|
|
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).
|
|
7
|
|
|
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
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,5
|
|
COMMENTS
|
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
|
|
|
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}]];
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|