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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. 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

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.

LINKS

Table of n, a(n) for n=1..44.

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.

EXAMPLE

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.

Sequence in context: A113355 A201201 A077102 * A259051 A192722 A300141

Adjacent sequences:  A258149 A258150 A258151 * A258153 A258154 A258155

KEYWORD

nonn,easy,tabf

AUTHOR

Wolfdieter Lang, Jun 17 2015

STATUS

approved

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Last modified October 16 06:34 EDT 2019. Contains 328051 sequences. (Running on oeis4.)