%I #38 Jan 06 2022 13:26:46
%S 1,1,1,1,3,1,1,6,6,1,1,10,21,10,1,1,15,55,55,15,1,1,21,120,215,120,21,
%T 1,1,28,231,665,665,231,28,1,1,36,406,1736,2835,1736,406,36,1,1,45,
%U 666,3990,9891,9891,3990,666,45,1,1,55,1035,8310,29505,45297,29505,8310,1035,55,1,1,66,1540,16005,77715,173712,173712,77715,16005,1540,66,1
%N Triangle read by rows: T(n,k) is the number of k-matchings of the corona K'(n) of the complete graph K(n) and the complete graph K(1); in other words, K'(n) is the graph constructed from K(n) by adding for each vertex v a new vertex v' and the edge vv'.
%C Row n has n+1 terms.
%C Row sums yield A005425.
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Barry4/barry142.html">On a Generalization of the Narayana Triangle</a>, J. Int. Seq., Vol. 14 (2011), Article 11.4.5.
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Barry2/barry281.html">Constructing Exponential Riordan Arrays from Their A and Z Sequences</a>, Journal of Integer Sequences, Vol. 17 (2014), Article 14.2.6.
%H Paul Barry, <a href="https://arxiv.org/abs/1804.05027">The Gamma-Vectors of Pascal-like Triangles Defined by Riordan Arrays</a>, arXiv:1804.05027 [math.CO], 2018.
%F T(n,k) = T(n,n-k).
%F E.g.f.: exp(z+t*z+t*z^2/2).
%F Row generating polynomial P[n] = [ -i*sqrt(t/2)]^n*H(n, i(1+t)/sqrt(2t)), where H(n, x) is a Hermite polynomial and i=sqrt(-1).
%F Row generating polynomials P[n] satisfy P[0]=1, P[n]=(1+t)P[n-1]+(n-1)tP[n-2].
%F From _Fabián Pereyra_, Jan 05 2022: (Start)
%F T(n,k) = T(n-1,k) + (n-1)*T(n-2,k-1) + T(n-1,k-1), n>=0, 0<=k<=n; T(0,0) = 1.
%F T(n,k) = Sum_{j=k..n} C(n,j)*B(j,j-k), where B are the Bessel numbers A100861. (End)
%e T(3,2)=6 because in the graph with vertex set {A,B,C,a,b,c} and edge set {AB,AC,BC,Aa,Bb,Cc} we have the following six 2-matchings: {Aa,BC},{Bb,AC},{Cc,AB},{Aa,Bb},{Aa,Cc} and {Bb,Cc}.
%e Triangle starts:
%e 1;
%e 1, 1;
%e 1, 3, 1;
%e 1, 6, 6, 1;
%e 1, 10, 21, 10, 1;
%e 1, 15, 55, 55, 15, 1;
%p P[0]:=1: for n from 1 to 11 do P[n]:=sort(expand((1+t)*P[n-1]+(n-1)*t*P[n-2])) od: for n from 0 to 11 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # gives the sequence in triangular form
%t P[0] = 1; P[1] = 1+t; P[n_] := P[n] = (1+t) P[n-1] + (n-1) t P[n-2];
%t Table[CoefficientList[P[n], t], {n, 0, 11}] // Flatten (* _Jean-François Alcover_, Jul 23 2018 *)
%o (PARI) {T(n,k)=local(X=x+x*O(x^n),Y=y+y*O(y^k)); n!*polcoeff(polcoeff(exp(X+Y*X^2/2+X*Y),n,x),k,y)} \\ _Paul D. Hanna_, Jul 18 2005
%Y Cf. A005425.
%Y Cf. A107102 (matrix inverse).
%K nonn,tabl
%O 0,5
%A _Emeric Deutsch_, Jan 08 2005