%I #17 Dec 12 2023 20:27:04
%S 1,0,1,0,0,2,0,0,1,5,0,0,0,6,14,0,0,0,3,28,42,0,0,0,1,28,120,132,0,0,
%T 0,0,20,180,495,429,0,0,0,0,10,195,990,2002,1430,0,0,0,0,4,165,1430,
%U 5005,8008,4862,0,0,0,0,1,117,1650,9009,24024,31824,16796,0,0,0,0,0,70,1617
%N Square table read by antidiagonals of number of ways of arranging n chords on a circle with k simple intersections (i.e., no intersections with 3 or more chords).
%C Row sums are A001147 (Double factorial).
%C Columns include A000108 (Catalan) for k=0 and A002694 for k=1.
%H H. Bottomley, <a href="/A002694/a002694.gif">Illustration for A000108, A001147, A002694, A067310 and A067311</a>
%H J. Touchard, <a href="http://dx.doi.org/10.4153/CJM-1952-001-8">Sur un problème de configurations et sur les fractions continues</a>, Canad. J. Math., 4 (1952), 2-25, g_n(x).
%H <a href="http://groups.google.com/groups?threadm=3c40c437.108548986%40news.btinternet.com&rnum=1">alt.math.recreational discussion</a>
%F T(n,k) = Sum_{j=0..n-1} (-1)^j * C((n-j)*(n-j+1)/2-1-k, n-1) * (C(2n, j) - C(2n, j-1)) where C(r,s)=binomial(r,s) if r>=s>=0 and 0 otherwise.
%e Rows start:
%e 1, 0, 0, 0, 0, 0, 0, ...;
%e 1, 0, 0, 0, 0, 0, 0, ...;
%e 2, 1, 0, 0, 0, 0, 0, ...;
%e 5, 6, 3, 1, 0, 0, 0, ...;
%e 14, 28, 28, 20, 10, 4, 1, ...; etc.,
%e i.e., there are 5 ways of arranging 3 chords with no intersections, 6 with one, 3 with two and 1 with three.
%Y A067311 has a different view of the same table.
%K nonn,tabl
%O 0,6
%A _Henry Bottomley_, Jan 14 2002