%I #44 Feb 23 2023 12:19:07
%S 1,1,1,3,3,1,13,14,6,1,73,84,41,10,1,501,609,325,95,15,1,4051,5155,
%T 2944,965,190,21,1,37633,49790,30023,10689,2415,343,28,1,394353,
%U 539616,340402,129220,32179,5348,574,36,1,4596553,6478521,4246842,1698374,455511,84567,10794,906,45,1
%N Triangular array read by rows: T(n,k) is the number of partial permutations of {1,2,...,n} that have exactly k cycles, 0<=k<=n.
%C A partial permutation on a set X is a bijection between two subsets of X.
%C Row sums are A002720.
%C First column (corresponding to k=0) is A000262.
%H Alois P. Heinz, <a href="/A216294/b216294.txt">Rows n = 0..50, flattened</a>
%H Philippe Flajolet and Robert Sedgewick, <a href="http://algo.inria.fr/flajolet/Publications/AnaCombi/anacombi.html">Analytic Combinatorics</a>, Cambridge Univ. Press, 2009, page 132.
%F E.g.f.: exp(x/(1-x))/(1-x)^y.
%F From _Peter Bala_, Aug 23 2013: (Start)
%F Exponential Riordan array [exp(x/(1-x)), log(1/(1-x))].
%F The row polynomials R(n,y), n > = 0, satisfy the 2nd order recurrence equation R(n,y) = (2*n + y - 1)*R(n-1,y) - (n - 1)*(n + y - 2)*R(n-2,y) with R(0,y) = 1 and R(1,y) = 1 + y.
%F Modulo variations in offset we have: R(n,0) = A000262, R(n,1) = A002720, R(n,2) = A000262, R(n,3) = A052852, R(n,4) = A062147, R(n,5) = A062266 and R(n,6) = A062192. In general, for fixed k, the sequence {R(n,k)}n>=1 gives the entries on a diagonal of the square array A088699. (End)
%e 1;
%e 1, 1;
%e 3, 3, 1;
%e 13, 14, 6, 1;
%e 73, 84, 41, 10, 1;
%e 501, 609, 325, 95, 15, 1;
%p gf := exp(x / (1 - x)) / (1 - x)^y:
%p serx := series(gf, x, 10): poly := n -> simplify(coeff(serx, x, n)):
%p seq(print(seq(n!*coeff(poly(n), y, k), k = 0..n)), n = 0..9); # _Peter Luschny_, Feb 23 2023
%t nn=10;t=Sum[n^(n-1)x^n/n!,{n,1,nn}];f[list_]:=Select[list,#>0&];Map[f,Range[0,nn]!CoefficientList[Series[Exp[ x/(1-x)]/(1-x)^y,{x,0,nn}],{x,y}]]//Flatten
%Y Cf. A021009. A000262, A002720, A052852, A062147, A062192, A062266, A088699.
%K nonn,tabl
%O 0,4
%A _Geoffrey Critzer_, Sep 04 2012