A238016 Number A(n,k) of partitions of n^k into parts that are at most n; square array A(n,k), n>=0, k>=0, read by antidiagonals.

0, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 1, 5, 12, 5, 1, 1, 1, 9, 75, 64, 7, 1, 1, 1, 17, 588, 2280, 377, 11, 1, 1, 1, 33, 5043, 123464, 106852, 2432, 15, 1, 1, 1, 65, 44652, 7566280, 55567352, 6889527, 16475, 22, 1

Offset: 0

Author

Alois P. Heinz, Feb 17 2014

Keywords

• nonn
• tabl

Comments

In general, for k>3, is column k asymptotic to exp(2*n) * n^((k-2)*n-k) / (2*Pi). For k=1 see A000041, for k=2 see A206226 and for k=3 see A238608. - Vaclav Kotesovec, May 25 2015

Conjecture: If f(n) >= O(n^4) then "number of partitions of f(n) into parts that are at most n" is asymptotic to f(n)^(n-1) / (n!*(n-1)!). See also A237998, A238000, A236810 or A258668-A258672. - Vaclav Kotesovec, Jun 07 2015

Examples

			A(3,1) = 3: 3, 21, 111.
A(3,2) = 12: 333, 3222, 3321, 22221, 32211, 33111, 222111, 321111, 2211111, 3111111, 21111111, 111111111.
A(2,3) = 5: 2222, 22211, 221111, 2111111, 11111111.
A(2,4) = 9: 22222222, 222222211, 2222221111, 22222111111, 222211111111, 2221111111111, 22111111111111, 211111111111111, 1111111111111111.
Square array A(n,k) begins:
  0, 1,   1,      1,        1,           1, ...
  1, 1,   1,      1,        1,           1, ...
  1, 2,   3,      5,        9,          17, ...
  1, 3,  12,     75,      588,        5043, ...
  1, 5,  64,   2280,   123464,     7566280, ...
  1, 7, 377, 106852, 55567352, 33432635477, ...
		

Links

• Alois P. Heinz, Antidiagonals n = 0..54, flattened
• A. V. Sills and D. Zeilberger, Formulae for the number of partitions of n into at most m parts (using the quasi-polynomial ansatz) arXiv:1108.4391 [math.CO], 2011.

Crossrefs

Columns k=0-10 give: A057427, A000041, A206226, A238608, A238609, A238610, A238611, A238612, A238613, A238614, A238615.

Rows n=0-10 give: A057427, A000012, A094373, A238630, A238631, A238632, A238633, A238634, A238635, A238636, A238637.

Main diagonal gives A238000.

Cf. A238010.

Programs

Mathematica

A[n_, k_] := SeriesCoefficient[Product[1/(1-x^j), {j, 1, n}], {x, 0, n^k}]; A[0, 0] = 0; Table[A[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Oct 11 2015 *)

Formula

A(n,k) = [x^(n^k)] Product_{j=1..n} 1/(1-x^j).