%I #23 Jun 25 2023 12:48:18
%S 1,1,1,1,6,1,1,21,18,1,1,60,150,40,1,1,155,900,650,75,1,1,378,4515,
%T 7000,2100,126,1,1,889,20286,59535,36750,5586,196,1,1,2040,84700,
%U 435120,486570,148176,12936,288,1,1,4599,335880,2864820,5358150,2876202,493920,27000,405,1
%N Triangular array read by rows. T(n,k) is the number of Green's H-classes of rank k in the semigroup of partial transformations, n >= 0, 0 <= k <= n.
%C Let H_f denote the H-class in the semigroup of partial transformations containing f. Then H_f contains an idempotent iff the image of f is a transversal for the kernel of f.
%C Let H_f ~ H_g iff the image of f is contained in the image of g and the kernel of f is more coarse than the kernel of g. Then ~ is a partial order on the H-classes, hence a preorder (quasi-order) on the semigroup. The poset is isomorphic to the Segre product of the Boolean lattice of rank n and the partition lattice of [n+1].
%D O. Ganyushkin and V. Mazorchuk, Classical Finite Transformation Semigroups, 2009, Chapter 4.4 - 4.6.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Green's_relations">Green's relations</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Transversal_(combinatorics)">Transversal (combinatorics)</a>
%F T(n,k) = A007318(n,k)*A008277(n+1,k+1).
%F Sum_{k=0..n} T(n,k)*k! = (n+1)^n = A000169(n+1).
%F T(n,1) = A101818(n,1) = A066524(n) = n*(2^n - 1). (Every partial function of rank 1 is idempotent.)
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 6, 1;
%e 1, 21, 18, 1;
%e 1, 60, 150, 40, 1;
%e 1, 155, 900, 650, 75, 1;
%e ...
%p T:= (n, k)-> binomial(n, k)*Stirling2(n+1, k+1):
%p seq(seq(T(n, k), k=0..n), n=0..10); # _Alois P. Heinz_, Jun 24 2023
%t Table[Table[Binomial[n, k] StirlingS2[n + 1, k + 1], {k, 0, n}], {n,0, 5}] // Grid
%Y Columns k=0-1 give: A000012, A066524.
%Y Row sums give A134055(n+1).
%Y T(n,n-1) gives A002411.
%Y Cf. A000169, A007318, A008277, A090683, A101818.
%K nonn,tabl
%O 0,5
%A _Geoffrey Critzer_, Jun 24 2023