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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A334433 Heinz numbers of all integer partitions sorted first by sum, then by length, and finally lexicographically.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 12, 16, 11, 15, 14, 18, 20, 24, 32, 13, 25, 21, 22, 27, 30, 28, 36, 40, 48, 64, 17, 35, 33, 26, 45, 50, 42, 44, 54, 60, 56, 72, 80, 96, 128, 19, 49, 55, 39, 34, 75, 63, 70, 66, 52, 81, 90, 100, 84, 88, 108, 120, 112, 144, 160, 192, 256
Offset: 0

Views

Author

Gus Wiseman, Apr 30 2020

Keywords

Comments

First differs from A334435 at a(75) = 99, A334435(75) = 98.
A permutation of the positive integers.
This is the Abramowitz-Stegun ordering of integer partitions when the parts are read in the usual (weakly decreasing) order. The case of reversed (weakly increasing) partitions is A185974.
The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.
As a triangle with row lengths A000041, the sequence starts {{1},{2},{3,4},{5,6,8},...}, so offset is 0.

Examples

			The sequence of terms together with their prime indices begins:
    1: {}            32: {1,1,1,1,1}       42: {1,2,4}
    2: {1}           13: {6}               44: {1,1,5}
    3: {2}           25: {3,3}             54: {1,2,2,2}
    4: {1,1}         21: {2,4}             60: {1,1,2,3}
    5: {3}           22: {1,5}             56: {1,1,1,4}
    6: {1,2}         27: {2,2,2}           72: {1,1,1,2,2}
    8: {1,1,1}       30: {1,2,3}           80: {1,1,1,1,3}
    7: {4}           28: {1,1,4}           96: {1,1,1,1,1,2}
    9: {2,2}         36: {1,1,2,2}        128: {1,1,1,1,1,1,1}
   10: {1,3}         40: {1,1,1,3}         19: {8}
   12: {1,1,2}       48: {1,1,1,1,2}       49: {4,4}
   16: {1,1,1,1}     64: {1,1,1,1,1,1}     55: {3,5}
   11: {5}           17: {7}               39: {2,6}
   15: {2,3}         35: {3,4}             34: {1,7}
   14: {1,4}         33: {2,5}             75: {2,3,3}
   18: {1,2,2}       26: {1,6}             63: {2,2,4}
   20: {1,1,3}       45: {2,2,3}           70: {1,3,4}
   24: {1,1,1,2}     50: {1,3,3}           66: {1,2,5}
Triangle begins:
   1
   2
   3   4
   5   6   8
   7   9  10  12  16
  11  15  14  18  20  24  32
  13  25  21  22  27  30  28  36  40  48  64
  17  35  33  26  45  50  42  44  54  60  56  72  80  96 128
This corresponds to the tetrangle:
                  0
                 (1)
               (2)(11)
             (3)(21)(111)
        (4)(22)(31)(211)(1111)
  (5)(32)(41)(221)(311)(2111)(11111)
		

Crossrefs

Row lengths are A000041.
Compositions under the same order are A124734 (triangle).
The version for reversed (weakly increasing) partitions is A185974.
The constructive version is A334301.
Ignoring length gives A334434, or A334437 for reversed partitions.
The dual version (sum/length/revlex) is A334438.
Lexicographically ordered reversed partitions are A026791.
Reversed partitions in Abramowitz-Stegun (sum/length/lex) order are A036036.
Partitions in increasing-length colexicographic order (sum/length/colex) are A036037.
Graded reverse-lexicographically ordered partitions are A080577.
Sorting reversed partitions by Heinz number gives A112798.
Graded lexicographically ordered partitions are A193073.
Graded Heinz numbers are A215366.
Sorting partitions by Heinz number gives A296150.
Partitions in increasing-length reverse-lexicographic order (sum/length/revlex) are A334439 (not A036037).

Programs

  • Mathematica
    Join@@Table[Times@@Prime/@#&/@Sort[IntegerPartitions[n]],{n,0,8}]

Formula

A001222(a(n)) = A036043(n).

A334302 Irregular triangle read by rows where row k is the k-th reversed integer partition, if reversed partitions are sorted first by sum, then by length, and finally reverse-lexicographically.

