%I #15 May 04 2021 10:42:47
%S 1,2,3,1,5,3,7,9,13,17,2,19,39,6,35,72,21,59,141,55,1,107,266,132,7,
%T 187,511,300,26,351,952,660,85,631,1827,1395,240,3,1181,3459,2901,636,
%U 15,2191,6595,5977,1554,67,4115,12604,12123,3698,228,7711,24173,24504
%N Sequence is an irregular triangle read by rows with zeros removed where T(n,k) is the number of compositions of n whose standard factorization into Lyndon words has k distinct factors.
%C Row sums are 2^(n-1). First column is A008965. A regular version is A299070.
%H Andrew Howroyd, <a href="/A299072/b299072.txt">Table of n, a(n) for n = 1..1196</a>
%e Triangle begins:
%e 1
%e 2
%e 3 1
%e 5 3
%e 7 9
%e 13 17 2
%e 19 39 6
%e 35 72 21
%e 59 141 55 1
%e 107 266 132 7
%e 187 511 300 26
%t LyndonQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And]&&Array[RotateRight[q,#]&,Length[q],1,UnsameQ];
%t qit[q_]:=If[#===Length[q],{q},Prepend[qit[Drop[q,#]],Take[q,#]]]&[Max@@Select[Range[Length[q]],LyndonQ[Take[q,#]]&]];
%t DeleteCases[Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[Union[qit[#]]]===k&]],{n,11},{k,n}],0,{2}]
%o (PARI) \\ here b(n) is A059966.
%o b(n)={sumdiv(n, d, moebius(n/d) * (2^d-1))/n}
%o A(n)=[Vecrev(p/y) | p<-Vec(prod(k=1, n, (1 - y + y/(1-x^k) + O(x*x^n))^b(k))-1)]
%o my(T=A(15)); for(n=1, #T, print(T[n])) \\ _Andrew Howroyd_, Dec 08 2018
%Y Cf. A001045, A001221, A008965, A059966, A116608, A146289, A185700, A296373, A299070.
%K nonn,tabf
%O 1,2
%A _Gus Wiseman_, Feb 01 2018