Báo cáo toán học: "Asymptotics of Permutations with Nearly Periodic Patterns of Rises and Falls" ppsx

Asymptotics of Permutations with Nearly Periodic Patterns of Rises and Falls Edward A. Bender Department of Mathematics University of California, San Diego La Jolla, CA 92093-0112 ebender@ucsd.edu William J. Helton ∗ Department of Mathematics University of California, San Diego La Jolla, CA 92093-0112 helton@ucsd.edu L. Bruce Richmond Department of Combinatorics and Optimization University of Waterloo Waterloo, Ontario CANADA N2L 3G1 lbrichmond@uwaterloo.ca MR Subject Classifications: 05A16, 45C05, 60C05 Submitted: May 20, 2003; Accepted: Oct 1, 2003; Published: Oct 23, 2003 Abstract Ehrenborg obtained asymptotic results for nearly alternating permutations and conjectured an asymptotic formula for the number of permutations that have a nearly periodic run pattern. We prove a generalization of this conjecture, rederive the fact that the asymptotic number of permutations with a periodic run pattern has the form Cr −n n!, and show how to compute the various constants. A reformulation in terms of iid random variables leads to an eigenvalue problem for a Fredholm integral equation. Tools from functional analysis establish the necessary properties. ∗ Partially supported by the NSF and the Ford Motor Company. the electronic journal of combinatorics 10 (2003), #R40 1 1 Introduction Definition 1 (words) A word is a sequence of symbols. If v and w are words, then vw is the concatenation and w k is the concatenation of k copies of w. The length |w| of w is the number of symbols in the sequence. The descent word of a sequence σ 1 , ,σ n of numbers is α = a 1 ···a n−1 ∈{d, u} n−1 where a i = d if σ i >σ i+1 and a i = u otherwise. If a permutation has descent word α, then its run word is a sequence L of positive integers where L i is the length of the ith run in α. The size L of a run word L is the sum of its parts plus 1. Thus, its size is one more than the length of the corresponding descent word. In other words, it is the size of the set being permuted. Let Run(N) be the number of permutations that begin with an ascent and have run word N. For example, the descent and run words of the permutation 3, 2, 7, 5, 1, 4, 6aredudduu and 1122, respectively, and 1122 = 7. Note that each run word corresponds to two descent words: just interchange the roles of d and u. Thus the total number of permutations with run word N is 2 Run(N). We prove the following generalization of Ehrenborg’s Conjecture 7.1 [3]. Theorem 1 Let L 0 , ,L k be (possibly empty) run words and let M 1 , ,M k be nonempty run words. There are nonzero constants B 0 , ,B k such that Run(L 0 M a 1 1 L 1 M a 2 2 ···M a k k L k ) L 0 M a 1 1 L 1 M a 2 2 ···M a k k L k ! ∼ B 0 ···B k Run(M a 1 1 M a k k ) M a 1 1 M a k k ! as min(a 1 , a k ) →∞.TheB i are given by B i =                        lim n→∞ Run(L 0 M n 1 ) M n 1 ! Run(M n 1 ) L 0 M n 1 ! , if i =0, lim n→∞ Run(M n i L i M n i+1 ) M n i M n i+1 ! Run(M n i M n i+1 ) M n i L i M n i+1 ! , if 0 <i<k, lim n→∞ Run(M n k L k ) M n k ! Run(M n k ) M n k L k ! , if i = k. We also rederive Theorem 2 [6]ForarunpatternL there are constants C(L) and λ(L) such that the fraction of permutations with run pattern L n is asymptotic to C(L) λ(L) n . Since L n −1=n(L−1), the theorem can be rewritten Run(L n ) ∼ C ∗ (L)(λ ∗ (L)) L n  L n !, (1.1) the electronic journal of combinatorics 10 (2003), #R40 2 where λ ∗ = λ 1/(L−1) and C ∗ = C/λ ∗ . When L = 1, Run(L n ) counts alternating permutations of size n + 1 and so we obtain the Euler numbers: 1 Run(1 n )=E n+1 ∼ 2(2/π) n+2 (n +1)!. Thus C(1) = 8/π 2 and λ(1) = 2/π (1.2) in Theorem 2. Permutations with L = t1 (in other notation, u t d) were studied by Leeming and MacLeod [5]. They proved that (Run(L n−1 t)) 1/n(t+1) ∼ n(t+1) e |p t+1 | ,wherep  is the zero of  ∞ n=0 z n /(n)! of smallest modulus. It follows that λ(L) ∼ (Run(L n )) 1/n ∼ ((n(t +1)+1)!) 