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 1-10 of 14 results. Next

A211100 Number of factors in Lyndon factorization of binary expansion of n.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Mar 31 2012

Keywords

Comments

Any binary word has a unique factorization as a product of nonincreasing Lyndon words (see Lothaire). a(n) = number of factors in Lyndon factorization of binary expansion of n.
It appears that a(n) = k for the first time when n = 2^(k-1)+1.
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. - Gus Wiseman, Nov 12 2019

Examples

			n=25 has binary expansion 11001, which has Lyndon factorization (1)(1)(001) with three factors, so a(25) = 3.
Here are the Lyndon factorizations for small values of n:
.0.
.1.
.1.0.
.1.1.
.1.0.0.
.1.01.
.1.1.0.
.1.1.1.
.1.0.0.0.
.1.001.
.1.01.0.
.1.011.
.1.1.0.0.
...
		

References

  • M. Lothaire, Combinatorics on Words, Addison-Wesley, Reading, MA, 1983. See Theorem 5.1.5, p. 67.
  • G. Melançon, Factorizing infinite words using Maple, MapleTech Journal, vol. 4, no. 1, 1997, pp. 34-42

Crossrefs

Cf. A001037 (number of Lyndon words of length m); A102659 (list thereof).
A211095 and A211096 give information about the smallest (or rightmost) factor. Cf. A211097, A211098, A211099.
Row-lengths of A329314.
The "co-" version is A329312.
Positions of 2's are A329327.
The reversed version is A329313.
The inverted version is A329312.
Ignoring the first digit gives A211097.

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,#]]&]]]];
    Table[Length[lynfac[IntegerDigits[n,2]]],{n,0,30}] (* Gus Wiseman, Nov 12 2019 *)

A329312 Length of the co-Lyndon factorization of the binary expansion of n.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Nov 10 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).
Also the length of the Lyndon factorization of the inverted binary expansion of n, where the inverted digits are 1 minus the binary digits.

Examples

			The binary indices of 1..20 together with their co-Lyndon factorizations are:
   1:     (1) = (1)
   2:    (10) = (10)
   3:    (11) = (1)(1)
   4:   (100) = (100)
   5:   (101) = (10)(1)
   6:   (110) = (110)
   7:   (111) = (1)(1)(1)
   8:  (1000) = (1000)
   9:  (1001) = (100)(1)
  10:  (1010) = (10)(10)
  11:  (1011) = (10)(1)(1)
  12:  (1100) = (1100)
  13:  (1101) = (110)(1)
  14:  (1110) = (1110)
  15:  (1111) = (1)(1)(1)(1)
  16: (10000) = (10000)
  17: (10001) = (1000)(1)
  18: (10010) = (100)(10)
  19: (10011) = (100)(1)(1)
  20: (10100) = (10100)
		

Crossrefs

The non-"co" version is A211100.
Positions of 1's are A275692.
The reversed version is A329326.

Programs

  • Mathematica
    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[IntegerDigits[n,2]]],{n,100}]

A329313 Length of the Lyndon factorization of the reversed binary expansion of n.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Nov 11 2019

Keywords

Comments

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. 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).

Examples

			The sequence of reversed binary expansions of the nonnegative integers together with their Lyndon factorizations begins:
   0:      () = ()
   1:     (1) = (1)
   2:    (01) = (01)
   3:    (11) = (1)(1)
   4:   (001) = (001)
   5:   (101) = (1)(01)
   6:   (011) = (011)
   7:   (111) = (1)(1)(1)
   8:  (0001) = (0001)
   9:  (1001) = (1)(001)
  10:  (0101) = (01)(01)
  11:  (1101) = (1)(1)(01)
  12:  (0011) = (0011)
  13:  (1011) = (1)(011)
  14:  (0111) = (0111)
  15:  (1111) = (1)(1)(1)(1)
  16: (00001) = (00001)
  17: (10001) = (1)(0001)
  18: (01001) = (01)(001)
  19: (11001) = (1)(1)(001)
  20: (00101) = (00101)
		

Crossrefs

The non-reversed version is A211100.
Positions of 1's are A328596.
The "co" version is A329326.
Binary Lyndon words are counted by A001037 and ranked by A102659.
Numbers whose reversed binary expansion is a necklace are A328595.
Numbers whose reversed binary expansion is a aperiodic are A328594.
Length of the co-Lyndon factorization of the binary expansion is A329312.

