A102316 - OEIS

A102316

Triangle, read by rows, where T(n,k) = T(n,k-1) + (k+1)*T(n-1,k) for n>k>0, T(n,0)=1 and T(n,n) = T(n,n-1) for n>=0.

1, 1, 1, 1, 3, 3, 1, 7, 16, 16, 1, 15, 63, 127, 127, 1, 31, 220, 728, 1363, 1363, 1, 63, 723, 3635, 10450, 18628, 18628, 1, 127, 2296, 16836, 69086, 180854, 311250, 311250, 1, 255, 7143, 74487, 419917, 1505041, 3683791, 6173791, 6173791, 1, 511, 21940

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OFFSET

0,5

COMMENTS

Main diagonal is A082161 (with offset). Row sums give A102317.

T(n,k) = number of column-marked subdiagonal paths of steps east (1,0) and north (0,1) from the origin to (n,k). Subdiagonal means that the path never rises above the diagonal line y=x and column-marked means that for 1 <= i <= n, one unit square directly below the i-th east step and above the line y=-1 is marked. - David Callan, Feb 04 2006

LINKS

Table of n, a(n) for n=0..47.

FORMULA

T(n, k) = Sum_{j=0..k} (j+1)*T(n-1, j) for n>k>0, T(n, 0)=1 for n>=0. T(n, n) = A082161(n) for n>0; A082161(n+1) = Sum_{k=0..n} (k+1)*T(n, k).

EXAMPLE

T(5,2) = 220 = 1*1 + 2*15 + 3*63 = 1*T(4,0) + 2*T(4,1) + 3*T(4,2).

T(5,2) = 220 = 31 + 3*63 = T(5,1) + (2+1)*T(4,2).

T(5,3) = 728 = 220 + 4*127 = T(5,2) + (3+1)*T(4,3).

Rows begin:

[1],

[1,1],

[1,3,3],