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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A275527 Number of distinct classes of permutations of length n under reversal and complement to n+1.

Original entry on oeis.org

1, 1, 1, 4, 12, 64, 360, 2544, 20160, 181632
Offset: 1

Views

Author

Olivier Gérard, Jul 31 2016

Keywords

Comments

Let us consider two permutations to be equivalent if they can be obtained from each other by cyclic rotation (12345->(23451,34512,45123,51234) or n+1-complement (31254->35412), or a combination of those two transformations (they commute with each other). a(n) is the number of classes.
We obtain the same number of classes if the transformations are (addition of a constant modulo n and reversal (12345->54321)) but not the same set of representatives.
It seems probable that a(2n+1) = (2n)!/2
This sequence may be related to A113247 (and A113248) as they share a common dissection 1, 4, 64, 2544, 181632. The fact that they count permutation classes for the major index is a further indication.
Number of path necklaces, defined as equivalence classes of (labeled, undirected) Hamiltonian paths under rotation of the vertices. The cycle version is A000939. - Gus Wiseman, Mar 02 2019

Examples

			Examples of permutation representatives. The representative is chosen to be the first of the class in lexicographic order.
n=4 case addition mod n and reversal
1234, 1243, 1324, 1423.
n=4 case rotation and complement
1234, 1243, 1324, 1342.
.
n=5 case addition mod n and reversal
12345, 12354, 12435, 12453, 12534, 13245, 13425, 13452, 13524, 14235, 14523, 15234.
n=5 case rotation and complement
12345, 12354, 12435, 12453, 12534, 13245, 13425, 13452, 13524, 14235, 14325, 14352.
		

Crossrefs

Cf. A000939, A000940, A002619, A089066, A262480 (other symmetry classes of permutations).
Cf. A193651 (inspiration for a(2n)).

Programs

  • Mathematica
    rotgra[g_,m_]:=Sort[Sort/@(g/.k_Integer:>If[k==m,1,k+1])];
    Table[Length[Select[Union[Sort[Sort/@Partition[#,2,1]]&/@Permutations[Range[n]]],#==First[Sort[Table[Nest[rotgra[#,n]&,#,j],{j,n}]]]&]],{n,8}] (* Gus Wiseman, Mar 02 2019 *)

Formula

(Conjecture). If n odd a(n)=((n - 1))!/2. If n even a(n)= 1/2 (n - 2)!! (1 + ( n - 1)!!).

A323866 Number of aperiodic toroidal necklaces of positive integers summing to n.

Original entry on oeis.org

1, 1, 1, 3, 5, 12, 18, 42, 72, 145, 262, 522, 960, 1879, 3531, 6831, 13013, 25148, 48177, 93186, 179507, 347509, 671955, 1303257, 2527162, 4910681, 9545176, 18579471, 36183505, 70540861, 137603801, 268655547, 524842088, 1026067205, 2007118657, 3928564113
Offset: 0

Views

Author

Gus Wiseman, Feb 04 2019

Keywords

Comments

The 1-dimensional (Lyndon word) case is A059966.
We define a toroidal necklace to be an equivalence class of matrices under all possible rotations of the sequence of rows and the sequence of columns. An n X k matrix is aperiodic if all n * k rotations of its sequence of rows and its sequence of columns are distinct.

Examples

			Inequivalent representatives of the a(6) = 18 toroidal necklaces:
  [6] [1 5] [2 4] [1 1 4] [1 2 3] [1 3 2] [1 1 1 3] [1 1 2 2] [1 1 1 1 2]
.
  [1] [2] [1 1]
  [5] [4] [1 3]
.
  [1] [1] [1]
  [1] [2] [3]
  [4] [3] [2]
.
  [1] [1]
  [1] [1]
  [1] [2]
  [3] [2]
.
  [1]
  [1]
  [1]
  [1]
  [2]
		

