%I #13 Jul 10 2018 10:59:31
%S 1,1,1,1,1,3,7,10,10,1,6,25,75,175,280,280,1,10,65,315,1225,3780,9100,
%T 15400,15400,1,15,140,980,5565,26145,102025,323400,800800,1401400,
%U 1401400,1,21,266,2520,19425,125895,695695,3273270,12962950
%N Triangle in A144385 with rows left-adjusted.
%C Row n has 2n+1 terms.
%H Alois P. Heinz, <a href="/A144399/b144399.txt">Rows n = 0..100, flattened</a>
%H Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, <a href="http://arxiv.org/abs/1701.08394">Analysis of the Gift Exchange Problem</a>, arXiv:1701.08394, 2017.
%H David Applegate and N. J. A. Sloane, <a href="http://arxiv.org/abs/0907.0513">The Gift Exchange Problem</a> (arXiv:0907.0513, 2009)
%e Triangle begins:
%e 1
%e 1, 1, 1
%e 1, 3, 7, 10, 10
%e 1, 6, 25, 75, 175, 280, 280
%e 1, 10, 65, 315, 1225, 3780, 9100, 15400, 15400
%p b:= proc(n) option remember; expand(`if`(n=0, 1, add(
%p b(n-j)*binomial(n-1, j-1), j=1..min(3, n))*x))
%p end:
%p T:= (n, k)-> coeff(b(k), x, n):
%p seq(seq(T(n, k), k=n..3*n), n=0..6); # _Alois P. Heinz_, May 31 2018
%t b[n_] := b[n] = Expand[If[n == 0, 1, Sum[b[n - j]*Binomial[n - 1, j - 1], {j, 1, Min[3, n]}]*x]];
%t T[n_, k_] := Coefficient[b[k], x, n];
%t Table[T[n, k], {n, 0, 6}, { k, n, 3n}] // Flatten (* _Jean-François Alcover_, Jul 10 2018, after _Alois P. Heinz_ *)
%Y Cf. A144385. Row sums give A144416.
%K nonn,tabf
%O 0,6
%A _David Applegate_ and _N. J. A. Sloane_, Dec 07 2008