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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 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.)