Set Partitions with Successions and Separations

Partitions of the set {1,2,…,n} are classified as having successions if a block contains consecutive integers, and separated otherwise. This paper constructs enumeration formulas for such set partitions and some variations using Stirling numbers of the second kind.


Definition 1.2. A partition of [m]
is said to have r successions (r ≥ 0) if it contains r pairs of consecutive integers, where each pair of consecutive integers is counted within one class, and a string of more than two consecutive integers in a class are considered two at a time. The set of k-partitions of [m] with r successions will be denoted by C r (m,k).
Our terminology is consistent with that of [6, page 11] which deals with combinations of elements of [m] with a prescribed number of successions. We define a distinguished subset of C r (m,k).
The next two sections are devoted to statements and proofs of essential results. The last section examines certain generalizations of them.

Statement of results
The first theorem gives the recurrence equation satisfied by the number h t (m,k) of tseparated k-partitions of [m], and its solution.
where the last equality follows from Definition 1.2. This implies The following theorem gives the solution of (2.4).
The following corollary is immediate from Theorem 2.3.
Remark 2.6. Corollary 2.5 suggests the summation of (2.10) over r to obtain the total number B(m) of partitions of [m]. This leads to the recurrence for the Bell numbers as follows: Thus by (2.11) we obtain a natural interpretation of the summands in the Bell recurrence where N denotes the integer part of N, and the last equality in The solution of (2.12) is given in the next theorem. (2.14) (ii) By using (2.14) in (2.6), which may be iterated to The following corollary follows from Theorem 2.8.
A nice generalization of the numbers d r (m,k) is given in Theorem 4.1 below.

Proofs of theorems
We give the proofs of Theorems 2.1, 2.2, 2.3, 2.7, and 2.8. We recall the basic recurrence for the Stirling set numbers s2(m,k) [3, page 245], [7]: has t − 1 elements which must belong to t − 1 different classes, the latter case gives rise to (k − t + 1)h t (m − 1,k) partitions. Thus the result follows: The starting value h t (m,t) = 1 counts the unique partition of [m] into a complete set of residue classes modulo t. The bounds (1,1). Assume that (ii) holds for all positive integers up to m. Then part (i) gives where the second equality is the inductive hypothesis and the third follows from (3.1).
Hence (ii) is true for m + 1, and the proof is complete.
(iii) This follows from (ii) and the definition of the Bell number B(m) [9, page 20]: Proof of Theorem 2.2, that is, (2.4).
Proof of Theorem 2.3, that is, (2.5). We apply induction on m. The following results show that (2.5) holds for m = 1,2: Assume that (2.5) holds for all positive integers up to m. Then Theorem 2.2 gives The second equality is the inductive hypothesis and the fourth follows from ( (ii) We can put the integer m into a class of a p ∈ D r (m − 1,k) which does not contain m − 1 to get a total of (k − 1)(d r (m − 1,k)) partitions.
(iii) Lastly we count the p ∈ D r (m,k) in which m and m − 1 belong to a class. These are obtained by putting m into the class containing m − 1 in any q ∈ D r−1 (m − 1,k) in which m − 1 is not part of a succession. The latter partitions are counted by d r−1 (m − 2,k − 1) + (k − 1)d r−1 (m − 2,k), where r > 0. Adding all the partitions in (i), (ii), and (iii) gives the main result, namely (2.12). For the starting values note that d 0 (m,k) = c 0 (m,k) which is s2(m − 1,k − 1) by Theorem 2.2. The range of r follows from the fact that [m] contains exactly m/2 disjoint 2-subsets of consecutive integers; the range of k is specified as in the proof of Theorem 2.2. The condition on the last inequality follows from the fact that m < 2r ⇒ m − 2r < 0 ⇒ [m] contains less than r distinct pairs of consecutive integers which implies d r (m,m − r) = 0.
Proof of Theorem 2.8, that is, (2.13). We apply induction on m. Since d r (m,k) = c r (m,k) for r = 0,1, it follows from (the proof of) Theorem 2.3 that (2.13) holds for m = 1,2. Assume that (2.13) holds for all positive integers up to m. Then Theorem 2.7 gives The second equality is the inductive hypothesis, the fourth follows from (3.1), and the last follows from the Pascal triangle of binomial coefficients. Hence the proof of Theorem 2.8 follows by mathematical induction.

