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A296658
Length of the standard Lyndon word factorization of the first n terms of A000002.
15
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
1,4
LINKS
Frédérique Bassino, Julien Clement, and Cyril Nicaud, The standard factorization of Lyndon words: an average point of view, Discrete Mathematics, 290-1 (2005), 1-25.
EXAMPLE
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
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Gus Wiseman, Dec 18 2017
STATUS
approved