%I
%S 1,1,1,1,2,2,1,3,6,6,1,4,12,24,20,1,5,20,60,100,70,1,6,30,120,300,420,
%T 252,1,7,42,210,700,1470,1764,924,1,8,56,336,1400,3920,7056,7392,3432,
%U 1,9,72,504,2520,8820,21168,33264,30888,12870,1,10,90,720,4200,17640,52920,110880,154440,128700,48620
%N Triangle read by rows: number of orderpreserving partial transformations (of an nelement chain) of width and waist both equal to r (width(alpha) = Dom(alpha) and waist(alpha) = max(Im(alpha)).
%C Whenever the formula generates a zero, it is changed to a one in the sequence.  _Harvey P. Dale_, Sep 06 2013
%H Laradji, A. and Umar, <a href="http://dx.doi.org/10.1016/j.jalgebra.2003.10.023">A. Combinatorial results for semigroups of orderpreserving partial transformations</a>, Journal of Algebra 278, (2004), 342359.
%H Laradji, A. and Umar, A. <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL7/Umar/um.html">Combinatorial results for semigroups of orderdecreasing partial transformations</a>, J. Integer Seq. 7 (2004), 04.3.8
%F G(n,k) = binomial(n,k)*binomial(2*k2,k1), n >= k > 0.
%e G(3,2)=6 because there are exactly 6 orderpreserving partial transformations (on a 3element chain) of both width and waist equal to 2, namely: (1,2)>(1,2),(1,2)>(2,2),(1,3)>(1,2),(1,3)>(2,2),(2,3)>(1,2),(2,3)>(2,2)
%e 1;
%e 1,1;
%e 1,2,2;
%e 1,3,6,6;
%e 1,4,12,24,20;
%e 1,5,20,60,100,70;
%e 1,6,30,120,300,420,252;
%e 1,7,42,210,700,1470,1764,924;
%e 1,8,56,336,1400,3920,7056,7392,3432;
%p A110858 := proc(n,k) if k = 0 then 1; else binomial(n,k)*binomial(2*k2,k1) ; end if; end proc: # _R. J. Mathar_, Jun 20 2011
%t Flatten[Table[Binomial[n,k]Binomial[2k2,k1],{n,0,10},{k,0,n}]/.{0>1}] (* _Harvey P. Dale_, Sep 04 2013 *)
%K nonn,tabl
%O 0,5
%A _Abdullahi Umar_, Aug 25 2008
