A331160 Array read by antidiagonals: A(n,k) is the number of nonnegative integer matrices with k distinct columns and any number of distinct nonzero rows with column sums n and rows in decreasing lexicographic order.

1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 4, 2, 1, 0, 1, 27, 15, 2, 1, 0, 1, 266, 317, 44, 3, 1, 0, 1, 3599, 12586, 2763, 120, 4, 1, 0, 1, 62941, 803764, 390399, 21006, 319, 5, 1, 0, 1, 1372117, 75603729, 103678954, 10074052, 147296, 804, 6, 1

Offset: 0

Author

Andrew Howroyd, Jan 10 2020

Keywords

• nonn
• tabl

Comments

The condition that the rows be in decreasing order is equivalent to considering nonequivalent matrices with distinct rows up to permutation of rows.

Examples

			Array begins:
===================================================================
n\k | 0 1   2      3          4              5                6
----+--------------------------------------------------------------
  0 | 1 1   0      0          0              0                0 ...
  1 | 1 1   1      1          1              1                1 ...
  2 | 1 1   4     27        266           3599            62941 ...
  3 | 1 2  15    317      12586         803764         75603729 ...
  4 | 1 2  44   2763     390399      103678954      46278915417 ...
  5 | 1 3 120  21006   10074052    10679934500   21806685647346 ...
  6 | 1 4 319 147296  232165926   956594630508 8717423133548684 ...
  7 | 1 5 804 967829 4903530137 76812482919237 ...
      ...
The A(2,2) = 4 matrices are:
   [2 1]   [2 0]   [1 2]   [1 1]
   [0 1]   [0 2]   [1 0]   [1 0]
                           [0 1]
		

Links

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

Crossrefs

Rows n=1..3 are A000012, A331316, A331344

Columns k=0..2 are A000012, A000009, A331317.

Cf. A219585, A331039, A331126, A331161, A331315, A331318.

Programs

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]]++); binomial(EulerT(v)[n], k)*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, k<=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)}

Formula

A(n, k) = Sum_{j=0..k} Stirling1(k, j)*A219585(n, j).

A331318(n) = Sum_{d|n} A(n/d, d).