Original entry on oeis.org

1, 2, 1, 1, 3, 1, 2, 1, 1, 1, 4, 2, 2, 1, 3, 1, 1, 2, 1, 1, 1, 1, 5, 2, 3, 1, 4, 1, 2, 2, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 1, 6, 3, 3, 2, 4, 1, 5, 2, 2, 2, 1, 2, 3, 1, 1, 4, 1, 1, 2, 2, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 7, 3, 4, 2, 5, 1, 6, 2, 2, 3
Offset: 0

Views

Author

Gus Wiseman, Apr 30 2020

Keywords

Examples

			The sequence of all reversed partitions begins:
  ()         (1,4)        (1,1,1,1,2)
  (1)        (1,2,2)      (1,1,1,1,1,1)
  (2)        (1,1,3)      (7)
  (1,1)      (1,1,1,2)    (3,4)
  (3)        (1,1,1,1,1)  (2,5)
  (1,2)      (6)          (1,6)
  (1,1,1)    (3,3)        (2,2,3)
  (4)        (2,4)        (1,3,3)
  (2,2)      (1,5)        (1,2,4)
  (1,3)      (2,2,2)      (1,1,5)
  (1,1,2)    (1,2,3)      (1,2,2,2)
  (1,1,1,1)  (1,1,4)      (1,1,2,3)
  (5)        (1,1,2,2)    (1,1,1,4)
  (2,3)      (1,1,1,3)    (1,1,1,2,2)
This sequence can also be interpreted as the following triangle, whose n-th row is itself a finite triangle with A000041(n) rows.
                            0
                           (1)
                        (2) (1,1)
                    (3) (1,2) (1,1,1)
            (4) (2,2) (1,3) (1,1,2) (1,1,1,1)
  (5) (2,3) (1,4) (1,2,2) (1,1,3) (1,1,1,2) (1,1,1,1,1)
Showing partitions as their Heinz numbers (see A334435) gives:
   1
   2
   3   4
   5   6   8
   7   9  10  12  16
  11  15  14  18  20  24  32
  13  25  21  22  27  30  28  36  40  48  64
  17  35  33  26  45  50  42  44  54  60  56  72  80  96 128
		

Crossrefs

Row lengths are A036043.
Lexicographically ordered reversed partitions are A026791.
The dual ordering (sum/length/lex) of reversed partitions is A036036.
Reverse-lexicographically ordered partitions are A080577.
Sorting reversed partitions by Heinz number gives A112798.
Lexicographically ordered partitions are A193073.
Graded Heinz numbers are A215366.
Ignoring length gives A228531.
Sorting partitions by Heinz number gives A296150.
The version for compositions is A296774.
The dual ordering (sum/length/lex) of non-reversed partitions is A334301.
Taking Heinz numbers gives A334435.
The version for regular (non-reversed) partitions is A334439 (not A036037).

Programs

  • Mathematica
    revlensort[f_,c_]:=If[Length[f]!=Length[c],Length[f]
    				

A368109 Number of ways to choose a binary index of each binary index of n.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 4, 4, 4, 4, 8, 8, 8, 8, 3, 3, 3, 3, 6, 6, 6, 6, 3, 3, 3, 3, 6, 6, 6, 6, 6, 6, 6, 6, 12, 12, 12
Offset: 0

Views

Author

Gus Wiseman, Dec 12 2023

Keywords

Comments

First differs from A367912 at a(52) = 8, A367912(52) = 7.
A binary index of n (row n of A048793) is any position of a 1 in its reversed binary expansion. For example, 18 has reversed binary expansion (0,1,0,0,1) and binary indices {2,5}.
Run-lengths are all 4 or 8.

Examples

			The binary indices of binary indices of 20 are {{1,2},{1,3}}, with choices (1,1), (1,3), (2,1), (2,3), so a(20) = 4.
The binary indices of binary indices of 52 are {{1,2},{1,3},{2,3}}, with choices (1,1,1), (1,1,3), (1,3,2), (1,3,3), (2,1,2), (2,1,3), (2,3,2), (2,3,3), so a(52) = 8.
		

Crossrefs

All entries appear to belong to A003586.
Positions of ones are A253317.
The version for prime indices is A355741, for multisets A355744.
Choosing a multiset (not sequence) gives A367912, firsts A367913.
Positions of first appearances are A368111, sorted A368112.
A048793 lists binary indices, length A000120, sum A029931.
A058891 counts set-systems, covering A003465, connected A323818.
A070939 gives length of binary expansion.
A096111 gives product of binary indices.