1/n  n(t +1) e |p t+1 |  t+1 and so λ(t1) = |p t+1 | −(t+1) . (1.3) Lemma 2 and Theorem 4 in the next section provide the tools for calculating all the con- stants in Theorems 1 and 2. In Section 3, we illustrate by rederiving (1.3) and computing the associated C(t1) in (3.6). In giving proofs, we find it more convenient to work with descent words and then translate those results into run-word terms. Our proofs are based on the probabilistic approach of Ehrenborg, Levin and Readdy [4]. This leads us to the study of a Fredholm integral equation. In the next section, we introduce the probabilistic approach and state the relevant probabilistic theorems. In Sections 4–9, we prove the various theorems. 2 A Probabilistic Formulation Definition 2 (ends of descent words) The lengths of the longest initial and final constant strings in a descent word α are denoted by A(α) and Z(α), respectively. These are the initial and final integers in the run word corresponding to α. We now define the probability distributions and a measure of deviation from independence that play a central role in our approach. Definition 3 (some probability) If α ∈{d, u} n−1 , then f(x, y, α) is the probability den- sity function for the event that the sequence X 1 , ,X n of iid random variables with the uniform distribution on [0, 1] has X 1 = x, X n = y and descent word α. Also, f(x, y | α) is the conditional density function. We replace x and/or y with ∗ to indicate marginal distributions. For example, f(x, ∗,α)=  1 0 f(x, y, α) dy. Let α 1 ,α 2 , be a sequence of descent words with |α n |→∞. We call the sequence asymptotically independent if either 1 This works for both odd and even n since 1 2k corresponds to (ud) k and 1 2k+1 corresponds to (ud) k u. the electronic journal of combinatorics 10 (2003), #R40 3 (a) lim n→∞ A(α n )=∞, (b) lim n→∞ Z(α n )=∞,or (c) A(α n ) and Z(α n ) are bounded and lim n→∞  sup (x,y)    f(x, y | α n ) − f(x, ∗|α n )f(∗,y | α n )     =0. (2.1) We call the sequence stable if lim n→∞ f(x, ∗|α n ) and lim n→∞ f(∗,y | α n ) exist or are delta functions. Clearly any infinite subsequence of an asymptotically independent or stable sequence also has that property. The following lemma, noted in [4], connects the probability distributions with permu- tations. Lemma 1 If X 1 , ,X n are independent identically distributed (iid) random variables with a continuous density function, then the probability that the sequence X 1 , ,X n has descent word α equals the probability that a random permutation of {1, ,n} has descent word α. In other words, the number of permutations with descent word α is (1 + |α|)! f(∗, ∗,α). Due to the lemma, we may study permutations via the probability distributions. Stability and asymptotic independence imply a result needed to prove Theorem 1: Theorem 3 Fix k>0. Suppose that, for each 1 ≤ i ≤ k, the sequence α i,1 ,α i,2 , is stable and asymptotically independent. Suppose that β i are possibly empty descent words for 0 ≤ i ≤ k.Let δ n = β 0 α 1,n β 1 ···α k,n β k . Let a(β) and z(β) be the first and last letters in β, respectively. If β i is not empty, assume both • that Z(α i,n a(β i )) is bounded for all n when 0 <i≤ k and • that A(z(β i )α i+1,n ) is bounded for all n when 0 ≤ i<k. If β i is empty and 0 <i<k, assume either • that Z(α i,n ) and A(α i+1,n ) are bounded for all n or • that z(α i,n ) = a(α i+1,n ) for all n. the electronic journal of combinatorics 10 (2003), #R40 4 Then f(∗, ∗,δ n )  k i=1 f(∗, ∗,α i,n ) ∼ k  i=0 C i , (2.2) where the C i are nonzero and given by C i =                        