%I
%S 1,1,1,2,2,2,3,6,6,3,6,12,24,12,6,10,30,60,60,30,10,20,60,180,180,180,
%T 60,20,35,140,420,630,630,420,140,35,70,280,1120,1680,2520,1680,1120,
%U 280,70,126,630,2520,5040,7560,7560,5040,2520,630,126,252,1260,6300
%N Triangle read by rows: T(n,k) = binomial(n,k)*binomial(k,floor(k/2))*binomial(nk,floor((nk)/2)) (0<=k<=n).
%C T(n,k) is the number of paths in the first quadrant, starting from the origin, with unit steps up, down, right, or left, having a total of n steps, exactly k of which are vertical (up or down). Example: T(3,2)=6 because we have NNE, NEN, ENN, NSE, ENS and NES. [_Emeric Deutsch_, Nov 22 2008]
%H David M. Bloom et al., <a href="https://www.jstor.org/stable/3647977">A Convolution of Middle Binomial Coefficients: Problem 10921</a>, Amer. Math. Monthly 110, (2003), 958959.
%H E. Deutsch and D. Lovit, <a href="https://www.jstor.org/stable/27643000">Problem 1739</a>, Math. Magazine, vol. 80, No. 1, 2007, p. 80. [_Emeric Deutsch_, Nov 22 2008]
%F T(n, k) = binomial(n, k)*binomial(k, floor(k/2))*binomial(nk, floor((nk)/2)) (0<=k<=n).
%p T:=(n,k)>binomial(n,k)*binomial(k,floor(k/2))*binomial(nk,floor((nk)/2)): for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
%Y Row sums yield A005566. T(n, 0)=T(n, n)=A001405(n).
%Y Cf. A005566, A001405.
%K nonn,tabl
%O 0,4
%A _Emeric Deutsch_, Apr 23 2005