Crossrefs

Programs

  • GAP
    List([0..30], A323866); # See A323861 for code; Andrew Howroyd, Aug 21 2019
  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
    ptnmats[n_]:=Union@@Permutations/@Select[Union@@(Tuples[Permutations/@#]&/@Map[primeMS,facs[n],{2}]),SameQ@@Length/@#&];
    apermatQ[m_]:=UnsameQ@@Join@@Table[RotateLeft[m,{i,j}],{i,Length[m]},{j,Length[First[m]]}];
    neckmatQ[m_]:=m==First[Union@@Table[RotateLeft[m,{i,j}],{i,Length[m]},{j,Length[First[m]]}]];
    Table[If[n==0,1,Length[Union@@Table[Select[ptnmats[k],And[apermatQ[#],neckmatQ[#]]&],{k,Times@@Prime/@#&/@IntegerPartitions[n]}]]],{n,0,10}]

Extensions

Terms a(16) and beyond from Andrew Howroyd, Aug 21 2019

A323869 Number of aperiodic matrices of size n whose entries cover an initial interval of positive integers.

Original entry on oeis.org

1, 4, 24, 212, 1080, 18672, 94584, 2182752, 21261708, 408988080, 3245265144, 168549358368, 1053716696760, 42565371692592, 921132763909200, 26578273403903040, 260741534058271800, 20313207979498492344, 185603174638656822264, 16066126777465282744800, 324499299994016295338064
Offset: 1

Views

Author

Gus Wiseman, Feb 04 2019

Keywords

Comments

An n X k matrix is aperiodic if all n * k rotations of its sequence of rows and its sequence of columns are distinct.

Examples

			The a(3) = 24 matrices:
  [123][132][213][312][231][321][122][211][112][221][121][212]
.
  [1][1][2][3][2][3][1][2][1][2][1][2]
  [2][3][1][1][3][2][2][1][1][2][2][1]
  [3][2][3][2][1][1][2][1][2][1][1][2]
		

Crossrefs

Programs

  • GAP
    List([1..30], A323869); # See A323861 for code; Andrew Howroyd, Aug 21 2019
  • Mathematica
    sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
    nrmmats[n_]:=Join@@Table[Table[Table[Position[stn,{i,j}][[1,1]],{i,d},{j,n/d}],{stn,Join@@Permutations/@sps[Tuples[{Range[d],Range[n/d]}]]}],{d,Divisors[n]}];
    apermatQ[m_]:=UnsameQ@@Join@@Table[RotateLeft[m,{i,j}],{i,Length[m]},{j,Length[First[m]]}];
    Table[Length[Select[nrmmats[n],apermatQ]],{n,6}]

Formula

a(n) = n*A323871(n). - Andrew Howroyd, Aug 21 2019

Extensions

Terms a(9) and beyond from Andrew Howroyd, Aug 21 2019

A329316 Irregular triangle read by rows where row n gives the sequence of lengths of components of the Lyndon factorization of the reversed first n terms of A000002.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 4, 1, 1, 4, 1, 2, 4, 1, 1, 2, 4, 1, 1, 1, 2, 4, 1, 3, 2, 4, 1, 1, 3, 2, 4, 1, 1, 1, 3, 2, 4, 1, 3, 3, 2, 4, 1, 9, 4, 1, 1, 9, 4, 1, 2, 9, 4, 1, 16, 1, 1, 16, 1, 1, 1, 16, 1, 3, 16, 1, 1, 3, 16, 1, 5, 16, 1, 6, 16, 1, 1, 6, 16, 1, 2, 6
Offset: 0

Views

Author

Gus Wiseman, Nov 11 2019

Keywords

Comments

There are no repeated rows, as row n has sum n.
We define the Lyndon product of two or more finite sequences to be the lexicographically maximal sequence obtainable by shuffling the sequences together. For example, the Lyndon product of (231) with (213) is (232131), the product of (221) with (213) is (222131), and the product of (122) with (2121) is (2122121). A Lyndon word is a finite sequence that is prime with respect to the Lyndon product. Equivalently, a Lyndon word is a finite sequence that is lexicographically strictly less than all of its cyclic rotations. Every finite sequence has a unique (orderless) factorization into Lyndon words, and if these factors are arranged in lexicographically decreasing order, their concatenation is equal to their Lyndon product. For example, (1001) has sorted Lyndon factorization (001)(1).
It appears that some numbers (such as 10) never appear in the sequence.

