cp's OEIS Frontend

Let F be a family of nonempty sets on an n-element set A with |F| > 1 such that every pair of distinct elements of F have the same nonempty intersection and there are no two distinct elements of F such that one is a subset of the other. Then a(n) = the total number of such families.

Proof: Let binomial(n,k) denote the binomial coefficient (the number of ways to choose k elements from an n-element set) and StirlingS2(n,k) the Stirling numbers of the second kind (the number of ways to partition an n-element set A into k nonempty parts, the union of which is A). As is well known, StirlingS2(n+1,k+1) = StirlingS2(n,k) + k*StirlingS2(n,k+1), where we assume StirlingS2(0,0) = 1, StirlingS2(n,0) = StirlingS2(0,n) = 0, and StirlingS2(n,k) = 0 when n < k, for n > 0.

Enumerate the elements of F in the following manner. Begin by partitioning the elements of A into either 1) k or 2) k+1 parts. For case 1, there are StirlingS2(n,k) possible partitions. From such a partition, select k-1 of the k parts to assign to the elements of F (where k >= 3). The remaining part constitutes the nonempty intersection. So this enumeration can be accomplished in binomial(k,k-1)*binomial(1,1)*StirlingS2(n,k) = k*StirlingS2(n,k) ways. Here we have that the size of the union of the elements of F equals |A|.

For case 2, there are StirlingS2(n,k+1) partitions. From such a partition, select k-1 of the k+1 parts to assign to the elements of F (where, again, k >= 3). Then select 1 of the 2 remaining parts to constitute the nonempty intersection. So this enumeration can be accomplished in binomial(k+1,k-1)*binomial(2,1)*StirlingS2(n,k+1) = k*(k+1)*StirlingS2(n,k+1) ways. Here we have that the size of the union of the elements of F is less than |A|. So the 2 cases cover both possibilities, i.e., the union of the elements of F is either equal to |A| or less than |A|.

Multiplying the above recurrence by k, we have k*StirlingS2(n+1,k+1) = k*StirlingS2(n,k) + k*(k+1)*StirlingS2(n,k+1), and the claim follows by summing over this for 3 <= k <= n. (Observe that n >= 3 as for n = 1, say [n] = {1}, there is only 1 subset of [n], and for n = 2, say [n] = {1,2}, the subsets of n are {},{1},{2},{1,2}, so that there are no pairs here that have a nonempty intersection and for which neither is a subset of the other. By similar reasoning, k >= 3, as we need at least 2 distinct sets in F, and we need at least 1 element of A not in either of these sets to add to them to create their common nonempty intersection.)

The families F counted here are very close in definition to sunflowers = delta-systems.

The families F counted here could be described perhaps more clearly as intersecting Sperner families such that every pair of distinct elements of F have the same nonempty intersection and |F| > 1.

Columns 0,1,2,3 respectively correspond to relations R_3, R_4, R_0, R_1 defined in La Haye paper listed below.

T(n,3): continued fraction expansion of e - 1.

T(n,1): continued fraction expansion of coth(1).

T(n,2): continued fraction expansion of tan(1) = cot(pi/2 - 1).