Programs

  • Mathematica
    bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n, 2]],1];
    Table[Length[Tuples[bpe/@bpe[n]]], {n,0,100}]

Formula

a(n) = Product_{k in A048793(n)} A000120(k).

A133457 Irregular triangle read by rows: row n gives exponents in expression for n as a sum of powers of 2.

Original entry on oeis.org

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

Views

Author

Masahiko Shin, Nov 27 2007

Keywords

Comments

This sequence contains every increasing finite sequence. For example, the finite sequence {0,2,3,5} arises from n = 45.
Essentially A030308(n,k)*k, then entries removed where A030308(n,k)=0. - R. J. Mathar, Nov 30 2007
In the corresponding irregular triangle {a(n)+1}, the m-th row gives all positive integer roots m_i of polynomial {m,k}. - see link [Shevelev]; see also A264613. - Vladimir Shevelev, Dec 13 2015

Examples

			1 = 2^0.
2 = 2^1.
3 = 2^0 + 2^1.
4 = 2^2.
5 = 2^0 + 2^2.
etc. and reading the exponents gives the rows of the triangle.
		

Crossrefs

Cf. A073642 (row sums), A272011 (rows reversed).

Programs

  • Haskell
    a133457 n k = a133457_tabf !! (n-1) !! n
    a133457_row n = a133457_tabf !! (n-1)
    a133457_tabf = map (fst . unzip . filter ((> 0) . snd) . zip [0..]) $
                       tail a030308_tabf
    -- Reinhard Zumkeller, Oct 28 2013, Feb 06 2013
  • Maple
    A133457 := proc(n) local a,bdigs,i ; a := [] ; bdigs := convert(n,base,2) ; for i from 1 to nops(bdigs) do if op(i,bdigs) <> 0 then a := [op(a),i-1] ; fi ; od: a ; end: seq(op(A133457(n)),n=1..80) ; # R. J. Mathar, Nov 30 2007
  • Mathematica
    Array[Join @@ Position[#, 1] - 1 &@ Reverse@ IntegerDigits[#, 2] &, 41] // Flatten (* Michael De Vlieger, Oct 08 2017 *)

Formula

a(n) = A048793(n) - 1.

Extensions

More terms from R. J. Mathar, Nov 30 2007

A335373 Numbers k such that the k-th composition in standard order (A066099) is not unimodal.

Original entry on oeis.org

22, 38, 44, 45, 46, 54, 70, 76, 77, 78, 86, 88, 89, 90, 91, 92, 93, 94, 102, 108, 109, 110, 118, 134, 140, 141, 142, 148, 150, 152, 153, 154, 155, 156, 157, 158, 166, 172, 173, 174, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 198
Offset: 1

Views

Author

Gus Wiseman, Jun 03 2020

Keywords

Comments

A sequence of integers is unimodal if it is the concatenation of a weakly increasing and a weakly decreasing sequence.
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 sequence together with the corresponding compositions begins:
  22: (2,1,2)
  38: (3,1,2)
  44: (2,1,3)
  45: (2,1,2,1)
  46: (2,1,1,2)
  54: (1,2,1,2)
  70: (4,1,2)
  76: (3,1,3)
  77: (3,1,2,1)
  78: (3,1,1,2)
  86: (2,2,1,2)
  88: (2,1,4)
  89: (2,1,3,1)
  90: (2,1,2,2)
  91: (2,1,2,1,1)
  92: (2,1,1,3)
  93: (2,1,1,2,1)
  94: (2,1,1,1,2)
		

Crossrefs

The dual version (non-co-unimodal compositions) is A335374.
The case that is not co-unimodal either is A335375.
Unimodal compositions are A001523.
Unimodal normal sequences are A007052.
Unimodal permutations are A011782.
Non-unimodal permutations are A059204.
Non-unimodal compositions are A115981.
Non-unimodal normal sequences are A328509.
Numbers with non-unimodal unsorted prime signature are A332282.
Partitions with non-unimodal 0-appended first differences are A332284.
Non-unimodal permutations of the multiset of prime indices of n are A332671.

Programs

  • Mathematica
    unimodQ[q_]:=Or[Length[q]<=1,If[q[[1]]<=q[[2]],unimodQ[Rest[q]],OrderedQ[Reverse[q]]]];
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,200],!unimodQ[stc[#]]&]

A372429 Sum of binary indices of prime(n). Sum of positions of ones in the reversed binary expansion of prime(n).

