%I #5 Jan 17 2013 10:32:27
%S 1,1,3,3,2,10,10,8,4,35,35,30,20,8,126,126,112,84,48,16,462,462,420,
%T 336,224,112,32,1716,1716,1584,1320,960,576,256,64,6435,6435,6006,
%U 5148,3960,2640,1440,576,128,24310,24310,22880,20020,16016,11440,7040,3520,1280
%N T(n,k)=Binomial(2n-k,n)*2^(k-1) for n>=1,0<=k<=n
%C T(n,k) is the number of 2n digit binary sequences in which the (n+1)th zero occurs in the (2n-k+1)th position. T(n,k)/2^(2n-1) is the probability sought in Banach's matchbox problem. Row sum is 2^(2n-1) T(n,0)=T(n,1)=A001700(n)
%e T(2,1)=3 because there are 3 length 4 binary sequences in which the third zero appears in the fourth position: {0,0,1,0},{0,1,0,0},{1,0,0,0}.
%e Triangle begins
%e 1, 1
%e 3, 3, 2
%e 10, 10, 8, 4
%e 35, 35, 30, 20, 8
%e 126, 126, 112, 84, 48, 16
%t Table[Table[Binomial[2 n - k, n]*2^(k - 1), {k, 0, n}], {n, 1, 9}] // Grid
%K nonn,tabf
%O 1,3
%A _Geoffrey Critzer_, Aug 23 2009