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.

Showing 1-6 of 6 results.

A333217 Numbers k such that the k-th composition in standard order covers an initial interval of positive integers.

Original entry on oeis.org

0, 1, 3, 5, 6, 7, 11, 13, 14, 15, 21, 22, 23, 26, 27, 29, 30, 31, 37, 38, 41, 43, 44, 45, 46, 47, 50, 52, 53, 54, 55, 58, 59, 61, 62, 63, 75, 77, 78, 83, 85, 86, 87, 89, 90, 91, 92, 93, 94, 95, 101, 102, 105, 106, 107, 108, 109, 110, 111, 114, 116, 117, 118
Offset: 1

Views

Author

Gus Wiseman, Mar 15 2020

Keywords

Comments

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 of terms together with the corresponding compositions begins:
    0: ()              37: (3,2,1)           75: (3,2,1,1)
    1: (1)             38: (3,1,2)           77: (3,1,2,1)
    3: (1,1)           41: (2,3,1)           78: (3,1,1,2)
    5: (2,1)           43: (2,2,1,1)         83: (2,3,1,1)
    6: (1,2)           44: (2,1,3)           85: (2,2,2,1)
    7: (1,1,1)         45: (2,1,2,1)         86: (2,2,1,2)
   11: (2,1,1)         46: (2,1,1,2)         87: (2,2,1,1,1)
   13: (1,2,1)         47: (2,1,1,1,1)       89: (2,1,3,1)
   14: (1,1,2)         50: (1,3,2)           90: (2,1,2,2)
   15: (1,1,1,1)       52: (1,2,3)           91: (2,1,2,1,1)
   21: (2,2,1)         53: (1,2,2,1)         92: (2,1,1,3)
   22: (2,1,2)         54: (1,2,1,2)         93: (2,1,1,2,1)
   23: (2,1,1,1)       55: (1,2,1,1,1)       94: (2,1,1,1,2)
   26: (1,2,2)         58: (1,1,2,2)         95: (2,1,1,1,1,1)
   27: (1,2,1,1)       59: (1,1,2,1,1)      101: (1,3,2,1)
   29: (1,1,2,1)       61: (1,1,1,2,1)      102: (1,3,1,2)
   30: (1,1,1,2)       62: (1,1,1,1,2)      105: (1,2,3,1)
   31: (1,1,1,1,1)     63: (1,1,1,1,1,1)    106: (1,2,2,2)
		

Crossrefs

Sequences covering an initial interval are counted by A000670.
Composition in standard order are A066099.
The case of strictly increasing initial intervals is A164894.
The case of strictly decreasing initial intervals is A246534.
The case of permutations is A333218.
The weakly increasing version is A333379.
The weakly decreasing version is A333380.

Programs

  • Mathematica
    normQ[m_]:=Or[m=={},Union[m]==Range[Max[m]]];
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,100],normQ[stc[#]]&]

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

Views

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

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

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. 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.

A333256 Numbers k such that the k-th composition in standard order is strictly decreasing.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 16, 17, 18, 32, 33, 34, 37, 64, 65, 66, 68, 69, 128, 129, 130, 132, 133, 137, 256, 257, 258, 260, 261, 264, 265, 274, 512, 513, 514, 516, 517, 520, 521, 529, 530, 549, 1024, 1025, 1026, 1028, 1029, 1032, 1033, 1040, 1041, 1042, 1058, 1061
Offset: 1

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 sequence of positive terms together with the corresponding compositions begins:
     1: (1)         128: (8)         517: (7,2,1)
     2: (2)         129: (7,1)       520: (6,4)
     4: (3)         130: (6,2)       521: (6,3,1)
     5: (2,1)       132: (5,3)       529: (5,4,1)
     8: (4)         133: (5,2,1)     530: (5,3,2)
     9: (3,1)       137: (4,3,1)     549: (4,3,2,1)
    16: (5)         256: (9)        1024: (11)
    17: (4,1)       257: (8,1)      1025: (10,1)
    18: (3,2)       258: (7,2)      1026: (9,2)
    32: (6)         260: (6,3)      1028: (8,3)
    33: (5,1)       261: (6,2,1)    1029: (8,2,1)
    34: (4,2)       264: (5,4)      1032: (7,4)
    37: (3,2,1)     265: (5,3,1)    1033: (7,3,1)
    64: (7)         274: (4,3,2)    1040: (6,5)
    65: (6,1)       512: (10)       1041: (6,4,1)
    66: (5,2)       513: (9,1)      1042: (6,3,2)
    68: (4,3)       514: (8,2)      1058: (5,4,2)
    69: (4,2,1)     516: (7,3)      1061: (5,3,2,1)
		

Crossrefs

Strictly increasing runs are counted by A124768.
The normal case is A246534.
The weakly decreasing version is A114994.
The weakly increasing version is A225620.
The unequal version is A233564.
The equal version is A272919.
The strictly increasing version is A333255.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,1000],Greater@@stc[#]&]

A124760 Number of rises for compositions in standard order.

