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 10 results.

A296658 Length of the standard Lyndon word factorization of the first n terms of A000002.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Dec 18 2017

Keywords

Examples

			The standard Lyndon word factorization of (12211212212211211) is (122)(112122122)(112)(1)(1), so a(17) = 5.
The standard factorizations of initial terms of A000002:
1
12
122
122,1
122,1,1
122,112
122,112,1
122,11212
122,112122
122,112122,1
122,11212212
122,112122122
122,112122122,1
122,112122122,1,1
122,112122122,112
122,112122122,112,1
122,112122122,112,1,1
122,112122122,112,112
122,112122122,1121122
122,112122122,1121122,1
		

Crossrefs

Programs

  • Mathematica
    LyndonQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And]&&Array[RotateRight[q,#]&,Length[q],1,UnsameQ];
    qit[q_]:=If[#===Length[q],{q},Prepend[qit[Drop[q,#]],Take[q,#]]]&[Max@@Select[Range[Length[q]],LyndonQ[Take[q,#]]&]];
    kolagrow[q_]:=If[Length[q]<2,Take[{1,2},Length[q]+1],Append[q,Switch[{q[[Length[Split[q]]]],Part[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]]];
    Table[Length[qit[Nest[kolagrow,1,n]]],{n,150}]

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}]

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}]

A332273 Sizes of maximal weakly decreasing subsequences of A000002.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Mar 08 2020

Keywords

Examples

			The weakly decreasing subsequences begin: (1), (2,2,1,1), (2,1), (2,2,1), (2,2,1,1), (2,1,1), (2,2,1), (2,1,1), (2,1), (2,2,1,1), (2,1,1), (2,1), (2,2,1), (2,2,1,1).
		

Crossrefs

The number of runs in the first n terms of A000002 is A156253.
The weakly increasing version is A332875.

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]:=Nest[kolagrow,{1},n-1];
    Length/@Split[kol[40],#1>=#2&]

Formula

a(n) = A000002(2*n - 2) + A000002(2*n - 1) for n > 1.

A332875 Sizes of maximal weakly increasing subsequences of A000002.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Mar 08 2020

Keywords

Examples

			The weakly increasing subsequences begin: (1,2,2), (1,1,2), (1,2,2), (1,2,2), (1,1,2), (1,1,2,2), (1,2), (1,1,2), (1,2,2), (1,1,2), (1,1,2), (1,2,2), (1,2,2).
		

Crossrefs

The number of runs in the first n terms of A000002 is A156253.
The weakly decreasing version is A332273.

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]:=Nest[kolagrow,{1},n-1];
    Length/@Split[kol[40],#1<=#2&]

Formula

a(n) = A000002(2*n - 1) + A000002(2*n).

A329355 The binary expansion of a(n) is the second through n-th terms of A000002 - 1.

Original entry on oeis.org

0, 1, 3, 6, 12, 25, 50, 101, 203, 406, 813, 1627, 3254, 6508, 13017, 26034, 52068, 104137, 208275, 416550, 833101, 1666202, 3332404, 6664809, 13329618, 26659237, 53318475, 106636950, 213273900, 426547801, 853095602, 1706191204, 3412382409, 6824764818
Offset: 1

Views

Author

Gus Wiseman, Nov 12 2019

Keywords

Examples

			a(11) = 813 has binary expansion q = {1, 1, 0, 0, 1, 0, 1, 1, 0, 1}, and q + 1 is {2, 2, 1, 1, 2, 1, 2, 2, 1, 2}, which is the second through 11th terms of A000002.
		

Crossrefs

Replacing "A000002 - 1" with "2 - A000002" gives A329356.
Partial sums of A000002 are A054353.
Initial subsequences of A000002 are A329360.

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]];
    Table[FromDigits[kol[n]-1,2],{n,30}]

A329356 The binary expansion of a(n) is the first n terms of 2 - A000002.

Original entry on oeis.org

0, 1, 2, 4, 9, 19, 38, 77, 154, 308, 617, 1234, 2468, 4937, 9875, 19750, 39501, 79003, 158006, 316012, 632025, 1264050, 2528101, 5056203, 10112406, 20224813, 40449626, 80899252, 161798505, 323597011, 647194022, 1294388045, 2588776091, 5177552182, 10355104365
Offset: 0

Views

Author

Gus Wiseman, Nov 12 2019

Keywords

Examples

			a(7) = 77 has binary expansion q = {1, 0, 0, 1, 1, 0, 1}, and 2 - q is {1, 2, 2, 1, 1, 2, 1}, which is the first 7 terms of A000002.
		

Crossrefs

Replacing "2 - A000002" with "A000002 - 1" gives A329355.
Initial subsequences of A000002 are A329360.

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]];
    Table[FromDigits[2-kol[n],2],{n,0,30}]

Formula

a(n) = floor((1-c/2)*2^n), where c = A118270 is the Kolakoski constant. - Lorenzo Sauras Altuzarra, Jan 01 2023

A329360 The decimal expansion of a(n) is the first n terms of A000002.

Original entry on oeis.org

0, 1, 12, 122, 1221, 12211, 122112, 1221121, 12211212, 122112122, 1221121221, 12211212212, 122112122122, 1221121221221, 12211212212211, 122112122122112, 1221121221221121, 12211212212211211, 122112122122112112, 1221121221221121122, 12211212212211211221
Offset: 0

Views

Author

Gus Wiseman, Nov 12 2019

Keywords

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]];
    Table[FromDigits[kol[n]],{n,0,30}]

A329361 a(n) = Sum_{i = 1..n} 2^(n - i) * A000002(i).

Original entry on oeis.org

0, 1, 4, 10, 21, 43, 88, 177, 356, 714, 1429, 2860, 5722, 11445, 22891, 45784, 91569, 183139, 366280, 732562, 1465125, 2930252, 5860505, 11721011, 23442024, 46884049, 93768100, 187536202, 375072405, 750144811, 1500289624, 3000579249, 6001158499, 12002317000
Offset: 0

Views

Author

Gus Wiseman, Nov 12 2019

Keywords

Examples

			The first 5 terms of A000002 are {1, 2, 2, 1, 1}, so a(5) = 2^4 * 1 + 2^3 * 2 + 2^2 * 2 + 2^1 * 1 + 2^0 * 1 = 43.
		

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]];
    Table[FromDigits[kol[n],2],{n,0,30}]

Formula

a(n + 1) = A000002(n) + 2 a(n).

A333229 First sums of the Kolakoski sequence A000002.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Mar 18 2020

Keywords

Crossrefs

Positions of 3's are A054353.
Positions of 2's are A074262.
Positions of 4's are A074263.
The number of runs in the first n terms of A000002 is A156253(n).
Even-indexed terms are A332273 (without the first term).
Odd-indexed terms are A332875.

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]:=Nest[kolagrow,{1},n-1];
    Table[kol[n][[-1]]+kol[n+1][[-1]],{n,30}]

Formula

a(n) = A000002(n) + A000002(n + 1).
Showing 1-10 of 10 results.