Some generalizations
Let a partition of [m] be said to have r t-successions (r ≥ 0,t ≥ 1) if it contains exactly r t-strings of consecutive integers, where each t-string of consecutive integers is counted within one class, and a string of more than t consecutive integers in a class are considered t at a time. Denote the set of k-partitions of [m] with r t-successions by C t (m,k,r). As before let c t (m,k,r) = |C t (m,k,r)|. It follows that c r (m,k) = c 2 (m,k,r) and c 0 (m,k) = c 1 (m,k,m). Similarly, we generalize D r (m,k) by letting D t (m,k,r) represent the set of k-partitions of [m] with r t-successions in which every string of consecutive integers appearing in a class has length 1 or t. Thus D r (m,k) = D 2 (m,k,r); d t (m,k,r) = |D t (m,k,r)|. It turns out that there is an easy closed formula for d t (m,k,r) whereas the one for c t (m,k,r) remains inscrutable.
The proofs of the first two parts of the following theorem are analogous to those of Theorems 2.7, 2.8, respectively. The details are omitted.
It follows from Theorem 4.1(ii) that where d 1 (m,k,r) = m r d 1 (m,k,m),d 0 (m,k) = d 1 (m,k,m). Let W t (m,k,r) represent the set of k-partitions of [m] with r t-successions in which every string of consecutive integers appearing in a class has length at most t. Then the set difference E t (m,k,r) = C t (m,k,r) − W t (m,k,r) consists of those partitions in C t (m,k,r) in which at least one class in each partition contains a string of t + 1 consecutive integers, provided that both t and r are greater than 1. It follows that D t (m,k,r) ⊆ W t (m,k,r) ⊆ C t (m,k,r), with D 2 (m,k,r) = W 2 (m,k,r), and W t (m,k,r) = C t (m,k,r) for (t,r) = (1,r), (t,1). As usual let w t (m,k,r) = |W t (m,k,r)|.
We are unable to find a concise formula for w t (m,k,r) (t > 2) (and hence c t (m,k,r)), not even when t = 3 and r = 1. However, we have the following computational result which is established by extending the inclusion-exclusion-type reasoning used in the proof of Theorem 2.2.
Corollary 4.2. w t (m,k,r) satisfies the following recurrence: Proof. There are three ways to find an element of W t (m,k,r).
(1) We can insert the singleton {m} into any p ∈ W t (m − 1,k − 1,r) in w t (m − 1,k − 1,r) possible ways. (2) We can put the integer m into a class of p ∈ W t (m − 1,k,r) under two situations: , then put m into any class of p which does not contain m − 1; (ii) else put m into any class of p. We note that (ii) requires only those p ∈ W t (m − 1,k,r) in which the integer m − 1 is part of a u-succession for 1 ≤ u ≤ t − 2. Thus the total number of partitions in which m − 1 is part of a u-succession: treating the u numbers m − u, m − u + 1,...,m − 1 as an integer N, then a needed p ∈ W t (m − 1,k,r) can be formed by inserting {N } into a q ∈ W t (m − u − 1,k − 1,r) in w t (m − u,k − 1,r) possible ways, or by putting N into any of k − 1 classes of each q ∈ W t (m − u − 1,k,r) which does not contain m − u − 1, which can happen in (k − 1)w t (m − u − 1,k,r) ways. Thus the total number of partitions from (i) is  Adding all the partitions from (1), (2), and (3) gives the desired result: It is clear that w t (m,k,0) = m/(t−1) j=0 w t−1 (m,k, j). This completes the proof of Corollary 4.2.

Remark 4.3.
In the proof of Corollary 4.2, the apparent simplification suggested by enumerating the partitions from 2(i) first (since there are only two summands), and then obtaining those from 2(ii) by complementation, leads to the following three-group contribution instead of (4.5): (4.8) However, (4.8) will not always give correct results because it is inconsistent with the cumulative "origin" w t (m,k,0). This is easily verified by a few actual computations.
Corollary 4.2 is a special case of the following result.
Theorem 4.4. c t (m,k,r) satisfies the following recurrence: Before sketching the proof of Theorem 4.4 we state the special case of t = 3.         For starting values, we have c t (m,k,0) = m/(t−1) j=0 w t−1 (m,k, j) since a partition without a t-succession has successions of length at most t − 1. In particular if t = 3, then c 3 (m,k,0)= m/2 j=0 w 2 (m,k, j), which equals m/2 j=0 m− j j s2(m−1 − j,k − 1), by Theorem 2.3. This completes the proof of Theorem 4.4.