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,#1]]&]]]];
    Table[If[n==0,0,Length[lynfac[Reverse[IntegerDigits[n,2]]]]],{n,0,30}]

A329326 Length of the co-Lyndon factorization of the reversed binary expansion of n.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Nov 11 2019

Keywords

Comments

First differs from A211100 at a(77) = 3, A211100(77) = 2. The reversed binary expansion of 77 is (1011001), with co-Lyndon factorization (10)(1100)(1), while the binary expansion is (1001101), with Lyndon factorization of (1)(001101).
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 certain order, their concatenation is equal to their co-Lyndon product. For example, (1001) has sorted co-Lyndon factorization (1)(100).

Examples

			The reversed binary expansion of each positive integer together with their co-Lyndon factorizations begins:
   1:     (1) = (1)
   2:    (01) = (0)(1)
   3:    (11) = (1)(1)
   4:   (001) = (0)(0)(1)
   5:   (101) = (10)(1)
   6:   (011) = (0)(1)(1)
   7:   (111) = (1)(1)(1)
   8:  (0001) = (0)(0)(0)(1)
   9:  (1001) = (100)(1)
  10:  (0101) = (0)(10)(1)
  11:  (1101) = (110)(1)
  12:  (0011) = (0)(0)(1)(1)
  13:  (1011) = (10)(1)(1)
  14:  (0111) = (0)(1)(1)(1)
  15:  (1111) = (1)(1)(1)(1)
  16: (00001) = (0)(0)(0)(0)(1)
  17: (10001) = (1000)(1)
  18: (01001) = (0)(100)(1)
  19: (11001) = (1100)(1)
  20: (00101) = (0)(0)(10)(1)
		

Crossrefs

The non-"co" version is A211100.
Positions of 2's are A329357.
Numbers whose binary expansion is co-Lyndon are A275692.
Length of the co-Lyndon factorization of the binary expansion is A329312.

Programs

  • Mathematica
    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[Reverse[IntegerDigits[n,2]]]],{n,100}]

A211097 Number of factors in Lyndon factorization of binary vectors of lengths 1, 2, 3, ...

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Apr 01 2012

Keywords

Comments

Any binary word has a unique factorization as a product of nonincreasing Lyndon words (see Lothaire). Here we look at the Lyndon factorizations of the binary vectors 0,1, 00,01,10,11, 000,001,010,011,100,101,110,111, 0000,...
For the largest (or leftmost) factor see A211098, A211099.
The smallest (or rightmost) factors are given by A211095 and A211096, offset by 2.

Examples

			Here are the Lyndon factorizations of the first few binary vectors:
.0.
.1.
.0.0.
.01.
.1.0.
.1.1.
.0.0.0.
.001.
.01.0. <- this means that the factorization is (01)(0), for example
.011.
.1.0.0.
.1.01.
.1.1.0.
.1.1.1.
.0.0.0.0.
...
		

References

  • M. Lothaire, Combinatorics on Words, Addison-Wesley, Reading, MA, 1983. See Theorem 5.1.5, p. 67.
  • G. Melançon, Factorizing infinite words using Maple, MapleTech Journal, vol. 4, no. 1, 1997, pp. 34-42

Crossrefs

A211098 and A211099 give information about the largest (or leftmost) factor.
Row-lengths of A329325.
The "co" version is A329400.
Retaining the first digit gives A211100.
Binary Lyndon words are counted by A001037 and constructed by A102659.
Numbers whose reversed binary expansion is Lyndon are A328596.

