A233564 -id:A233564 - cp's OEIS frontend

A124762 Number of levels for compositions in standard order.

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

Franklin T. Adams-Watters, Nov 06 2006

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

			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

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.
Cf. A000120, A029931, A048793, A059893, A070939, A114994, A225620, A228351.

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 *)

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

Omar E. Pol, Aug 03 2013

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

			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)

Column 1 is A000079. Row n has length A000041(n). Right border gives A000225.
Cf. A026792, A065609, A135010, A141285, A186114, A194446, A194546, A206437, A207779, A211978, A225600, A225610, A228351.
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.
Cf. A000120, A029931, A048793, A070939, A124766, A233249, A333219, A333220.

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

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

A333219 Heinz number of the n-th composition in standard order.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 6, 8, 7, 10, 9, 12, 10, 12, 12, 16, 11, 14, 15, 20, 15, 18, 18, 24, 14, 20, 18, 24, 20, 24, 24, 32, 13, 22, 21, 28, 25, 30, 30, 40, 21, 30, 27, 36, 30, 36, 36, 48, 22, 28, 30, 40, 30, 36, 36, 48, 28, 40, 36, 48, 40, 48, 48, 64, 17, 26, 33, 44

Offset: 1

Gus Wiseman, Mar 16 2020

nonn

Includes all positive integers.
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 Heinz number of a composition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).

			The sequence of terms together with their prime indices begins:
   1: {}           15: {2,3}          25: {3,3}
   2: {1}          20: {1,1,3}        30: {1,2,3}
   3: {2}          15: {2,3}          30: {1,2,3}
   4: {1,1}        18: {1,2,2}        40: {1,1,1,3}
   5: {3}          18: {1,2,2}        21: {2,4}
   6: {1,2}        24: {1,1,1,2}      30: {1,2,3}
   6: {1,2}        14: {1,4}          27: {2,2,2}
   8: {1,1,1}      20: {1,1,3}        36: {1,1,2,2}
   7: {4}          18: {1,2,2}        30: {1,2,3}
  10: {1,3}        24: {1,1,1,2}      36: {1,1,2,2}
   9: {2,2}        20: {1,1,3}        36: {1,1,2,2}
  12: {1,1,2}      24: {1,1,1,2}      48: {1,1,1,1,2}
  10: {1,3}        24: {1,1,1,2}      22: {1,5}
  12: {1,1,2}      32: {1,1,1,1,1}    28: {1,1,4}
  12: {1,1,2}      13: {6}            30: {1,2,3}
  16: {1,1,1,1}    22: {1,5}          40: {1,1,1,3}
  11: {5}          21: {2,4}          30: {1,2,3}
  14: {1,4}        28: {1,1,4}        36: {1,1,2,2}

The length of the k-th composition in standard order is A000120(k).
The sum of the k-th composition in standard order is A070939(k).
The maximum of the k-th composition in standard order is A070939(k).
A partial inverse is A333220. See also A233249.
Cf. A048793, A056239, A066099, A112798, A114994, A124767, A213925, A225620, A228351, A233564, A272919, A333218.

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

A056239(a(n)) = A070939(n).

A374249 Numbers k such that the k-th composition in standard order has its equal parts contiguous.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 26, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 47, 48, 50, 52, 56, 58, 60, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 78, 79, 80, 81, 83, 84, 85

Offset: 1

Gus Wiseman, Jul 13 2024

nonn

These are compositions avoiding the patterns (1,2,1) and (2,1,2).

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 terms together with their standard compositions begin:
   0: ()
   1: (1)
   2: (2)
   3: (1,1)
   4: (3)
   5: (2,1)
   6: (1,2)
   7: (1,1,1)
   8: (4)
   9: (3,1)
  10: (2,2)
  11: (2,1,1)
  12: (1,3)
  14: (1,1,2)
  15: (1,1,1,1)
  16: (5)
See A374253 for the complement: 13, 22, 25, 27, 29, ...

Gus Wiseman, Statistics, classes, and transformations of standard compositions

The strict (also anti-run) case is A233564, counted by A032020.
Compositions of this type are counted by A274174.
Permutations of prime indices of this type are counted by A333175.
The complement is A374253 (anti-run A374254), counted by A335548.
A003242 counts anti-run compositions, ranks A333489.
A011782 counts compositions.
A066099 lists compositions in standard order.
A124767 counts runs in standard compositions, anti-runs A333381.
A333755 counts compositions by number of runs.
A335454 counts patterns matched by standard compositions.
A335462 counts (1,2,1)- and (2,1,2)-matching permutations of prime indices.
- A335470 counts (1,2,1)-matching compositions, ranks A335466.
- A335471 counts (1,2,1)-avoiding compositions, ranks A335467.
- A335472 counts (2,1,2)-matching compositions, ranks A335468.
- A335473 counts (2,1,2)-avoiding compositions, ranks A335469.
Cf. A106356, A124762, A238130, A238279, A261982, A272919, A333382, A335450, A335460, A335524, A335525.

stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Select[Range[0,100],UnsameQ@@First/@Split[stc[#]]&]

Equals A335467 /\ A335469.

