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.

A107428 Number of gap-free compositions of n.

Original entry on oeis.org

1, 2, 4, 6, 11, 21, 39, 71, 141, 276, 542, 1070, 2110, 4189, 8351, 16618, 33134, 66129, 131937, 263483, 526453, 1051984, 2102582, 4203177, 8403116, 16800894, 33593742, 67174863, 134328816, 268624026, 537192064, 1074288649, 2148414285, 4296543181, 8592585289
Offset: 1

Views

Author

N. J. A. Sloane, May 26 2005

Keywords

Comments

A gap-free composition contains all the parts between its smallest and largest part. a(5)=11 because we have: 5, 3+2, 2+3, 2+2+1, 2+1+2, 1+2+2, 2+1+1+1, 1+2+1+1, 1+1+2+1, 1+1+1+2, 1+1+1+1+1. - Geoffrey Critzer, Apr 13 2014

Examples

			From _Gus Wiseman_, Oct 04 2022: (Start)
The a(0) = 1 through a(5) = 11 gap-free compositions:
  ()  (1)  (2)   (3)    (4)     (5)
           (11)  (12)   (22)    (23)
                 (21)   (112)   (32)
                 (111)  (121)   (122)
                        (211)   (212)
                        (1111)  (221)
                                (1112)
                                (1121)
                                (1211)
                                (2111)
                                (11111)
(End)
		

Crossrefs

The unordered version (partitions) is A034296, ranked by A073491.
The initial case is A107429, unordered A000009, ranked by A333217.
The unordered complement is counted by A239955, ranked by A073492.
These compositions are ranked by A356841.
The complement is counted by A356846, ranked by A356842
A356230 ranks gapless factorization lengths, firsts A356603.
A356233 counts factorizations into gapless numbers.

Programs

  • Maple
    b:= proc(n, i, t) option remember; `if`(n=0, t!,
          `if`(i<1 or n add(b(n, i, 0), i=1..n):
    seq(a(n), n=1..40);  # Alois P. Heinz, Apr 14 2014
  • Mathematica
    Table[Length[Select[Level[Map[Permutations,IntegerPartitions[n]],{2}],Length[Union[#]]==Max[#]-Min[#]+1&]],{n,1,20}] (* Geoffrey Critzer, Apr 13 2014 *)
    b[n_, i_, t_] := b[n, i, t] = If[n == 0, t!, If[i < 1 || n < i, 0, Sum[b[n - i*j, i - 1, t + j]/j!, {j, 1, n/i}]]]; a[n_] := Sum[b[n, i, 0], {i, 1, n}]; Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Aug 30 2016, after Alois P. Heinz *)

Formula

a(n) ~ 2^(n-2). - Alois P. Heinz, Dec 07 2014
G.f.: Sum_{j>0} Sum_{k>=j} C({j..k},x) where C({s},x) = Sum_{i in {s}} (C({s}-{i},x)*x^i)/(1 - Sum_{i in {s}} (x^i)) is the g.f. for compositions such that the set of parts equals {s} with C({},x) = 1. - John Tyler Rascoe, Jun 01 2024

Extensions

More terms from Vladeta Jovovic, May 26 2005

A356841 Numbers k such that the k-th composition in standard order covers an interval of positive integers (gapless).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 13, 14, 15, 16, 18, 20, 21, 22, 23, 26, 27, 29, 30, 31, 32, 36, 37, 38, 41, 42, 43, 44, 45, 46, 47, 50, 52, 53, 54, 55, 58, 59, 61, 62, 63, 64, 68, 72, 74, 75, 77, 78, 82, 83, 84, 85, 86, 87, 89, 90, 91, 92, 93, 94, 95, 101
Offset: 1

Views

Author

Gus Wiseman, Aug 31 2022

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The terms and their corresponding 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)
  10: (2,2)
  11: (2,1,1)
  13: (1,2,1)
  14: (1,1,2)
  15: (1,1,1,1)
  16: (5)
  18: (3,2)
  20: (2,3)
  21: (2,2,1)
		

Crossrefs

See link for sequences related to standard compositions.
An unordered version is A073491, complement A073492.
These compositions are counted by A107428.
The complement is A356842.
The non-initial case is A356843, unordered A356845.
A356230 ranks gapless factorization lengths, firsts A356603.
A356233 counts factorizations into gapless numbers.
A356844 ranks compositions with at least one 1.

Programs

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

A356844 Numbers k such that the k-th composition in standard order contains at least one 1. Numbers that are odd or whose binary expansion contains at least two adjacent 1's.

Original entry on oeis.org

1, 3, 5, 6, 7, 9, 11, 12, 13, 14, 15, 17, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 35, 37, 38, 39, 41, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 67, 69, 70, 71, 73, 75, 76, 77, 78, 79, 81, 83, 85, 86, 87
Offset: 1

Views

Author

Gus Wiseman, Sep 02 2022

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The terms, binary expansions, and standard compositions:
   1:       1  (1)
   3:      11  (1,1)
   5:     101  (2,1)
   6:     110  (1,2)
   7:     111  (1,1,1)
   9:    1001  (3,1)
  11:    1011  (2,1,1)
  12:    1100  (1,3)
  13:    1101  (1,2,1)
  14:    1110  (1,1,2)
  15:    1111  (1,1,1,1)
  17:   10001  (4,1)
  19:   10011  (3,1,1)
  21:   10101  (2,2,1)
  22:   10110  (2,1,2)
  23:   10111  (2,1,1,1)
  24:   11000  (1,4)
  25:   11001  (1,3,1)
  26:   11010  (1,2,2)
  27:   11011  (1,2,1,1)
  28:   11100  (1,1,3)
  29:   11101  (1,1,2,1)
  30:   11110  (1,1,1,2)
  31:   11111  (1,1,1,1,1)
		

Crossrefs

See link for sequences related to standard compositions.
The case beginning with 1 is A004760, complement A004754.
The complement is A022340.
These compositions are counted by A099036, complement A212804.
The case covering an initial interval is A333217.
The gapless but non-initial version is A356843, unordered A356845.

