%I #15 Jan 02 2015 11:40:15
%S 1,2,1,10,2,1,50,10,2,1,258,50,10,2,1,1362,258,50,10,2,1,7306,1362,
%T 258,50,10,2,1,39650,7306,1362,258,50,10,2,1,217090,39650,7306,1362,
%U 258,50,10,2,1,1196834,217090,39650,7306,1362,258,50,10,2,1,6634890,1196834
%N Triangle read by rows: T(n,k) (0<=k<=n) is the number of Delannoy paths of length n that start with exactly k (1,1) steps.
%C A Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1).
%C Row sums are the central Delannoy numbers (A001850). Column 0 yields A110170 (first differences of the central Delannoy numbers). sum(k*T(n,k),k=0..n)=A089165(n-1) (n>=1; partial sums of the central Delannoy numbers).
%H Robert A. Sulanke, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Sulanke/delannoy.html">Objects Counted by the Central Delannoy Numbers</a>, Journal of Integer Sequences, Volume 6, 2003, Article 03.1.5.
%F T(n, k) = A001850(n-k)-A001850(n-k-1) for k<n; T(n, n)=1.
%F T(n, k) = P_{n-k}(3)-P_{n-k-1}(3) for k<n; T(n, n)=1, where P_j is j-th Legendre polynomial.
%F G.f.: (1-z)/((1-t*z)*sqrt(1-6*z+z^2)).
%e T(3,2)=2 because we have DDNE and DDEN.
%e Triangle starts:
%e 1;
%e 2,1;
%e 10,2,1;
%e 50,10,2,1;
%e 258,50,10,2,1;
%p with(orthopoly): S:=proc(n,k) if k<n then P(n-k,3)-P(n-k-1,3) elif k=n then 1 else 0 fi end: for n from 0 to 10 do seq(S(n,k),k=0..n) od; # yields sequence in triangular form
%Y Cf. A001850, A110170, A089165.
%K nonn,tabl
%O 0,2
%A _Emeric Deutsch_, Jul 14 2005
%E Keyword tabf changed to tabl by _Michel Marcus_, Apr 09 2013