Original entry on oeis.org

2, 3, 4, 6, 7, 8, 6, 8, 11, 13, 15, 10, 11, 13, 16, 15, 18, 19, 10, 13, 12, 17, 15, 17, 14, 17, 19, 20, 21, 19, 28, 11, 13, 15, 17, 19, 21, 17, 20, 22, 22, 23, 29, 16, 19, 21, 23, 30, 24, 25, 26, 31, 27, 33, 10, 15, 17, 19, 18, 19, 21, 19, 23, 26, 25, 28, 23
Offset: 1

Views

Author

Gus Wiseman, May 02 2024

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.
Do 2, 3, 4, 7, 12, 14 appear just once?
Are 1, 5, 9 missing?
The above questions hold true up to n = 10^6. - John Tyler Rascoe, May 21 2024

Examples

			The primes together with their binary expansions and binary indices begin:
   2:      10 ~ {2}
   3:      11 ~ {1,2}
   5:     101 ~ {1,3}
   7:     111 ~ {1,2,3}
  11:    1011 ~ {1,2,4}
  13:    1101 ~ {1,3,4}
  17:   10001 ~ {1,5}
  19:   10011 ~ {1,2,5}
  23:   10111 ~ {1,2,3,5}
  29:   11101 ~ {1,3,4,5}
  31:   11111 ~ {1,2,3,4,5}
  37:  100101 ~ {1,3,6}
  41:  101001 ~ {1,4,6}
  43:  101011 ~ {1,2,4,6}
  47:  101111 ~ {1,2,3,4,6}
  53:  110101 ~ {1,3,5,6}
  59:  111011 ~ {1,2,4,5,6}
  61:  111101 ~ {1,3,4,5,6}
  67: 1000011 ~ {1,2,7}
  71: 1000111 ~ {1,2,3,7}
  73: 1001001 ~ {1,4,7}
  79: 1001111 ~ {1,2,3,4,7}
		

Crossrefs

The number instead of sum of binary indices is A014499.
Restriction of A029931 (sum of binary indices) to the primes A000040.
The maximum instead of sum of binary indices is A035100, see also A023506.
Row-sums of A372471.
A019565 gives Heinz number of binary indices, adjoint A048675.
A029837 gives greatest binary index, least A001511.
A048793 lists binary indices, length A000120, reverse A272020.
A056239 adds up prime indices.
A070939 gives length of binary expansion.
A096111 gives product of binary indices.
A326031 gives weight of the set-system with BII-number n.
A372427 lists numbers whose binary and prime indices have the same sum.

Programs

  • Mathematica
    bix[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
    Table[Total[bix[Prime[n]]],{n,100}]

Formula

a(n) = A029931(prime(n)).

A326701 BII-numbers of set partitions.

Original entry on oeis.org

0, 1, 2, 3, 4, 8, 9, 10, 11, 12, 16, 18, 32, 33, 64, 128, 129, 130, 131, 132, 136, 137, 138, 139, 140, 144, 146, 160, 161, 192, 256, 258, 264, 266, 288, 512, 513, 520, 521, 528, 1024, 1032, 2048, 2049, 2050, 2051, 2052, 4096, 4098, 8192, 8193, 16384, 32768, 32769
Offset: 1

Views

Author

Gus Wiseman, Jul 21 2019

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. We define the set-system with BII-number n to be obtained by taking the binary indices of each binary index of n. For example, 18 has reversed binary expansion (0,1,0,0,1), and since the binary indices of 2 and 5 are {2} and {1,3} respectively, and {{2},{1,3}} is a set partition, it follows that 18 belongs to the sequence.

Examples

			The sequence of all set partitions together with their BII numbers begins:
    0: {}
    1: {{1}}
    2: {{2}}
    3: {{1},{2}}
    4: {{1,2}}
    8: {{3}}
    9: {{1},{3}}
   10: {{2},{3}}
   11: {{1},{2},{3}}
   12: {{1,2},{3}}
   16: {{1,3}}
   18: {{2},{1,3}}
   32: {{2,3}}
   33: {{1},{2,3}}
   64: {{1,2,3}}
  128: {{4}}
  129: {{1},{4}}
  130: {{2},{4}}
  131: {{1},{2},{4}}
  132: {{1,2},{4}}
  136: {{3},{4}}
		

Crossrefs

MM-numbers of set partitions are A302521.
BII-numbers of chains of nonempty sets are A326703.
BII-numbers of antichains of nonempty sets are A326704.