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,#]]&]]]];
    Table[Length[lynfac[Rest[IntegerDigits[n,2]]]],{n,2,50}] (* Gus Wiseman, Nov 14 2019 *)

A281013 Tetrangle T(n,k,i) = i-th part of k-th prime composition of n.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Jan 12 2017

Keywords

Comments

The *-product of two or more finite sequences is defined to be the lexicographically minimal sequence obtainable by shuffling them together. Every finite positive integer sequence has a unique *-factorization using prime compositions P = {(1), (2), (21), (3), (211), ...}. See A060223 and A228369 for details.
These are co-Lyndon compositions, ordered first by sum and then lexicographically. - Gus Wiseman, Nov 15 2019

Examples

			The prime factorization of (1, 1, 4, 2, 3, 1, 5, 5) is: (11423155) = (1)*(1)*(5)*(5)*(4231). The prime factorizations of the initial terms of A000002 are:
             (1) = (1)
            (12) = (1)*(2)
           (122) = (1)*(2)*(2)
          (1221) = (1)*(221)
         (12211) = (1)*(2211)
        (122112) = (1)*(2)*(2211)
       (1221121) = (1)*(221121)
      (12211212) = (1)*(2)*(221121)
     (122112122) = (1)*(2)*(2)*(221121)
    (1221121221) = (1)*(221)*(221121)
   (12211212212) = (1)*(2)*(221)*(221121)
  (122112122122) = (1)*(2)*(2)*(221)*(221121).
Read as a sequence:
(1), (2), (21), (3), (211), (31), (4), (2111), (221), (311), (32), (41), (5).
Read as a triangle:
(1)
(2)
(21), (3)
(211), (31), (4)
(2111), (221), (311), (32), (41), (5).
Read as a sequence of triangles:
1    2    2 1    2 1 1    2 1 1 1    2 1 1 1 1    2 1 1 1 1 1
          3      3 1      2 2 1      2 2 1 1      2 1 2 1 1
                 4        3 1 1      3 1 1 1      2 2 1 1 1
                          3 2        3 1 2        2 2 2 1
                          4 1        3 2 1        3 1 1 1 1
                          5          4 1 1        3 1 1 2
                                     4 2          3 1 2 1
                                     5 1          3 2 1 1
                                     6            3 2 2
                                                  3 3 1
                                                  4 1 1 1
                                                  4 1 2
                                                  4 2 1
                                                  4 3
                                                  5 1 1
                                                  5 2
                                                  6 1
                                                  7.
		

Crossrefs

The binary version is A329318.
The binary non-"co" version is A102659.
A sequence listing all Lyndon compositions is A294859.
Numbers whose binary expansion is co-Lyndon are A328596.
Numbers whose binary expansion is co-Lyndon are A275692.
Binary Lyndon words are A001037.
Lyndon compositions are A059966.
Normal Lyndon words are A060223.

Programs

  • Mathematica
    colynQ[q_]:=Array[Union[{RotateRight[q,#],q}]=={RotateRight[q,#],q}&,Length[q]-1,1,And];
    lexsort[f_,c_]:=OrderedQ[PadRight[{f,c}]];
    Table[Sort[Select[Join@@Permutations/@IntegerPartitions[n],colynQ],lexsort],{n,5}] (* Gus Wiseman, Nov 15 2019 *)

Formula

Row lengths are A059966(n) = number of prime compositions of n.

A329315 Irregular triangle read by rows where row n gives the sequence of lengths of components of the Lyndon factorization of the first n terms of A000002.

Original entry on oeis.org

1, 2, 3, 3, 1, 3, 1, 1, 3, 3, 3, 3, 1, 3, 5, 3, 6, 3, 6, 1, 3, 8, 3, 9, 3, 9, 1, 3, 9, 1, 1, 3, 9, 3, 3, 9, 3, 1, 3, 9, 3, 1, 1, 3, 9, 3, 3, 3, 9, 7, 3, 9, 7, 1, 3, 9, 9, 3, 9, 9, 1, 3, 9, 9, 1, 1, 3, 9, 9, 3, 3, 9, 9, 3, 1, 3, 9, 14, 3, 9, 15, 3, 9, 15, 1, 3
Offset: 1

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 4) never appear in the sequence.

