A296372 Triangle read by rows: T(n,k) is the number of normal sequences of length n whose standard factorization into Lyndon words (aperiodic necklaces) has k factors.

1, 1, 2, 4, 5, 4, 18, 31, 18, 8, 108, 208, 153, 56, 16, 778, 1700, 1397, 616, 160, 32, 6756, 15980, 14668, 7197, 2196, 432, 64, 68220, 172326, 171976, 93293, 31564, 7208, 1120, 128

Offset: 1

Author

Gus Wiseman, Dec 11 2017

Keywords

• nonn
• tabl

Comments

A finite sequence is normal if its union is an initial interval of positive integers.

Examples

			The T(3,2) = 5 normal sequences are {2,1,2}, {1,2,1}, {2,1,3}, {2,3,1}, {3,1,2}.
Triangle begins:
     1;
     1,     2;
     4,     5,     4;
    18,    31,    18,     8;
   108,   208,   153,    56,    16;
   778,  1700,  1397,   616,   160,    32;
  6756, 15980, 14668,  7197,  2196,   432,    64;
		

Links

• Andrew Howroyd, Rows n = 1..50 of triangle, flattened
• Wikipedia, Lyndon word: Standard factorization

Crossrefs

Row sums are A000670.

First column is A060223.

Cf. A000740, A001045, A008965, A019536, A059966, A074650, A185700, A228369, A232472, A277427, A281013, A296373.

Programs

Mathematica

neckQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And];
aperQ[q_]:=UnsameQ@@Table[RotateRight[q,k],{k,Length[q]}];
qit[q_]:=If[#===Length[q],{q},Prepend[qit[Drop[q,#]],Take[q,#]]]&[Max@@Select[Range[Length[q]],neckQ[Take[q,#]]&&aperQ[Take[q,#]]&]];
allnorm[n_]:=Function[s,Array[Count[s,y_/;y<=#]+1&,n]]/@Subsets[Range[n-1]+1];
Table[Length[Select[Join@@Permutations/@allnorm[n],Length[qit[#]]===k&]],{n,5},{k,n}]

PARI

\\ here U(n,k) is A074650(n,k).
EulerMT(u)={my(n=#u, p=x*Ser(u), vars=variables(p)); Vec(exp( sum(i=1, n, substvec(p + O(x*x^(n\i)), vars, apply(v->v^i,vars))/i ))-1)}
U(n,k)={sumdiv(n, d, moebius(n/d) * k^d)/n}
A(n)={[Vecrev(p/y) | p<-sum(k=1, n, EulerMT(vector(n, n, y*U(n,k)))*sum(j=k, n, (-1)^(k-j)*binomial(j,k)))]}
{ my(T=A(10)); for(n=1, #T, print(T[n])) } \\ Andrew Howroyd, Dec 08 2018

Extensions

Example and program corrected by Gus Wiseman, Dec 08 2018