A124766 Number of monotonically increasing runs for compositions in standard order.

Original entry on oeis.org

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

Offset: 0

Franklin T. Adams-Watters, Nov 06 2006

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

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

Cf. A066099, A124761, A011782 (row lengths).
Compositions of n with k strict descents are A238343.
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 (this sequence).
- 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.
- Initial intervals are A246534.
- Constant compositions are A272919.
- Normal compositions are A333217.
- Permutations are A333218.
- Strictly decreasing compositions are A333255.
- Strictly increasing compositions are A333256.
- Runs-resistance is A333628.
Cf. A124760, A124763, A124764, A333213, A333220, A333379.

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

a(0) = 0, a(n) = A124761(n) + 1 for n > 0.

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

Gus Wiseman, Mar 24 2020

nonn

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

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

Wikipedia, Longest increasing subsequence

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.
Cf. A000005, A000120, A003242, A029931, A048793, A059893, A070939, A114994, A225620, A228351, A238424.

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}]

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

A333218 Numbers k such that the k-th composition in standard order is a permutation (of an initial interval).

Original entry on oeis.org

0, 1, 5, 6, 37, 38, 41, 44, 50, 52, 549, 550, 553, 556, 562, 564, 581, 582, 593, 600, 610, 616, 649, 652, 657, 664, 708, 712, 786, 788, 802, 808, 836, 840, 16933, 16934, 16937, 16940, 16946, 16948, 16965, 16966, 16977, 16984, 16994, 17000, 17033, 17036, 17041

Offset: 1

Gus Wiseman, Mar 16 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 sequence of terms together with their corresponding compositions begins:
        0: ()             593: (3,2,4,1)      16937: (5,4,2,3,1)
        1: (1)            600: (3,2,1,4)      16940: (5,4,2,1,3)
        5: (2,1)          610: (3,1,4,2)      16946: (5,4,1,3,2)
        6: (1,2)          616: (3,1,2,4)      16948: (5,4,1,2,3)
       37: (3,2,1)        649: (2,4,3,1)      16965: (5,3,4,2,1)
       38: (3,1,2)        652: (2,4,1,3)      16966: (5,3,4,1,2)
       41: (2,3,1)        657: (2,3,4,1)      16977: (5,3,2,4,1)
       44: (2,1,3)        664: (2,3,1,4)      16984: (5,3,2,1,4)
       50: (1,3,2)        708: (2,1,4,3)      16994: (5,3,1,4,2)
       52: (1,2,3)        712: (2,1,3,4)      17000: (5,3,1,2,4)
      549: (4,3,2,1)      786: (1,4,3,2)      17033: (5,2,4,3,1)
      550: (4,3,1,2)      788: (1,4,2,3)      17036: (5,2,4,1,3)
      553: (4,2,3,1)      802: (1,3,4,2)      17041: (5,2,3,4,1)
      556: (4,2,1,3)      808: (1,3,2,4)      17048: (5,2,3,1,4)
      562: (4,1,3,2)      836: (1,2,4,3)      17092: (5,2,1,4,3)
      564: (4,1,2,3)      840: (1,2,3,4)      17096: (5,2,1,3,4)
      581: (3,4,2,1)    16933: (5,4,3,2,1)    17170: (5,1,4,3,2)
      582: (3,4,1,2)    16934: (5,4,3,1,2)    17172: (5,1,4,2,3)

A superset of A164894.
Also a superset of A246534.
Not requiring the parts to be distinct gives A333217.
Cf. A000120, A000142, A048793, A066099, A070939, A114994, A225620, A233564, A272919, A333219, A333221, A333255, A333256.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Select[Range[0,1000],#==0||UnsameQ@@stc[#]&&Max@@stc[#]==Length[stc[#]]&]

A275692 Numbers k such that every rotation of the binary digits of k is less than k.

Original entry on oeis.org

0, 1, 2, 4, 6, 8, 12, 14, 16, 20, 24, 26, 28, 30, 32, 40, 48, 50, 52, 56, 58, 60, 62, 64, 72, 80, 84, 96, 98, 100, 104, 106, 108, 112, 114, 116, 118, 120, 122, 124, 126, 128, 144, 160, 164, 168, 192, 194, 196, 200, 202, 208, 210, 212, 216, 218, 224, 226, 228

