A112340 Triangle read by rows of numbers b_{n,k}, n>=1, 1<=k<=n such that Product_{n,k} 1/(1-q^n t^k)^{b_{n,k}} = 1 + Sum_{i,j>=1} S_{i,j} q^i t^j where S_{i,j} are entries in the table A008277 (the inverse Euler transformation of the table of Stirling numbers of the second kind).

1, 1, 0, 1, 2, 0, 1, 5, 3, 0, 1, 13, 16, 4, 0, 1, 28, 67, 34, 5, 0, 1, 60, 249, 229, 65, 6, 0, 1, 123, 853, 1265, 609, 107, 7, 0, 1, 251, 2787, 6325, 4696, 1376, 168, 8, 0, 1, 506, 8840, 29484, 31947, 14068, 2772, 244, 9, 0, 1, 1018, 27503, 131402, 199766, 124859, 36252

Offset

1,5

Comments

Row sums equal to A085686, second column = A084174 - 1

The number of set partitions of size n length k which are 'Lyndon,' that is, since all set partitions are isomorphic to sequences of atomic set partitions (A087903), those which are smallest of all rotations of these sequences in lex order (with respect to some ordering on the atomic set partitions) are Lyndon. 1; 1, 0; 1, 2, 0; 1, 5, 3, 0; 1, 13, 16, 4, 0;

Links

Table of n, a(n) for n=1..62.

N. Bergeron, M. Zabrocki, The Hopf algebras of symmetric functions and quasisymmetric functions in non-commutative variables are free and cofree, arXiv:math/0509265 [math.CO], 2005.

M. Rosas and B. Sagan, Symmetric Functions in Noncommuting Variables, Transactions of the American Mathematical Society, 358 (2006), no. 1, 215-232.

M. C. Wolf, Symmetric functions for non-commutative elements, Duke Math. J., 2 (1936), 626-637.

Example

There are 6 set partitions of size 4 and length 3, {12|3|4}, {13|2|4}, {14|2|3}, {1|23|4}, {1|24|3}, {1|2|34} and the sequences the correspond to are ({12},{1},{1}), ({13|2}, {1}), ({14|2|3}), ({1},{12},{1}), ({1},{13|2}), ({1},{1},{12}). Now there are three {({12},{1},{1}), ({1},{12},{1}), ({1},{1},{12})} that are rotations of each other and ({1}, {1}, {12}) is the smallest of these, {({13|2}, {1}), ({1},{13|2})} are rotations of each other and ({1},{13|2}) is the smallest and ({14|2|3}) is atomic and all atomic s.p. are Lyndon. Hence {1|2|34}, {1|24|3}, {14|2|3} are Lyndon and a(4,3) = 3
Triangle begins:
  1;
  1,  0;
  1,  2,  0;
  1,  5,  3,  0;
  1, 13, 16,  4, 0;
  1, 28, 67, 34, 5, 0;
  ...

Maple

EULERitable:=proc(tbl) local ser, out, i, j, tmp; ser:=1+add(add(q^i*t^j*tbl[i][j], j=1..nops(tbl[i])), i=1..nops(tbl)); out:=[]; for i from 1 to nops(tbl) do tmp:=coeff(ser, q, i); ser:=expand(ser*mul(add((-q^i*t^j)^k*choose(abs(coeff(tmp, t, j)), k), k=0..nops(tbl)/i), j = 1..degree(tmp, t))); ser:=subs({seq(q^j=0, j=nops(tbl)+1..degree(ser, q))}, ser); out:=[op(out), [seq(abs(coeff(tmp, t, j)), j=1..degree(tmp, t))]]; end do; out; end: EULERitable([seq([seq(combinat[stirling2](n, k), k=1..n)], n=1..10)]);

Mathematica

nmax = 11; b[n_, k_] /; k < 1 || k > n = 0;
coes[m_] := Product[1/(1 - q^n t^k)^b[n, k], {n, 1, m}, {k, 1, m}] - 1 - Sum[ StirlingS2[i, j] q^i t^j, {i, 1, m}, {j, 1, m}] + O[t]^m + O[q]^m // Normal // CoefficientList[#, {t, q}]&;
sol[1] = {b[1, 1] -> 1};
Do[sol[m] = Solve[Thread[(coes[m] /. sol[m - 1]) == 0]], {m, 2, nmax + 1}];
bb = Flatten[Table[sol[m], {m, 1, nmax + 1}]];
Table[b[n, k] /. bb, {n, 1, nmax}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 11 2017 *)

Crossrefs

Cf. A008277, A085686, A112339.

Cf. A087903, A000110.

Sequence in context: A079134 A175528 A163940 * A356264 A037186 A343239

Adjacent sequences: A112337 A112338 A112339 * A112341 A112342 A112343

Keyword

nonn,tabl

Author

Mike Zabrocki, Sep 05 2005; Aug 06 2006

Status

approved