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
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
Programs
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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}]
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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
Comments