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%I #6 Nov 19 2017 12:44:23
%S 1,1,4,4,15,18,3,56,80,24,210,350,150,10,792,1512,840,120,3003,6468,
%T 4410,980,35,11440,27456,22176,6720,560,43758,115830,108108,41580,
%U 5670,126,167960,486200,514800,240240,46200,2520
%N Triangle read by rows: T(n,k) = binomial(n,k)*binomial(2*n-2*k,n-1), n>=1, 0<=k<=floor(n/2+1/2).
%C Row n contains floor(n/2+3/2) terms.
%C Row sums with alternate signs are 0.
%H D. Beckwith, <a href="https://www.jstor.org/stable/27642489">A Vanishing Alternating Sum: Problem 11212/11220</a>, Amer. Math. Monthly 115, (2008), p. 366.
%e Triangle starts:
%e 1,1;
%e 4,4;
%e 15,18,3;
%e 56,18,3;
%e 210,350,150,10;
%p T:=proc(n,r) options operator, arrow: binomial(n,r)*binomial(2*n-2*r,n-1) end proc: for n to 10 do seq(T(n,k),k=0..floor((1/2)*n+1/2)) end do; # yields sequence in triangular form
%K nonn,tabf
%O 1,3
%A _Emeric Deutsch_, Mar 30 2008