Examples

			Triangle begins:
   1: (1)
   2: (1,1)
   3: (1,1,1)
   4: (3,1)
   5: (4,1)
   6: (1,4,1)
   7: (2,4,1)
   8: (1,2,4,1)
   9: (1,1,2,4,1)
  10: (3,2,4,1)
  11: (1,3,2,4,1)
  12: (1,1,3,2,4,1)
  13: (3,3,2,4,1)
  14: (9,4,1)
  15: (1,9,4,1)
  16: (2,9,4,1)
  17: (16,1)
  18: (1,16,1)
  19: (1,1,16,1)
  20: (3,16,1)
For example, the reversed first 13 terms of A000002 are (1221221211221), with Lyndon factorization (122)(122)(12)(1122)(1), so row 13 is (3,3,2,4,1).
		

Crossrefs

Row lengths are A329317.
The non-reversed version is A329315.

Programs

  • Mathematica
    lynQ[q_]:=Array[Union[{q,RotateRight[q,#]}]=={q,RotateRight[q,#]}&,Length[q]-1,1,And];
    lynfac[q_]:=If[Length[q]==0,{},Function[i,Prepend[lynfac[Drop[q,i]],Take[q,i]]][Last[Select[Range[Length[q]],lynQ[Take[q,#]]&]]]];
    kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],q[[-2]],Last[q]},{1,1,1},0,{1,1,2},1,{1,2,1},2,{1,2,2},0,{2,1,1},2,{2,1,2},2,{2,2,1},1,{2,2,2},1]]]
    kol[n_Integer]:=Nest[kolagrow,{1},n-1];
    Table[Length/@lynfac[Reverse[kol[n]]],{n,100}]

A334265 Numbers k such that the k-th composition in standard order is a reversed Lyndon word.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 9, 11, 16, 17, 18, 19, 21, 23, 32, 33, 34, 35, 37, 39, 41, 43, 47, 64, 65, 66, 67, 68, 69, 71, 73, 74, 75, 77, 79, 81, 83, 85, 87, 91, 95, 128, 129, 130, 131, 132, 133, 135, 137, 138, 139, 141, 143, 145, 146, 147, 149, 151, 155, 159, 161, 163
Offset: 1

Views

Author

Gus Wiseman, Apr 22 2020

Keywords

Comments

Reversed Lyndon words are different from co-Lyndon words (A326774).
A Lyndon word is a finite sequence of positive integers that is lexicographically strictly less than all of its cyclic rotations.
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 of all reversed Lyndon words begins:
    0: ()            37: (3,2,1)         83: (2,3,1,1)
    1: (1)           39: (3,1,1,1)       85: (2,2,2,1)
    2: (2)           41: (2,3,1)         87: (2,2,1,1,1)
    4: (3)           43: (2,2,1,1)       91: (2,1,2,1,1)
    5: (2,1)         47: (2,1,1,1,1)     95: (2,1,1,1,1,1)
    8: (4)           64: (7)            128: (8)
    9: (3,1)         65: (6,1)          129: (7,1)
   11: (2,1,1)       66: (5,2)          130: (6,2)
   16: (5)           67: (5,1,1)        131: (6,1,1)
   17: (4,1)         68: (4,3)          132: (5,3)
   18: (3,2)         69: (4,2,1)        133: (5,2,1)
   19: (3,1,1)       71: (4,1,1,1)      135: (5,1,1,1)
   21: (2,2,1)       73: (3,3,1)        137: (4,3,1)
   23: (2,1,1,1)     74: (3,2,2)        138: (4,2,2)
   32: (6)           75: (3,2,1,1)      139: (4,2,1,1)
   33: (5,1)         77: (3,1,2,1)      141: (4,1,2,1)
   34: (4,2)         79: (3,1,1,1,1)    143: (4,1,1,1,1)
   35: (4,1,1)       81: (2,4,1)        145: (3,4,1)
		

