A258371 -id:A258371 - cp's OEIS frontend

A259051 Triangle T(n,m) for the number of ways to put n stones into an m X n square grid such that each of the m rows contains at least one stone.

1, 1, 4, 1, 18, 27, 1, 68, 288, 256, 1, 250, 2250, 5000, 3125, 1, 922, 15795, 65880, 97200, 46656, 1, 3430, 105987, 739508, 1932805, 2117682, 823543, 1, 12868, 696864, 7653632, 31539200, 59179008, 51380224, 16777216, 1, 48618, 4540968, 75687696, 461828790, 1320099444, 1919564892, 1377495072, 387420489

Offset: 1

Wolfdieter Lang, Jun 18 2015

This is the triangle A258371(n, m)/binomial(n, m).

For the corresponding partition array see A258152.

			The triangle T(n, m) begins:
n\k 1     2       3        4         5         6           7
1:  1
2:  1     4
3:  1    18      27
4:  1    68     288      256
5:  1   250    2250     5000      3125
6:  1   922   15795    65880     97200      46656
7:  1  3430  105987   739508   1932805    2117682     823543
...
8:  1 12868  696864  7653632  31539200 59179008 51380224
16777216,
9:  1 48618 4540968 75687696 461828790 1320099444 1919564892 1377495072 387420489.
a(4, 2) = 68 from the sum 32 + 36 of the n=4 row of A258152 which belong to the partitions of 4 with m=2 parts, namely (1, 3) and (2, 2).

Giovanni Resta, Table of n, a(n) for n = 1..1830 (first 60 rows)

Cf. A258152, A258371.

T[n_, k_]:= Sum[Multinomial@@ (Last/@ Tally[e]) * Times@@ Binomial[n, e], {e, IntegerPartitions[n, {k}]}]; Flatten@ Table[ T[n, k], {n, 9}, {k, n}] (* Giovanni Resta, Jun 18 2015 *)

T(n, m) = sum over the A258152(n, k) entries corresponding to partitions of n with m parts; n >= 1, m = 1,2, ..., n.

T(n, m) = A258371(n, m)/binomial(n, m).

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.

Original entry on oeis.org

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

Wolfdieter Lang, Jun 17 2015

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.

			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.

Cf. A258371, A259051.

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.

cp's OEIS Frontend

A259051 Triangle T(n,m) for the number of ways to put n stones into an m X n square grid such that each of the m rows contains at least one stone.

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