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
Examples
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; ...
Links
- 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.
Programs
-
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 *)
Comments