Crossrefs

The non-reversed version is A275692.
The generalization to necklaces is A333943.
The dual version (reversed co-Lyndon words) is A328596.
The case that is also co-Lyndon is A334266.
Binary Lyndon words are counted by A001037.
Lyndon compositions are counted by A059966.
Normal Lyndon words are counted by A060223.
Numbers whose prime signature is a reversed Lyndon word are A334298.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Necklaces are A065609.
- Sum is A070939.
- Reverse is A228351 (triangle).
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon words are A275692.
- Reversed Lyndon words are A334265 (this sequence).
- Co-Lyndon words are A326774.
- Reversed co-Lyndon words are A328596.
- Length of Lyndon factorization is A329312.
- Distinct rotations are counted by A333632.
- Lyndon factorizations are counted by A333940.
- Length of Lyndon factorization of reverse is A334297.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    lynQ[q_]:=Length[q]==0||Array[Union[{q,RotateRight[q,#1]}]=={q,RotateRight[q,#1]}&,Length[q]-1,1,And];
    Select[Range[0,100],lynQ[Reverse[stc[#]]]&]

A296975 Number of aperiodic normal sequences of length n.

Original entry on oeis.org

1, 2, 12, 72, 540, 4668, 47292, 545760, 7087248, 102247020, 1622632572, 28091562840, 526858348380, 10641342923148, 230283190977300, 5315654681435520, 130370767029135900, 3385534663249753392, 92801587319328411132, 2677687796244281955480, 81124824998504073834516
Offset: 1

Views

Author

Gus Wiseman, Dec 22 2017

Keywords

Comments

A finite sequence is normal if it spans an initial interval of positive integers. It is aperiodic if every cyclic rotation is different.

Examples

			The a(3) = 12 aperiodic normal sequences are 112, 121, 122, 123, 132, 211, 212, 213, 221, 231, 312, 321.
The 15 non-aperiodic normal sequences of length 6 are: 111111, 112112, 121121, 121212, 122122, 123123, 132132, 211211, 212121, 212212, 213213, 221221, 231231, 312312, 321321.
		

Crossrefs

Programs

  • Mathematica
    Table[DivisorSum[n,MoebiusMu[n/#]*Sum[k!*StirlingS2[#,k],{k,#}]&],{n,25}]
  • PARI
    \\ here b(n) is A000670.
    b(n)={polcoef(serlaplace(1/(2-exp(x+O(x*x^n)))),n)}
    a(n)={sumdiv(n, d, moebius(d)*b(n/d))} \\ Andrew Howroyd, Aug 29 2018

Formula

a(n) = n * A060223(n) = Sum_{d|n} mu(d) * A000670(n/d).

A329362 Length of the co-Lyndon factorization of the first n terms of A000002.

Original entry on oeis.org

0, 1, 2, 3, 2, 2, 3, 2, 3, 4, 3, 4, 5, 4, 3, 4, 3, 3, 4, 5, 4, 5, 3, 3, 4, 3, 4, 5, 4, 3, 4, 3, 3, 4, 3, 4, 5, 4, 5, 6, 5, 4, 5, 4, 5, 6, 5, 6, 4, 4, 5, 4, 4, 5, 6, 5, 6, 7, 6, 5, 6, 5, 6, 7, 6, 7, 8, 7, 6, 7, 6, 5, 6, 5, 6, 7, 6, 7, 5, 5, 6, 7, 6, 7, 8, 7, 6, 7
Offset: 0

Views

Author

Gus Wiseman, Nov 12 2019

Keywords

Comments

The co-Lyndon product of two or more finite sequences is defined to be the lexicographically minimal sequence obtainable by shuffling the sequences together. For example, the co-Lyndon product of (231) and (213) is (212313), the product of (221) and (213) is (212213), and the product of (122) and (2121) is (1212122). A co-Lyndon word is a finite sequence that is prime with respect to the co-Lyndon product. Equivalently, a co-Lyndon word is a finite sequence that is lexicographically strictly greater than all of its cyclic rotations. Every finite sequence has a unique (orderless) factorization into co-Lyndon words, and if these factors are arranged in a certain order, their concatenation is equal to their co-Lyndon product. For example, (1001) has sorted co-Lyndon factorization (1)(100).

