakly increasing version is this sequence.
- The weakly decreasing version is A114994.
- The strictly increasing version is A333255.
- The strictly decreasing version is A333256.
- The unequal version is A233564.
- The equal version is A272919.
- The case covering an initial interval is A333217.
- Initial intervals are ranked by A164894.
- Reversed initial intervals are ranked by A246534.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,100],LessEqual@@stc[#]&] (* Gus Wiseman, Apr 01 2020 *)

Formula

Conjecture: a(A000070(m) - k) = 2^m - A228354(k) for m > 0, 0 < k <= A000041(m). - Mikhail Kurkov, Oct 20 2023

A333382 Number of adjacent unequal parts in the n-th composition in standard-order.

Original entry on oeis.org

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

Author

Gus Wiseman, Mar 24 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.
For n > 0, a(n) is one fewer than the number of maximal runs of the n-th composition in standard-order.

Examples

			The 46th composition in standard order is (2,1,1,2), with maximal runs ((2),(1,1),(2)), so a(46) = 3 - 1 = 2.
		

Crossrefs

Indices of first appearances (not counting 0) are A113835.
Partitions whose 0-appended first differences are a run are A007862.
Partitions whose first differences are a run are A049988.
A triangle counting maximal anti-runs of compositions is A106356.
A triangle counting maximal runs of compositions is A238279.
All of the following pertain to compositions in standard order (A066099):
- Adjacent equal pairs are counted by A124762.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769.
- Strict compositions are ranked by A233564.
- Constant compositions are ranked by A272919.
- Normal compositions are ranked by A333217.
- Anti-runs are ranked by A333489.
- Anti-runs are counted by A333381.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Length[Select[Partition[stc[n],2,1],UnsameQ@@#&]],{n,0,100}]

Formula

For n > 0, a(n) = A124767(n) - 1.

A124768 Number of strictly increasing runs for compositions in standard order.

Original entry on oeis.org

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

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. a(n) is the number of maximal strictly increasing runs in this composition. Alternatively, a(n) is one plus the number of weak descents in the same composition. For example, the strictly increasing runs of the 1234567th composition are ((3),(2),(1,2),(2),(1,2,5),(1),(1),(1)), so a(1234567) = 8. The 7 weak descents together with the strict ascents are: 3 >= 2 >= 1 < 2 >= 2 >= 1 < 2 < 5 >= 1 >= 1 >= 1. - Gus Wiseman, Apr 08 2020

Examples

			Composition number 11 is 2,1,1; the strictly increasing runs are 2; 1; 1; so a(11) = 3.
The table starts:
  0
  1
  1 2
  1 2 1 3
  1 2 2 3 1 2 2 4
  1 2 2 3 1 3 2 4 1 2 2 3 2 3 3 5
  1 2 2 3 2 3 2 4 1 2 3 4 2 3 3 5 1 2 2 3 1 3 2 4 2 3 3 4 3 4 4 6
		

Crossrefs

Cf. A066099, A124763, A011782 (row lengths).
Compositions of n with k weak descents are A333213.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Partial sums from the right are A048793.
- Sum is A070939.
- Weakly decreasing compositions are A114994.
- Adjacent equal pairs are counted by A124762.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768 (this sequence).
- Strictly decreasing runs are counted by A124769.
- Weakly increasing compositions are A225620.
- Reverse is A228351 (triangle).
- Strict compositions are A233564.
- Initial intervals are A246534.
- Constant compositions are A272919.
- Normal compositions are A333217.
- Permutations are A333218.
- Heinz number is A333219.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
- Anti-runs are A333489.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Length[Split[stc[n],Less]],{n,0,100}] (* Gus Wiseman, Apr 08 2020 *)

Formula

a(0) = 0, a(n) = A124763(n) + 1 for n > 0.

A124765 Number of monotonically decreasing runs for compositions in standard order.

Original entry on oeis.org

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

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. a(n) is the number of maximal weakly decreasing runs in this composition. Alternatively, a(n) is one plus the number of strict ascents in the same composition. For example, the weakly decreasing runs of the 1234567th composition are ((3,2,1),(2,2,1),(2),(5,1,1,1)), so a(1234567) = 4. The 3 strict ascents together with the weak descents are: 3 >= 2 >= 1 < 2 >= 2 >= 1 < 2 < 5 >= 1 >= 1 >= 1. - Gus Wiseman, Apr 08 2020

Examples

			Composition number 11 is 2,1,1; the decreasing runs are 2,1,1; so a(11) = 1.
The table starts:
  0
  1
  1 1
  1 1 2 1
  1 1 1 1 2 2 2 1
  1 1 1 1 2 1 2 1 2 2 2 2 2 2 2 1
  1 1 1 1 1 1 2 1 2 2 1 1 2 2 2 1 2 2 2 2 3 2 3 2 2 2 2 2 2 2 2 1
		

Crossrefs

Cf. A066099, A124760, A011782 (row lengths).
Compositions of n with k strict ascents are A238343.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Sum is A070939.
- Weakly decreasing compositions are A114994.
- Weakly decreasing runs are counted by A124765.
- Weakly increasing runs are counted by A124766.
- Equal runs are counted by A124767.
- Strictly increasing runs are counted by A124768.
- Strictly decreasing runs are counted by A124769.
- Weakly increasing compositions are A225620.
- Reverse is A228351 (triangle).
- Strict compositions are A233564.
- Constant compositions are A272919.
- Normal compositions are A333217.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
- Anti-runs are counted by A333381.
- Adjacent unequal pairs are counted by A333382.
- Anti-runs are A333489.
- Runs-resistance is A333628.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Length[Split[stc[n],GreaterEqual]],{n,0,100}] (* Gus Wiseman, Apr 08 2020 *)

Formula

a(0) = 0, a(n) = A124760(n) + 1 for n > 0.
Showing 1-10 of 68 results. Next