Sequences A091057 through A091062 count equivalence classes of matrices or arrays. The matrices have a given number of rows: r, columns: c, and each cell in the matrix is filled with a value from a symbol set which has z elements. Two matrices are equivalent if one can be transformed to the other through any combination of permutations of the rows, columns or the symbol set. For example: consider the following 3×3 matrix with symbol set {1, 2, ..., 9}:

1	2	3
4	5	6
7	8	9

This matrix is equivalent to:

4	5	6
1	2	3
7	8	9

by permuting the first two rows. It's also equivalent to:

2	1	3
5	4	6
8	7	9

by permuting the first two columns. It's also equivalent to:

1	2	5
4	3	6
7	8	9

by permuting `3' and `5' in the symbol set. It is not equivalent to:

1	2	3
4	5	6
7	8	8

since `8' appears twice and no symbol appears twice in the original.
While there are 9⁹ = 387420489 different matrices there are only 777 equivalence classes as can be seen in A091057. There are 9 equivalence classes for a 2×2 with 4 symbols available. Those classes are:

1 1 1 1

1 1 1 2

1 1 2 2

1 2 1 2

1 2 2 1

1 1 2 3

1 2 1 3

1 2 3 1

1 2 3 4

To calculate the number of equivalence classes of such matrices, let S_r, S_c, and S_z be the sets of permutations of the rows, columns and symbols respectively. Each permutation p has a type (p₁, p₂, p₃, ...) where p₁ is the number of 1-cycles in the permutation, p₂ is the number of 2-cycles, etc. For example, the permutation (1234)(67)(89) in S₉ has type (1,2,0,1) for one 1-cycle (5), two 2-cycles (67) and (89), no 3-cycles and one 4-cycle (1234). The number of types of permutations in S_n is the number of partitions of n. (See A000041.) Let s, t and u be permutations from S_r, S_c, and S_z respectively. The value:
fix A[s₁, s₂, s₃, ...; t₁, t₂, t₃, ...; u₁, u₂, u₃, ...] = prod_i,j>=1 (sum_d|lcm(i,j) du_d)^{gcd(i,j)s_it_j}
is the number of matrices fixed or unmodified by a permutation of the rows of type (s₁, s₂, s₃, ...), a permutation of the columns of type (t₁, t₂, t₃, ...) and a permutation of the symbol set of type (u₁, u₂, u₃, ...). To calculate the number of equivalence classes, we must sum through each triplet of permutation types as follows:
a_r,c,z = sum_{1s1+2s2+...=r, 1t1+2t2+...=c, 1u1+2u2+...=z} fix A[s₁, s₂, ...; t₁, t₂, ...; u₁, u₂, ...] / (1^s₁s₁!2^s₂s₂!... 1^t₁t₁!2^t₂t₂!... 1^u₁u₁!2^u₂u₂!...)