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.

1, 1, 1, 3, 3, 2, 10, 10, 8, 4, 35, 35, 30, 20, 8, 126, 126, 112, 84, 48, 16, 462, 462, 420, 336, 224, 112, 32, 1716, 1716, 1584, 1320, 960, 576, 256, 64, 6435, 6435, 6006, 5148, 3960, 2640, 1440, 576, 128, 24310, 24310, 22880, 20020, 16016, 11440, 7040, 3520, 1280, 256

Offset

0,4

Comments

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).

Links

Table of n, a(n) for n=0..54.

Sean A. Irvine, Computing A382782, 2025.

Formula

Sum_{k=0..n} k * T(n,k) = A000531(n). - Alois P. Heinz, Apr 06 2025

Example

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}.
Triangle begins
   1;
   1,   1;
   3,   3,   2;
  10,  10,   8,  4;
  35,  35,  30, 20,  8;
 126, 126, 112, 84, 48, 16;
 ...

Maple

T:= (n, k)-> ceil(binomial(2*n-k, n)*2^(k-1)):
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Apr 06 2025

Mathematica

Table[Table[Binomial[2 n - k, n]*2^(k - 1), {k, 0, n}], {n, 0, 9}] // Grid

Crossrefs

Row sums give A081294.

Main diagonal gives A011782.

Cf. A000531, A088218.

Sequence in context: A250304 A330307 A256916 * A073754 A193229 A112458

Adjacent sequences: A164702 A164703 A164704 * A164706 A164707 A164708

Keyword

nonn,tabl

Author

Geoffrey Critzer, Aug 23 2009

Extensions

T(0,0)=1 prepended by Sean A. Irvine, Apr 05 2025

Status

approved