6 D ec 1 99 9 RESTRICTED PERMUTATIONS , CONTINUED FRACTIONS , AND CHEBYSHEV POLYNOMIALS

Horse paths, restricted 132-avoiding permutations, continued fractions, and Chebyshev polynomials

Several authors have examined connections among 132-avoiding permutations, continued fractions, and Chebyshev polynomials of the second kind. In this paper we find analogues for some of these results for permutations π avoiding 132 and 1223 (there is no occurrence πi < πj < πj+1 such that 1 ≤ i ≤ j − 2) and provide a combinatorial interpretation for such permutations in terms of lattice paths. ...

O ct 2 00 6 RESTRICTED MOTZKIN PERMUTATIONS , MOTZKIN PATHS , CONTINUED FRACTIONS , AND CHEBYSHEV POLYNOMIALS

We say that a permutation π is a Motzkin permutation if it avoids 132 and there do not exist a < b such that π a < π b < π b+1. We study the distribution of several statistics in Motzkin permutations, including the length of the longest increasing and decreasing subsequences and the number of rises and descents. We also enumerate Motzkin permutations with additional restrictions, and study the ...

0 Restricted Permutations , Continued Fractions , and Chebyshev Polynomials

RESTRICTED 132-AVOIDING k-ARY WORDS, CHEBYSHEV POLYNOMIALS, AND CONTINUED FRACTIONS

We study generating functions for the number of n-long k-ary words that avoid both 132 and an arbitrary `-ary pattern. In several interesting cases the generating function depends only on ` and is expressed via Chebyshev polynomials of the second kind and continued fractions. 1. Extended abstract 1.1. Permutations. Let Sn denote the set of permutations of [n] = {1, 2, . . . , n}, written in one...

Restricted Permutations, Continued Fractions, and Chebyshev Polynomials

Let fr n(k) be the number of 132-avoiding permutations on n letters that contain exactly r occurrences of 12 . . . k, and let Fr(x; k) and F (x, y; k) be the generating functions defined by Fr(x; k) = P n>0 f r n(k)x n and F (x, y; k) = P r>0 Fr(x; k)y r. We find an explicit expression for F (x, y; k) in the form of a continued fraction. This allows us to express Fr(x; k) for 1 6 r 6 k via Cheb...