Examples

			The co-Lyndon factorizations of the initial terms of A000002:
                      () = 0
                     (1) = (1)
                    (12) = (1)(2)
                   (122) = (1)(2)(2)
                  (1221) = (1)(221)
                 (12211) = (1)(2211)
                (122112) = (1)(2211)(2)
               (1221121) = (1)(221121)
              (12211212) = (1)(221121)(2)
             (122112122) = (1)(221121)(2)(2)
            (1221121221) = (1)(221121)(221)
           (12211212212) = (1)(221121)(221)(2)
          (122112122122) = (1)(221121)(221)(2)(2)
         (1221121221221) = (1)(221121)(221)(221)
        (12211212212211) = (1)(221121)(2212211)
       (122112122122112) = (1)(221121)(2212211)(2)
      (1221121221221121) = (1)(221121)(221221121)
     (12211212212211211) = (1)(221121)(2212211211)
    (122112122122112112) = (1)(221121)(2212211211)(2)
   (1221121221221121122) = (1)(221121)(2212211211)(2)(2)
  (12211212212211211221) = (1)(221121)(2212211211)(221)
		

Crossrefs

Programs

  • Mathematica
    kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],q[[-2]],Last[q]},{1,1,1},0,{1,1,2},1,{1,2,1},2,{1,2,2},0,{2,1,1},2,{2,1,2},2,{2,2,1},1,{2,2,2},1]]]
    kol[n_Integer]:=If[n==0,{},Nest[kolagrow,{1},n-1]];
    colynQ[q_]:=Array[Union[{RotateRight[q,#],q}]=={RotateRight[q,#],q}&,Length[q]-1,1,And];
    colynfac[q_]:=If[Length[q]==0,{},Function[i,Prepend[colynfac[Drop[q,i]],Take[q,i]]]@Last[Select[Range[Length[q]],colynQ[Take[q,#]]&]]];
    Table[Length[colynfac[kol[n]]],{n,0,100}]

A333939 Number of multisets of compositions that can be shuffled together to obtain the k-th composition in standard order.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 4, 2, 5, 4, 5, 1, 2, 2, 4, 2, 4, 5, 7, 2, 5, 4, 10, 4, 10, 7, 7, 1, 2, 2, 4, 2, 5, 5, 7, 2, 5, 3, 9, 5, 13, 11, 12, 2, 5, 5, 10, 5, 11, 13, 18, 4, 10, 9, 20, 7, 18, 12, 11, 1, 2, 2, 4, 2, 5, 5, 7, 2, 4, 4, 11, 5, 14, 11, 12, 2
Offset: 0

Views

Author

Gus Wiseman, Apr 15 2020

Keywords

Comments

Number of ways to deal out the k-th composition in standard order to form a multiset of hands.
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.

Examples

			The dealings for n = 1, 3, 7, 11, 13, 23, 43:
  (1)  (11)    (111)      (211)      (121)      (2111)        (2211)
       (1)(1)  (1)(11)    (1)(21)    (1)(12)    (11)(21)      (11)(22)
               (1)(1)(1)  (2)(11)    (1)(21)    (1)(211)      (1)(221)
                          (1)(1)(2)  (2)(11)    (2)(111)      (21)(21)
                                     (1)(1)(2)  (1)(1)(21)    (2)(211)
                                                (1)(2)(11)    (1)(1)(22)
                                                (1)(1)(1)(2)  (1)(2)(21)
                                                              (2)(2)(11)
                                                              (1)(1)(2)(2)
		

Crossrefs

Multisets of compositions are counted by A034691.
Combinatory separations of normal multisets are counted by A269134.
Dealings with total sum n are counted by A292884.
Length of co-Lyndon factorization of binary expansion is A329312.
Length of Lyndon factorization of reversed binary expansion is A329313.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Necklaces are A065609.
- Sum is A070939.
- Runs are counted by A124767.
- Rotational symmetries are counted by A138904.
- Strict compositions are A233564.
- Constant compositions are A272919.
- Lyndon words are A275692.
- Co-Lyndon words are A326774.
- Aperiodic compositions are A328594.
- Length of Lyndon factorization is A329312.
- Distinct rotations are counted by A333632.
- Co-Lyndon factorizations are counted by A333765.
- Lyndon factorizations are counted by A333940.
- Length of co-Lyndon factorization is A334029.
- Combinatory separations are A334030.

