A371568 Array read by ascending antidiagonals: A(n, k) is the number of paths of length k in Z^n from the origin to points such that x1+x2+...+xn = k with x1,...,xn > 0.

1, 0, 1, 0, 2, 1, 0, 0, 6, 1, 0, 0, 6, 14, 1, 0, 0, 0, 36, 30, 1, 0, 0, 0, 24, 150, 62, 1, 0, 0, 0, 0, 240, 540, 126, 1, 0, 0, 0, 0, 120, 1560, 1806, 254, 1, 0, 0, 0, 0, 1800, 8400, 5796, 510, 1

Offset: 1

Author

Shel Kaphan, Mar 28 2024

Keywords

• nonn
• tabl

Comments

T(n, k) can also be seen as the number of ordered partitions of k items into n nonempty buckets.

T(n, n) = n!, which is readily seen because to go from the origin to a point in Z^n a distance n away, with at least one step taken in each dimension, the first step can be in any of n dimensions, the second step in any of n-1 dimensions, and so on.

This array is the image of Pascal's triangle A007318 under the Akiyama-Tanigawa transformation. See the Python program. - Peter Luschny, Apr 19 2024

Examples

			 n\k 1 2 3  4   5    6     7      8       9       10
  --------------------------------------------------
 1|  1 1 1  1   1    1     1      1       1        1
 2|  0 2 6 14  30   62   126    254     510     1022
 3|  0 0 6 36 150  540  1806   5796   18150    55980
 4|  0 0 0 24 240 1560  8400  40824  186480   818520
 5|  0 0 0  0 120 1800 16800 126000  834120  5103000
 6|  0 0 0  0   0  720 15120 191520 1905120 16435440
 7|  0 0 0  0   0    0  5040 141120 2328480 29635200
 8|  0 0 0  0   0    0     0  40320 1451520 30240000
 9|  0 0 0  0   0    0     0      0  362880 16329600
10|  0 0 0  0   0    0     0      0       0  3628800
		

Crossrefs

Cf. A000918 (n=2), A001117 (n=3), A000919 (n=4), A001118 (n=5), A000920 (n=6).

Cf. A135456 (n=7), A133068 (n=8), A133360 (n=9), A133132 (n=10).

Cf. A371064, A007318.

See A019538 and A131689 for other versions.

Programs

Mathematica

A[n_,k_] := Sum[(-1)^(n-i) * i^k * Binomial[n,i], {i,1,n}]

Python

# The Akiyama-Tanigawa algorithm for the binomial generates the rows.
# Adds row(0) = 0^k and column(0) = 0^n.
from math import comb as binomial
def ATBinomial(n, len):
    A = [0] * len
    R = [0] * len
    for k in range(len):
        R[k] = binomial(k, n)
        for j in range(k, 0, -1):
            R[j - 1] = j * (R[j] - R[j - 1])
        A[k] = R[0]
    return A
for n in range(11): print([n], ATBinomial(n, 11))  # Peter Luschny, Apr 19 2024

Formula

A(n,k) = Sum_{i=1..n} (-1)^(n-i) * binomial(n,i) * i^k