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.

Previous Showing 11-13 of 13 results.

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).

A296659 Length of the final word in the standard Lyndon word factorization of the first n terms of A000002.

Original entry on oeis.org

1, 2, 3, 1, 1, 3, 1, 5, 6, 1, 8, 9, 1, 1, 3, 1, 1, 3, 7, 1, 9, 1, 1, 3, 1, 14, 15, 1, 1, 3, 1, 1, 3, 1, 8, 9, 1, 11, 12, 1, 1, 3, 1, 17, 18, 1, 20, 1, 1, 3, 1, 1, 3, 27, 1, 29, 30, 1, 1, 3, 1, 35, 36, 1, 38, 39, 1, 1, 3, 1, 1, 3, 1, 8, 9, 1, 11, 1, 1, 3, 15, 1
Offset: 1

Views

Author

Gus Wiseman, Dec 18 2017

Keywords

Examples

			The sequence of final words begins: 1, 12, 122, 1, 1, 112, 1, 11212, 112122, 1, 11212212, 112122122, 1, 1, 112, 1, 1, 112, 1121122, 1, 112112212, 1, 1, 112, 1, 11211221211212, 112112212112122, 1, 1, 112.
		

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[Last[qit[Nest[kolagrow,1,n]]]],{n,150}]
Previous Showing 11-13 of 13 results.