A258152 Partition array in Abramowitz-Stegun order for the number of ways of putting n stones into a rectangular m X n grid of squares such that each of the m rows contains at least one stone.

1, 1, 4, 1, 18, 27, 1, 32, 36, 288, 256, 1, 50, 200, 750, 1500, 5000, 3125, 1, 72, 450, 400, 1620, 10800, 3375, 17280, 48600, 97200, 46656, 1, 98, 882, 2450, 3087, 30870, 25725, 46305, 48020, 432180, 259308, 420175, 1512630, 2117682, 823543

Offset: 1

Author

Wolfdieter Lang, Jun 17 2015

Keywords

• nonn
• easy
• tabf

Comments

Motivated by A258371 by Adam J.T. Partridge.

The sequence for the row lengths is A000041(n).

The k-th partition of n in Abramowitz-Stegun (A-St) order is denoted by P(n, k) = [1^e(n,k,1), ..., n^e(n,k,n)], with nonnegative exponents summing to m, the number of parts. Here j^0 is not 1, the corresponding P(n, k) list entry is missing, that is, this part j does not appear.

If the k-th partition of n in A-St order has m parts (see A008284) then the irregular triangle entry a(n, k) gives the number of ways of putting n stones into a rectangular m X n grid of squares such that each of the m rows contains at least one stone.

The triangle version of this partition array, with the numbers belonging to the same number of parts m of the partition of n summed, is given in A259051.

Examples

			The irregular triangle a(n, k), with entries belong to the same number of parts m = 1, ..., n enclosed in brackets, begins:
n\k 1   2     3     4     5      6      7 ...
1  [1]
2: [1] [4]
3: [1] [18] [27]
4: [1] [32   36] [288] [256]
5: [1] [50  200] [750  1500] [5000] [3125]
...
n=6: [1] [72  450   400] [1620  10800  3375] [17280, 48600] [97200] [46656],
n=7: [1] [98  882  2450] [3087  30870  25725  46305] [48020 432180  259308] [420175 1512630] [2117682] [823543].
a(4, 3) = 36 because the third partition of 4 is (2^2), and m = 2, hence (2/(2!))*binomial(4,2)^2 = 6^2 = 36. Both of the m=2 rows of n squares are occupied with two stones, hence the binomial(4,2)^2.
		

Crossrefs

Cf. A258371, A259051.

Formula

a(n, k) = (m!/product_{j=1..n} e(n,k,j)!)* product_{j=1..n} binomial(n, j)^e(n,k,j), for n >= 1 and k = 1, 2, ..., A000041(n). Note that for e(n,k,j) = 0 the binomial does not contribute to the product.