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A164705 T(n,k) = binomial(2n-k,n) * 2^(k-1), T(0,0)=1; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

%I A164705 #27 Apr 06 2025 18:24:25
%S A164705 1,1,1,3,3,2,10,10,8,4,35,35,30,20,8,126,126,112,84,48,16,462,462,420,
%T A164705 336,224,112,32,1716,1716,1584,1320,960,576,256,64,6435,6435,6006,
%U A164705 5148,3960,2640,1440,576,128,24310,24310,22880,20020,16016,11440,7040,3520,1280,256
%N A164705 T(n,k) = binomial(2n-k,n) * 2^(k-1), T(0,0)=1; triangle T(n,k), n>=0, 0<=k<=n, read by rows.
%C A164705 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) = A088218(n).
%H A164705 Sean A. Irvine, <a href="/A382782/a382782.pdf">Computing A382782</a>, 2025.
%F A164705 Sum_{k=0..n} k * T(n,k) = A000531(n). - _Alois P. Heinz_, Apr 06 2025
%e A164705 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 A164705 Triangle begins
%e A164705    1;
%e A164705    1,   1;
%e A164705    3,   3,   2;
%e A164705   10,  10,   8,  4;
%e A164705   35,  35,  30, 20,  8;
%e A164705  126, 126, 112, 84, 48, 16;
%e A164705  ...
%p A164705 T:= (n, k)-> ceil(binomial(2*n-k, n)*2^(k-1)):
%p A164705 seq(seq(T(n, k), k=0..n), n=0..10);  # _Alois P. Heinz_, Apr 06 2025
%t A164705 Table[Table[Binomial[2 n - k, n]*2^(k - 1), {k, 0, n}], {n, 0, 9}] // Grid
%Y A164705 Row sums give A081294.
%Y A164705 Main diagonal gives A011782.
%Y A164705 Cf. A000531, A088218.
%K A164705 nonn,tabl
%O A164705 0,4
%A A164705 _Geoffrey Critzer_, Aug 23 2009
%E A164705 T(0,0)=1 prepended by _Sean A. Irvine_, Apr 05 2025