A219585 Number A(n,k) of k-partite partitions of {n}^k into distinct k-tuples; square array A(n,k), n>=0, k>=0, read by antidiagonals.

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 5, 2, 1, 1, 15, 40, 17, 2, 1, 1, 52, 457, 364, 46, 3, 1, 1, 203, 6995, 14595, 2897, 123, 4, 1, 1, 877, 136771, 937776, 407287, 21369, 323, 5, 1, 1, 4140, 3299218, 88507276, 107652681, 10200931, 148257, 809, 6, 1

Offset: 0

Author

Alois P. Heinz, Nov 23 2012

Keywords

• nonn
• tabl

Comments

A(n,k) is the number of factorizations of m^n into distinct factors where m is a product of k distinct primes. A(2,2) = 5: (2*3)^2 = 36 has 5 factorizations into distinct factors: 36, 3*12, 4*9, 2*18, 2*3*6.

Examples

			A(1,3) = 5: [(1,1,1)], [(1,1,0),(0,0,1)], [(1,0,1),(0,1,0)], [(1,0,0),(0,1,0),(0,0,1)], [(0,1,1),(1,0,0)].
A(3,2) = 17: [(3,3)], [(3,0),(0,3)], [(3,2),(0,1)], [(2,3),(1,0)], [(3,1),(0,2)], [(2,2),(1,1)], [(1,3),(2,0)], [(2,1),(1,2)], [(2,1),(1,1),(0,1)], [(3,0),(0,2),(0,1)], [(2,2),(1,0),(0,1)], [(2,1),(0,2),(1,0)], [(1,2),(2,0),(0,1)], [(1,2),(1,1),(1,0)], [(0,3),(2,0),(1,0)], [(2,0),(1,1),(0,2)], [(2,0),(0,2),(1,0),(0,1)].
Square array A(n,k) begins:
  1,  1,   1,      1,          1,            1,         1, ...
  1,  1,   2,      5,         15,           52,       203, ...
  1,  1,   5,     40,        457,         6995,    136771, ...
  1,  2,  17,    364,      14595,       937776,  88507276, ...
  1,  2,  46,   2897,     407287,    107652681,  ...
  1,  3, 123,  21369,   10200931,  10781201973,  ...
  1,  4, 323, 148257,  233051939,  ...
  1,  5, 809, 970246, 4909342744,  ...
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..209

Crossrefs

Columns k=0..5 give: A000012, A000009, A219554, A219560, A219561, A219565.

Rows n=0..3 give: A000012, A000110, A094574, A319591.

Cf. A188445, A219727 (partitions of {n}^k into k-tuples), A318286.

Programs

Mathematica

f[n_, k_] := f[n, k] = 1/2 Product[Sum[O[x[j]]^(n+1), {j, 1, k}]+1+ Product[x[j]^i[j], {j, 1, k}], Evaluate[Sequence @@ Table[{i[j], 0, n}, {j, 1, k}]]];
a[0, ] = a[, 0] = 1; a[n_, k_] := SeriesCoefficient[f[n, k], Sequence @@ Table[{x[j], 0, n}, {j, 1, k}]];
Table[Print[a[n-k, k]]; a[n-k, k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 11 2013, updated Sep 16 2019 *)

PARI

EulerT(v)={Vec(exp(x*Ser(dirmul(v,vector(#v,n,1/n))))-1, -#v)}
D(p, n, k)={my(v=vector(n)); for(i=1, #p, v[p[i]]++); EulerT(v)[n]^k/prod(i=1, #v, i^v[i]*v[i]!)}
T(n, k)={my(m=n*k+1, q=Vec(exp(intformal(O(x^m) - x^n/(1-x)))/(1+x))); if(n==0, 1, (-1)^m*sum(j=0, m, my(s=0); forpart(p=j, s+=(-1)^#p*D(p, n, k), [1, n]); s*q[#q-j])/2)} \\ Andrew Howroyd, Dec 16 2018

Formula

A(n,k) = [(Product_{j=1..k} x_j)^n] 1/2 * Product_{i_1,...,i_k>=0} (1+Product_{j=1..k} x_j^i_j).