Examples

			Triangle begins:
   1: (1)
   2: (2)
   3: (3)
   4: (3,1)
   5: (3,1,1)
   6: (3,3)
   7: (3,3,1)
   8: (3,5)
   9: (3,6)
  10: (3,6,1)
  11: (3,8)
  12: (3,9)
  13: (3,9,1)
  14: (3,9,1,1)
  15: (3,9,3)
  16: (3,9,3,1)
  17: (3,9,3,1,1)
  18: (3,9,3,3)
  19: (3,9,7)
  20: (3,9,7,1)
For example, the first 10 terms of A000002 are (1221121221), with Lyndon factorization (122)(112122)(1), so row 10 is (3,6,1).
		

Crossrefs

Row lengths are A296658.
The reversed version is A329316.

Programs

  • Mathematica
    lynQ[q_]:=Array[Union[{q,RotateRight[q,#1]}]=={q,RotateRight[q,#1]}&,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,#1]]&]]]];
    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[kol[n]],{n,100}]

A329325 Irregular triangle read by rows where row n gives the lengths of the components in the Lyndon factorization of the binary expansion of n with first digit removed.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Nov 11 2019

Keywords

Comments

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. 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).

Examples

			Triangle begins:
   1: ()        21: (22)       41: (23)       61: (1112)
   2: (1)       22: (31)       42: (221)      62: (11111)
   3: (1)       23: (4)        43: (5)        63: (11111)
   4: (11)      24: (1111)     44: (311)      64: (111111)
   5: (2)       25: (13)       45: (32)       65: (6)
   6: (11)      26: (121)      46: (41)       66: (51)
   7: (11)      27: (13)       47: (5)        67: (6)
   8: (111)     28: (1111)     48: (11111)    68: (411)
   9: (3)       29: (112)      49: (14)       69: (6)
  10: (21)      30: (1111)     50: (131)      70: (51)
  11: (3)       31: (1111)     51: (14)       71: (6)
  12: (111)     32: (11111)    52: (1211)     72: (3111)
  13: (12)      33: (5)        53: (122)      73: (33)
  14: (111)     34: (41)       54: (131)      74: (51)
  15: (111)     35: (5)        55: (14)       75: (6)
  16: (1111)    36: (311)      56: (11111)    76: (411)
  17: (4)       37: (5)        57: (113)      77: (6)
  18: (31)      38: (41)       58: (1121)     78: (51)
  19: (4)       39: (5)        59: (113)      79: (6)
  20: (211)     40: (2111)     60: (11111)    80: (21111)
For example, the trimmed binary expansion of 41 is (01001), with Lyndon factorization (01)(001), so row 41 is {2,3}.
		

Crossrefs

Row lengths are A211097.
Row sums are A000523.
Keeping the first digit gives A329314.
Positions of singleton rows are A329327.
Binary Lyndon words are counted by A001037 and ranked by A102659.
Numbers whose reversed binary expansion is a Lyndon word are A328596.
Length of the co-Lyndon factorization of the binary expansion is A329312.

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,#1]]&]]]];
    Table[Length/@lynfac[Rest[IntegerDigits[n,2]]],{n,100}]

A329317 Length of the Lyndon factorization of the reversed first n terms of A000002.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Nov 11 2019

Keywords

Comments

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).

Examples

			The sequence of Lyndon factorizations of the reversed initial terms of A000002 begins:
   1: (1)
   2: (2)(1)
   3: (2)(2)(1)
   4: (122)(1)
   5: (1122)(1)
   6: (2)(1122)(1)
   7: (12)(1122)(1)
   8: (2)(12)(1122)(1)
   9: (2)(2)(12)(1122)(1)
  10: (122)(12)(1122)(1)
  11: (2)(122)(12)(1122)(1)
  12: (2)(2)(122)(12)(1122)(1)
  13: (122)(122)(12)(1122)(1)
  14: (112212212)(1122)(1)
  15: (2)(112212212)(1122)(1)
  16: (12)(112212212)(1122)(1)
  17: (1121122122121122)(1)
  18: (2)(1121122122121122)(1)
  19: (2)(2)(1121122122121122)(1)
  20: (122)(1121122122121122)(1)
For example, the reversed first 13 terms of A000002 are (1221221211221), with Lyndon factorization (122)(122)(12)(1122)(1), so a(13) = 5.
		

Crossrefs

Row-lengths of A329316.
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}]

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}]
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