Offset: 1

Robert Israel, Aug 05 2016

nonn

0, and terms of A065609 that are not in A121016.
Number of terms with d binary digits is A001037(d).
Take the binary representation of a(n), reverse it, add 1 to each digit.  The result is the decimal representation of A102659(n).
From Gus Wiseman, Apr 19 2020: (Start)
Also numbers k such that the k-th composition in standard order (row k of A066099) is a Lyndon word. For example, the sequence of all Lyndon words begins:
()            52: (1,2,3)        118: (1,1,2,1,2)
(1)           56: (1,1,4)        120: (1,1,1,4)
(2)           58: (1,1,2,2)      122: (1,1,1,2,2)
(3)           60: (1,1,1,3)      124: (1,1,1,1,3)
(1,2)         62: (1,1,1,1,2)    126: (1,1,1,1,1,2)
(4)           64: (7)            128: (8)
(1,3)         72: (3,4)          144: (3,5)
(1,1,2)       80: (2,5)          160: (2,6)
(5)           84: (2,2,3)        164: (2,3,3)
(2,3)         96: (1,6)          168: (2,2,4)
(1,4)         98: (1,4,2)        192: (1,7)
(1,2,2)      100: (1,3,3)        194: (1,5,2)
(1,1,3)      104: (1,2,4)        196: (1,4,3)
(1,1,1,2)    106: (1,2,2,2)      200: (1,3,4)
(6)          108: (1,2,1,3)      202: (1,3,2,2)
(2,4)        112: (1,1,5)        208: (1,2,5)
(1,5)        114: (1,1,3,2)      210: (1,2,3,2)
(1,3,2)      116: (1,1,2,3)      212: (1,2,2,3)
(End)

			6 is in the sequence because its binary representation 110 is greater than all the rotations 011 and 101.
10 is not in the sequence because its binary representation 1010 is unchanged under rotation by 2 places.
From _Gus Wiseman_, Oct 31 2019: (Start)
The sequence of terms together with their binary expansions and binary indices begins:
    1:       1 ~ {1}
    2:      10 ~ {2}
    4:     100 ~ {3}
    6:     110 ~ {2,3}
    8:    1000 ~ {4}
   12:    1100 ~ {3,4}
   14:    1110 ~ {2,3,4}
   16:   10000 ~ {5}
   20:   10100 ~ {3,5}
   24:   11000 ~ {4,5}
   26:   11010 ~ {2,4,5}
   28:   11100 ~ {3,4,5}
   30:   11110 ~ {2,3,4,5}
   32:  100000 ~ {6}
   40:  101000 ~ {4,6}
   48:  110000 ~ {5,6}
   50:  110010 ~ {2,5,6}
   52:  110100 ~ {3,5,6}
   56:  111000 ~ {4,5,6}
   58:  111010 ~ {2,4,5,6}
(End)

Robert Israel, Table of n, a(n) for n = 1..9868

A similar concept is A328596.
Numbers whose binary expansion is aperiodic are A328594.
Numbers whose reversed binary expansion is a necklace are A328595.
Binary necklaces are A000031.
Binary Lyndon words are A001037.
Lyndon compositions are A059966.
Length of Lyndon factorization of binary expansion is A211100.
Length of co-Lyndon factorization of binary expansion is A329312.
Length of Lyndon factorization of reversed binary expansion is A329313.
Length of co-Lyndon factorization of reversed binary expansion is A329326.
Cf. A000031, A000740, A008965, A027375, A102659, A121016.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Necklaces are A065609.
- Sum is A070939.
- Rotational symmetries are counted by A138904.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon compositions are A275692 (this sequence).
- Co-Lyndon compositions are A326774.
- Rotational period is A333632.
- Co-necklaces are A333764.
- Co-Lyndon factorizations are counted by A333765.
- Lyndon factorizations are counted by A333940.
- Reversed necklaces are A333943.
Cf. A034691, A060223, A124767, A269134, A292884.

filter:= proc(n) local L, k;
  L:= convert(convert(n,binary),string);
  for k from 1 to length(L)-1 do
    if lexorder(L,StringTools:-Rotate(L,k)) then return false fi;
  od;
  true
end proc:
select(filter, [$0..1000]);

filterQ[n_] := Module[{bits, rr}, bits = IntegerDigits[n, 2]; rr = NestList[RotateRight, bits, Length[bits]-1] // Rest; AllTrue[rr, FromDigits[#, 2] < n&]];
Select[Range[0, 1000], filterQ] (* Jean-François Alcover, Apr 29 2019 *)

def ok(n):
    b = bin(n)[2:]
    return all(b[i:] + b[:i] < b for i in range(1, len(b)))
print([k for k in range(230) if ok(k)]) # Michael S. Branicky, May 26 2022

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

Franklin T. Adams-Watters, Nov 06 2006

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

			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

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.
Cf. A029931, A124760, A124761, A124764, A238343, A333220.

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 *)

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

Previous Showing 11-20 of 155 results. Next

cp's OEIS Frontend

A102726 Number of compositions of the integer n into positive parts that avoid a fixed pattern of three letters.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A124762 Number of levels for compositions in standard order.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A333219 Heinz number of the n-th composition in standard order.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A374249 Numbers k such that the k-th composition in standard order has its equal parts contiguous.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A124766 Number of monotonically increasing runs for compositions in standard order.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A333218 Numbers k such that the k-th composition in standard order is a permutation (of an initial interval).

Views

Author