One way: recursion over the flattened length of the table, i.e., to produce sequences of length *r**c*.

*Base case: e*(1,*n*), the contingency sequence with 1 element, contains just the grand total,*n*.*Step case:*to compute e(*l*,*n*) for*l*> 1, you need- 0 conjoined with each sequence produced by
*e*(*l*-1,*n*), i.e., all tables one size smaller with the originally desired grand total - 1 conjoined with each
*e*(*l*-1,*n*-1) - 2 conjoined with each
*e*(*l*-1,*n*-2) - …
*n*-1 conjoined with each*e*(*l*-1,1)*n*conjoined with each*e*(*l*-1,0)

- 0 conjoined with each sequence produced by

The number of 2×2 contingency tables seems to be given by the triangular pyramidal numbers, 1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, 455, 560, 680, 816, 969, β¦ which is a sum of triangular numbers.

There’s a bunch of work on the geometry of contingency tables.Β Fun stuff!Β (Ended up playing with this to generate stimuli for an experiment.)