Programs

  • Mathematica
    bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
    Select[Range[0,1000],UnsameQ@@Join@@bpe/@bpe[#]&]
  • Python
    from itertools import chain, count, combinations, islice
    from sympy.utilities.iterables import multiset_partitions
    def a_gen():
        yield 0
        for n in count(1):
            t = []
            for i in chain.from_iterable(combinations(range(1,n+1),r) for r in range(n+1)):
                if n in i:
                    for j in multiset_partitions(i):
                        t.append(sum(2**(sum(2**(m-1) for m in k)-1) for k in j))
            yield from sorted(t)
    A326701_list = list(islice(a_gen(), 100)) # John Tyler Rascoe, May 24 2024

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

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. - 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)

A348612 Numbers k such that the k-th composition in standard order is not an anti-run, i.e., has adjacent equal parts.

Original entry on oeis.org

3, 7, 10, 11, 14, 15, 19, 21, 23, 26, 27, 28, 29, 30, 31, 35, 36, 39, 42, 43, 46, 47, 51, 53, 55, 56, 57, 58, 59, 60, 61, 62, 63, 67, 71, 73, 74, 75, 78, 79, 83, 84, 85, 86, 87, 90, 91, 92, 93, 94, 95, 99, 100, 103, 106, 107, 110, 111, 112, 113, 114, 115, 116
Offset: 1

Views

Author

Gus Wiseman, Nov 03 2021

Keywords

Comments

First differs from A345168 in lacking 37, corresponding to the composition (3,2,1).
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 terms and corresponding standard compositions begin:
     3: (1,1)          35: (4,1,1)        61: (1,1,1,2,1)
     7: (1,1,1)        36: (3,3)          62: (1,1,1,1,2)
    10: (2,2)          39: (3,1,1,1)      63: (1,1,1,1,1,1)
    11: (2,1,1)        42: (2,2,2)        67: (5,1,1)
    14: (1,1,2)        43: (2,2,1,1)      71: (4,1,1,1)
    15: (1,1,1,1)      46: (2,1,1,2)      73: (3,3,1)
    19: (3,1,1)        47: (2,1,1,1,1)    74: (3,2,2)
    21: (2,2,1)        51: (1,3,1,1)      75: (3,2,1,1)
    23: (2,1,1,1)      53: (1,2,2,1)      78: (3,1,1,2)
    26: (1,2,2)        55: (1,2,1,1,1)    79: (3,1,1,1,1)
    27: (1,2,1,1)      56: (1,1,4)        83: (2,3,1,1)
    28: (1,1,3)        57: (1,1,3,1)      84: (2,2,3)
    29: (1,1,2,1)      58: (1,1,2,2)      85: (2,2,2,1)
    30: (1,1,1,2)      59: (1,1,2,1,1)    86: (2,2,1,2)
    31: (1,1,1,1,1)    60: (1,1,1,3)      87: (2,2,1,1,1)
		

Crossrefs

Constant run compositions are counted by A000005, ranked by A272919.
Counting these compositions by sum and length gives A131044.
These compositions are counted by A261983.
The complement is A333489, counted by A003242.
The non-alternating case is A345168, complement A345167.
A011782 counts compositions, strict A032020.
A238279 counts compositions by sum and number of maximal runs.
A274174 counts compositions with equal parts contiguous.
A336107 counts non-anti-run permutations of prime factors.
A345195 counts non-alternating anti-runs, ranked by A345169.
For compositions in standard order (rows of A066099):
- Length is A000120.
- Sum is A070939
- Maximal runs are counted by A124767.
- Strict compositions are ranked by A233564.
- Maximal anti-runs are counted by A333381.
- Runs-resistance is A333628.

Programs

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

A228100 Triangle in which n-th row lists all partitions of n, such that partitions of n into m parts appear in lexicographic order previous to the partitions of n into k parts if k < m. (Fenner-Loizou tree.)

