Permutation Patterns and Continued Fractions

We find, in the form of a continued fraction, the generating function for the number of (132)-avoiding permutations that have a given number of (123) patterns, and show how to extend this to permutations that have exactly one (132) pattern. We also find some properties of the continued fraction, which is similar to, though more general than, those that were studied by Ramanujan. the electronic journal of combinatorics 6 (1999), #Rxx 2 A (132) pattern (resp. a (123) pattern) in a permutation π of |π| letters is a triple 1 ≤ i < j < k ≤ n of indices for which π(i) < π(k) < π(j) (resp. π(i) < π(j) < π(k)). Let fr(n) denote the number of permutations π of n letters that have no (132) patterns and exactly r (123) patterns. Our main result is the following. Theorem 1 The generating function for the {fr(n)} is ∑ r,n≥0 fr(n)z q = 1 1− z 1− z 1− zq 1− zq 3 1− zq 6 . . . (1) in which the nth numerator is zq( n−1 2 ). We think it is remarkable that such a continued fraction encodes information about (132)-avoiding permutations. We will first prove the theorem, and then study some consequences and generalizations. 1 The patterns Let the weight of a permutation π of |π| letters be z|π|q|123(π)|t|12(π)|, in which |123(π)| is the number of (123) patterns (rising triples) in π, and |12(π)| is the number of rising pairs in π. Let P (q, z, t) = ∑′ π weight(π), (2) where the sum extends over all (132)-avoiding permutations π. If π is a (132)-avoiding permutation on {1, 2, . . . , n}, (n > 0) and the largest element, n, is at the kth position, i.e., π(k) = n, then by letting π1 := {π(i)}k−1 1 and π2 := {π(i)}k+1, we have that every element in π1 must be larger than every element of π2, or else a (132) would be formed, with the n serving as the ‘3’ of the (132). Hence, π1 is a permutation of the set {n − k + 1, . . . , n − 1}, and π2 is a permutation of the set {1, . . . , n−k}. Furthermore, π1 and π2 are each (132)-avoiding. Conversely, if π1 and π2 are (132)-avoiding permutations on {n−k+1, . . . , n−1} and {1, . . . , n−k} respectively (for some k, 1 ≤ k ≤ n), then (π1nπ2) is a nonempty (132)-avoiding permutation. Thus we have |123(π)| = |123(π1)|+ |123(π2)|+ |12(π1)|, since a (123) pattern in π = def (π1nπ2) may either be totally immersed in the π1 part, or wholly immersed in the π2 part, or may be due to the n serving as the ‘3’ of the (123), the number of which is the number of (12) patterns in π1. the electronic journal of combinatorics 6 (1999), #Rxx 3 We also have |12(π)| = |12(π1)|+ |12(π2)|+ |π1|, and, of course |π| = |π1|+ |π2|+ 1. Hence, weight(π)(q, z, t) := q|123(π)|z|π|t|12(π)| = q|123(π1)|+|123(π2)|+|12(π1)|z|π1|+|π2|+1t|12(π1)|+|12(π2)|+|π1| = zq|123(π1)|(qt)|12(π1)|(zt)|π1|q|123(π2)|t|12(π2)|z|π2| = zweight(π1)(q, zt, tq)weight(π2)(q, z, t). Now sum over all possible (132)-avoiding permutations π, to get the functional equation P (q, z, t) = 1 + zP (q, zt, tq)P (q, z, t), (3) in which the 1 corresponds to the empty permutation. Next let Q(q, z, t) be the sum of all the weights of all permutations with exactly one (132) pattern. By adapting the argument from Miklós Bóna’s paper [?] we easily see that Q(q, z, t) satisfies Q(q, z, t) = zP (q, zt, qt)Q(q, z, t) + zQ(q, zt, qt)P (q, z, t) + tzP (q, zt, qt)(P (q, z, t)− 1). (4) This holds since our sole (132) pattern can either appear in the elements (a) before n, (b) after n, or (c) with n as the ‘3’ in the (132) pattern. The term zP (q, zt, qt)Q(q, z, t) corresponds to (a), zQ(q, zt, qt)P (q, z, t) corresponds to (b), and tzP (q, zt, qt)(P (q, z, t) − 1) corresponds to (c). We see that case (c) follows since π = (π1, n − k, n, π2), where π1 is a permutation of [n − k + 2, . . . , n − 1], π2 is a permutation of [1, . . . , n− k − 1] ∪ {n− k + 1}, and k 6= n.