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