Original entry on oeis.org

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

Views

Author

Peter Luschny, Aug 10 2013

Keywords

Comments

First differs from A193073 at a(58). - Omar E. Pol, Sep 22 2013
The partition lengths appear to be A331581. - Gus Wiseman, May 12 2020

Examples

			The sixth row is:
[1, 1, 1, 1, 1, 1]
[2, 1, 1, 1, 1]
[2, 2, 1, 1]
[3, 1, 1, 1]
[2, 2, 2]
[3, 2, 1]
[4, 1, 1]
[3, 3]
[4, 2]
[5, 1]
[6]
From _Gus Wiseman_, May 10 2020: (Start)
The triangle with partitions shown as Heinz numbers (A333485) begins:
    1
    2
    4   3
    8   6   5
   16  12   9  10   7
   32  24  18  20  15  14  11
   64  48  36  40  27  30  28  25  21  22  13
  128  96  72  80  54  60  56  45  50  42  44  35  33  26  17
(End)
		

References

  • T. I. Fenner, G. Loizou: A binary tree representation and related algorithms for generating integer partitions. The Computer J. 23(4), 332-337 (1980)
  • D. E. Knuth: The Art of Computer Programming. Generating all combinations and partitions, vol. 4, fasc. 3, 7.2.1.4, exercise 10.
  • K. Yamanaka, Y. Otachi, Sh. Nakano: Efficient enumeration of ordered trees with k leaves. In: WALCOM: Algorithms and Computation, Lecture Notes in Computer Science Volume 5431, 141-150 (2009)
  • S. Zaks, D. Richards: Generating trees and other combinatorial objects lexicographically. SIAM J. Comput. 8(1), 73-81 (1979)
  • A. Zoghbi, I. Stojmenovic': Fast algorithms for generating integer partitions. Int. J. Comput. Math. 70, 319-332 (1998)

Crossrefs

See A036036 for the Hindenburg (graded reflected colexicographic) ordering.
See A036037 for the graded colexicographic ordering.
See A080576 for the Maple (graded reflected lexicographic) ordering.
See A080577 for the Mathematica (graded reverse lexicographic) ordering.
See A182937 the Fenner-Loizou (binary tree in preorder traversal) ordering.
See A193073 for the graded lexicographic ordering.
The version for compositions is A296773.
Taking Heinz numbers gives A333485.
Lexicographically ordered reversed partitions are A026791.
Sorting partitions by Heinz number gives A296150, or A112798 for reversed partitions.
Reversed partitions under the (sum/length/revlex) ordering are A334302.

Programs

  • Maple
    b:= proc(n, i) b(n, i):= `if`(n=0 or i=1, [[1$n]], [b(n, i-1)[],
          `if`(i>n, [], map(x-> [i, x[]], b(n-i, i)))[]])
        end:
    T:= n-> map(h-> h[], sort(b(n$2), proc(x, y) local i;
            if nops(x)<>nops(y) then return nops(x)>nops(y) else
            for i to nops(x) do if x[i]<>y[i] then return x[i]Alois P. Heinz, Aug 13 2013
  • Mathematica
    row[n_] := Flatten[Reverse[Sort[#]]& /@ SplitBy[Sort[IntegerPartitions[n] ], Length], 1] // Reverse; Array[row, 8] // Flatten (* Jean-François Alcover, Dec 05 2016 *)
    ralensort[f_,c_]:=If[Length[f]!=Length[c],Length[f]>Length[c],OrderedQ[{f,c}]];
    Join@@Table[Sort[IntegerPartitions[n],ralensort],{n,0,8}] (* Gus Wiseman, May 10 2020 *)
  • Sage
    from collections import deque
    def GeneratePartitions(n, visit):
        p = ([], 0, n)
        queue = deque()
        queue.append(p)
        visit(p)
        while len(queue) > 0 :
            (phead, pheadLen, pnum1s) = queue.popleft()
            if pnum1s != 1 :
                head = phead[:pheadLen] + [2]
                q = (head, pheadLen + 1, pnum1s - 2)
                if 1 <= q[2] : queue.append(q)
                visit(q)
            if pheadLen == 1 or (pheadLen > 1 and \
                          (phead[pheadLen - 1] != phead[pheadLen - 2])) :
                head = phead[:pheadLen]
                head[pheadLen - 1] += 1
                q = (head, pheadLen, pnum1s - 1)
                if 1 <= q[2] : queue.append(q)
                visit(q)
    def visit(q): print(q[0] + [1 for i in range(q[2])])
    for n in (1..7): GeneratePartitions(n, visit)
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