Original entry on oeis.org

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

Views

Author

Keywords

Comments

The standard order of compositions is given by A066099.
A114994 seems to give the positions of zeros. - Antti Karttunen, Jul 09 2017
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 one fewer than the number of maximal weakly decreasing runs in this composition. Alternatively, a(n) is 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 - 1 = 3. 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; 2>=1>=1, so a(11) = 0.
The table starts:
  0
  0
  0 0
  0 0 1 0
  0 0 0 0 1 1 1 0
  0 0 0 0 1 0 1 0 1 1 1 1 1 1 1 0
  0 0 0 0 0 0 1 0 1 1 0 0 1 1 1 0 1 1 1 1 2 1 2 1 1 1 1 1 1 1 1 0
		

Crossrefs

Cf. A066099, A124761, A124762, A124763, A124764, A011782 (row lengths), A045883 (row sums), A233249, A333213, A333380.
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[Select[Partition[stc[n],2,1],Less@@#&]],{n,0,100}] (* Gus Wiseman, Apr 08 2020 *)
  • PARI
    A066099row(n) = {my(v=vector(n), j=0, k=0); while(n>0, k++; if(n%2==1, v[j++]=k; k=0); n\=2);  vector(j, i, v[j-i+1]); } \\ Returns empty for n=0. - From code of Franklin T. Adams-Watters in A066099.
    A124760(n) = { my(v=A066099row(n), r=0); for(i=2,length(v),r += (v[i]>v[i-1])); (r); }; \\ Antti Karttunen, Jul 09 2017

Formula

For a composition b(1),...,b(k), a(n) = Sum_{i = 1 .. k-1} [b(i+1) > b(i)], where [ ] is Iverson bracket, giving in this case 1 only if b(i+1) > b(i), and 0 otherwise. - Formula clarified by Antti Karttunen, Jul 10 2017
For n > 0, a(n) = A124765(n) - 1. - Gus Wiseman, Apr 08 2020

A333379 Numbers k such that the k-th composition in standard order is weakly increasing and covers an initial interval of positive integers.

Original entry on oeis.org

0, 1, 3, 6, 7, 14, 15, 26, 30, 31, 52, 58, 62, 63, 106, 116, 122, 126, 127, 212, 234, 244, 250, 254, 255, 420, 426, 468, 490, 500, 506, 510, 511, 840, 852, 932, 938, 980, 1002, 1012, 1018, 1022, 1023, 1700, 1706, 1864, 1876, 1956, 1962, 2004, 2026, 2036, 2042
Offset: 1

Views

Author

Gus Wiseman, Mar 21 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 of terms together with the corresponding compositions begins:
    0: ()               127: (1,1,1,1,1,1,1)
    1: (1)              212: (1,2,2,3)
    3: (1,1)            234: (1,1,2,2,2)
    6: (1,2)            244: (1,1,1,2,3)
    7: (1,1,1)          250: (1,1,1,1,2,2)
   14: (1,1,2)          254: (1,1,1,1,1,1,2)
   15: (1,1,1,1)        255: (1,1,1,1,1,1,1,1)
   26: (1,2,2)          420: (1,2,3,3)
   30: (1,1,1,2)        426: (1,2,2,2,2)
   31: (1,1,1,1,1)      468: (1,1,2,2,3)
   52: (1,2,3)          490: (1,1,1,2,2,2)
   58: (1,1,2,2)        500: (1,1,1,1,2,3)
   62: (1,1,1,1,2)      506: (1,1,1,1,1,2,2)
   63: (1,1,1,1,1,1)    510: (1,1,1,1,1,1,1,2)
  106: (1,2,2,2)        511: (1,1,1,1,1,1,1,1,1)
  116: (1,1,2,3)        840: (1,2,3,4)
  122: (1,1,1,2,2)      852: (1,2,2,2,3)
  126: (1,1,1,1,1,2)    932: (1,1,2,3,3)
		

Crossrefs

Sequences covering an initial interval are counted by A000670.
Compositions in standard order are A066099.
Weakly increasing runs are counted by A124766.
Removing the covering condition gives A225620.
Removing the ordering condition gives A333217.
The strictly increasing case is A164894.
The strictly decreasing version is A246534.
The unequal version is A333218.
The weakly decreasing version is A333380.

Programs

  • Mathematica
    normQ[m_]:=Or[m=={},Union[m]==Range[Max[m]]];
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,1000],normQ[stc[#]]&&LessEqual@@stc[#]&]

Formula

Intersection of A333217 and A225620.
Showing 1-6 of 6 results.