OFFSET
1,5
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
FORMULA
A(n,k) = Sum_{i=1..n} (-1)^(n-i) * binomial(n,i) * i^k
EXAMPLE
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
MATHEMATICA
A[n_, k_] := Sum[(-1)^(n-i) * i^k * Binomial[n, i], {i, 1, n}]
PROG
(Python)
# The Akiyama-Tanigawa algorithm for the binomial generates the rows.
# Adds row(0) = 0^k and column(0) = 0^n.
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
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Shel Kaphan, Mar 28 2024
STATUS
approved