Programs

  • Mathematica
    Select[Range[0,100],OddQ[#]||MatchQ[IntegerDigits[#,2],{_,1,1,_}]&]

Formula

Union of A005408 and A004780.

A356845 Odd numbers with gapless prime indices.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 23, 25, 27, 29, 31, 35, 37, 41, 43, 45, 47, 49, 53, 59, 61, 67, 71, 73, 75, 77, 79, 81, 83, 89, 97, 101, 103, 105, 107, 109, 113, 121, 125, 127, 131, 135, 137, 139, 143, 149, 151, 157, 163, 167, 169, 173, 175, 179, 181, 191
Offset: 1

Views

Author

Gus Wiseman, Sep 03 2022

Keywords

Comments

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.
A sequence is gapless if it covers an interval of positive integers.

Examples

			The terms together with their prime indices begin:
    1: {}
    3: {2}
    5: {3}
    7: {4}
    9: {2,2}
   11: {5}
   13: {6}
   15: {2,3}
   17: {7}
   19: {8}
   23: {9}
   25: {3,3}
   27: {2,2,2}
   29: {10}
   31: {11}
   35: {3,4}
   37: {12}
   41: {13}
   43: {14}
		

Crossrefs

Consists of the odd terms of A073491.
These partitions are counted by A264396.
The strict case is A294674, counted by A136107.
The version for compositions is A356843, counted by A251729.
A001221 counts distinct prime factors, sum A001414.
A056239 adds up prime indices, row sums of A112798, lengths A001222.
A356069 counts gapless divisors, initial A356224 (complement A356225).
A356230 ranks gapless factorization lengths, firsts A356603.
A356233 counts factorizations into gapless numbers.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    nogapQ[m_]:=Or[m=={},Union[m]==Range[Min[m],Max[m]]];
    Select[Range[1,100,2],nogapQ[primeMS[#]]&]

A356842 Numbers k such that the k-th composition in standard order does not cover an interval of positive integers (not gapless).

Original entry on oeis.org

9, 12, 17, 19, 24, 25, 28, 33, 34, 35, 39, 40, 48, 49, 51, 56, 57, 60, 65, 66, 67, 69, 70, 71, 73, 76, 79, 80, 81, 88, 96, 97, 98, 99, 100, 103, 104, 112, 113, 115, 120, 121, 124, 129, 130, 131, 132, 133, 134, 135, 137, 138, 139, 140, 141, 142, 143, 144, 145
Offset: 1

Views

Author

Gus Wiseman, Sep 01 2022

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The terms and their corresponding standard compositions begin:
   9: (3,1)
  12: (1,3)
  17: (4,1)
  19: (3,1,1)
  24: (1,4)
  25: (1,3,1)
  28: (1,1,3)
  33: (5,1)
  34: (4,2)
  35: (4,1,1)
  39: (3,1,1,1)
  40: (2,4)
  48: (1,5)
  49: (1,4,1)
  51: (1,3,1,1)
  56: (1,1,4)
  57: (1,1,3,1)
  60: (1,1,1,3)
		

Crossrefs

See link for sequences related to standard compositions.
An unordered version is A073492, complement A073491.
These compositions are counted by the complement of A107428.
The complement is A356841.
The gapless but non-initial version is A356843, unordered A356845.
A356230 ranks gapless factorization lengths, firsts A356603.
A356233 counts factorizations into gapless numbers.
A356844 ranks compositions with at least one 1.

Programs

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

A356956 Numbers k such that the k-th composition in standard order is a gapless interval (in increasing order).

Original entry on oeis.org

0, 1, 2, 4, 6, 8, 16, 20, 32, 52, 64, 72, 128, 256, 272, 328, 512, 840, 1024, 1056, 2048, 2320, 4096, 4160, 8192, 10512, 16384, 16512, 17440, 26896, 32768, 65536, 65792, 131072, 135232, 148512, 262144, 262656, 524288, 672800, 1048576, 1049600, 1065088, 1721376
Offset: 1

Views

Author

Gus Wiseman, Sep 24 2022

Keywords

Comments

An interval such as {3,4,5} is a set of positive integers with all differences of adjacent elements equal to 1.
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 terms and corresponding intervals begin:
        0: ()
        1: (1)
        2: (2)
        4: (3)
        6: (1,2)
        8: (4)
       16: (5)
       20: (2,3)
       32: (6)
       52: (1,2,3)
       64: (7)
       72: (3,4)
      128: (8)
      256: (9)
      272: (4,5)
      328: (2,3,4)
      512: (10)
      840: (1,2,3,4)
		

Crossrefs

See link for sequences related to standard compositions.
These compositions are counted by A001227.
An unordered version is A073485, non-strict A073491 (complement A073492).
The initial version is A164894, non-strict A356843 (unordered A356845).
The non-strict version is A356841, initial A333217, counted by A107428.
A066311 lists gapless numbers.
A356230 ranks gapless factorization lengths, firsts A356603.
A356233 counts factorizations into gapless numbers.
A356844 ranks compositions with at least one 1.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    chQ[y_]:=Length[y]<=1||Union[Differences[y]]=={1};
    Select[Range[0,1000],chQ[stc[#]]&]
Showing 1-6 of 6 results.