Programs

  • Mathematica
    nn=100;
    comps[0]:={{}};comps[n_]:=Join@@Table[Prepend[#,i]&/@comps[n-i],{i,n}];
    sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
    dealings[q_]:=Union[Function[ptn,Sort[q[[#]]&/@ptn]]/@sps[Range[Length[q]]]];
    Table[Length[dealings[stc[n]]],{n,0,nn}]

Formula

For n > 0, Sum_{k = 2^(n-1)..2^n-1} a(k) = A292884(n).

A324513 Number of aperiodic cycle necklaces with n vertices.

Original entry on oeis.org

1, 0, 0, 0, 2, 7, 51, 300, 2238, 18028, 164945, 1662067, 18423138, 222380433, 2905942904, 40864642560, 615376173176, 9880203467184, 168483518571789, 3041127459127222, 57926238289894992, 1161157775616335125, 24434798429947993043, 538583682037962702384
Offset: 1

Views

Author

Gus Wiseman, Mar 04 2019

Keywords

Comments

We define an aperiodic cycle necklace to be an equivalence class of (labeled, undirected) Hamiltonian cycles under rotation of the vertices such that all n of these rotations are distinct.

Crossrefs

Cf. A000740, A000939, A001037 (binary Lyndon words), A008965, A059966 (Lyndon compositions), A060223 (normal Lyndon words), A061417, A064852 (if cycle is oriented), A086675, A192332, A275527, A323866 (aperiodic toroidal arrays), A323871.

Programs

  • Mathematica
    rotgra[g_,m_]:=Sort[Sort/@(g/.k_Integer:>If[k==m,1,k+1])];
    Table[Length[Select[Union[Sort[Sort/@Partition[#,2,1,1]]&/@Permutations[Range[n]]],#==First[Sort[Table[Nest[rotgra[#,n]&,#,j],{j,n}]]]&&UnsameQ@@Table[Nest[rotgra[#,n]&,#,j],{j,n}]&]],{n,8}]
  • PARI
    a(n)={if(n<3, n==0||n==1, (if(n%2, 0, -(n/2-1)!*2^(n/2-2)) + sumdiv(n, d, moebius(n/d)*eulerphi(n/d)*(n/d)^d*d!/n^2))/2)} \\ Andrew Howroyd, Aug 19 2019

Formula

a(n) = A324512(n)/n.
a(2*n+1) = A064852(2*n+1)/2 for n > 0; a(2*n) = (A064852(2*n) - A002866(n-1))/2 for n > 1. - Andrew Howroyd, Aug 16 2019

Extensions

Terms a(10) and beyond from Andrew Howroyd, Aug 19 2019

A298941 Number of permutations of the multiset of prime factors of n > 1 that are Lyndon words.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Jan 29 2018

Keywords

Examples

			The a(90) = 3 Lyndon permutations are {2,3,3,5}, {2,3,5,3}, {2,5,3,3}.
		

Crossrefs

Programs

  • Maple
    with(combinat): with(numtheory):
    g:= l-> (n-> `if`(n=0, 1, add(mobius(j)*multinomial(n/j,
            (l/j)[]), j=divisors(igcd(l[])))/n))(add(i, i=l)):
    a:= n-> g(map(i-> i[2], ifactors(n)[2])):
    seq(a(n), n=2..150);  # Alois P. Heinz, Feb 09 2018
  • Mathematica
    primeMS[n_]:=If[n===1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    LyndonQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And]&&Array[RotateRight[q,#]&,Length[q],1,UnsameQ];
    Table[Length[Select[Permutations[primeMS[n]],LyndonQ]],{n,2,60}]
    (* Second program: *)
    multinomial[n_, k_List] := n!/Times @@ (k!);
    g[l_] := With[{n = Total[l]}, If[n == 0, 1, Sum[MoebiusMu[j] multinomial[ n/j, l/j], {j, Divisors[GCD @@ l]}]/n]];
    a[n_] := g[FactorInteger[n][[All, 2]]];
    a /@ Range[2, 150] (* Jean-François Alcover, Dec 15 2020, after Alois P. Heinz *)
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