lim n→∞ f(∗, ∗,β 0 α 1,n ) f(∗, ∗,α 1,n ) , if i =0, lim n→∞ f(∗, ∗,α i,n β i α i+1,n ) f(∗, ∗,α i,n ) f(∗, ∗,α i+1,n ) , if 0 <i<k, lim n→∞ f(∗, ∗,α k,n β k ) f(∗, ∗,α k,n ) , if i = k. Theorem 4 below proves stability and asymptotic independence for repeated descent pat- terns. Conjecture 1 While stability clearly depends on the form of the words α i,1 ,α i,2 , ,we conjecture that |α i,n |→∞implies asymptotic independence. We now provide the tools for calculating the constants in Theorems 1 and 2. Definition 4 (reversal of descent words) For any descent word α, define α R to be α read in reverse order and α to be α with the roles of d and u reversed. Lemma 2 Let α and β be arbitrary descent words, We have f(x, y, u)=  0, if x>y, 1, otherwise, (2.3) f(x, y, α)=f(y,x, α R )=f(1 − x, 1 − y,α) (2.4) f(x, y, αβ)=  1 0 f(x, t, α)f(t, y, β) dt (2.5) f(∗, ∗,α) ≥ f(∗, ∗,αβ) (2.6) We omit the proof of the lemma since it is simple and is essentially contained in Section 2 of [4]. Theorem 4 Let µ = m 1 m |µ| be a descent word containing both d and u. The sequence µ, µ 2 ,µ 3 , is asymptotically independent and stable. Let ω = e 2πi/|µ| . Define the |µ|×|µ| matrix M for 0 ≤ k, < |µ| by M k, =  ω k , if m k+1 = d , ω k exp(rω  ), if m k+1 = u. the electronic journal of combinatorics 10 (2003), #R40 5 Let r be the smallest magnitude number for which the matrix M is not invertible. Let U(µ) be the number of u’s in µ. Then, uniformly for (x, y) ∈ [0, 1] 2 , f(x, y, µ n )=C(µ)  φ(x, µ)φ(y, µ R )+o(1)  λ(µ) n , (2.7) where λ(µ)= (−1) U(µ) r |µ| , (2.8) φ(x, µ)= |µ|−1  t=0 D t exp(rω t x), (2.9) C(µ)= 1  1 0 φ(x, µ) φ(x, µ R ) dx , (2.10) and D =(D 0 , ,D |µ|−1 ) t is the solution of MD = 0 such that  1 0 φ(x, µ) dx =1.The value of φ(y, µ R ) is found by replacing x with y and µ with µ R . The values of λ and |r| are the same for µ and µ R . We may assume arg r =0if U(µ) is even, and arg r = π/|µ| otherwise. In particular, f(∗, ∗,µ n ) ∼ C(µ) λ(µ) n . Remark If (2.7) is integrated over x or y, we obtain Theorem 2 of Shapiro, Shapiro and Vainshtein [6], including the same formulas for calculating C, λ and φ.Theirmethod of proof differs from ours. If our Conjecture 1 were proved, then our Theorem 4 would follow from Theorem 2 [6]. Remark The second smallest magnitude r,sayr 2 , for which M is singular gives the “second largest eigenvalue” λ 2 =1/|r 2 | |µ| , which is discussed in later sections. This can be used to obtain information about rate of convergence because of (6.1). See also Section 8. Using the lemma, one can compute f(x, y, α) for any particular descent word α.We use (2.4) to convert results for d into results for u and results for the left end of α into results for the right, generally without comment. To study the asymptotics of something like f (∗, ∗,α k βµ  )ask,  →∞, one combines the lemma and theorem: f(∗, ∗,α k βµ  )=  1 0  1 0 f(∗,x,α k ) f(x, y, β) f(y, ∗,µ  ) dx dy ∼ C(α) C(µ) λ(α) k λ(µ)   1 0  1 0 φ(x, α R ) f(x, y, β) φ(y, µ) dx dy. 3 An Illustration: µ = ud −1 We now obtain equations for C, φ and λ when µ = ud −1 .Thevalueofλ is given implicitly and, since φ and C depend on λ, they are given implicitly as well. Of course, our equation for λ will be the same as Leeming and MacLeod’s result. Note that |µ| = . the electronic journal of combinatorics 10 (2003), #R40 6 The matrix equation MD = 0 in Theorem 4 is written as |µ| separate equations in (8.11). With ω = e 2πi/ ,theseare −1  t=0 ω kt D t = 0 for 1 ≤ k ≤  − 1. It is easily seen that these equations have the one parameter solution given by D 0 = D 1 = ···= D −1 The condition for k =0is 0= −1  t=0 D t exp(rω t )=D 0 −1  t=0 exp(rω t ). (3.1) Since we do not want the identically zero solution, (3.1) gives us the complex transcen- dental equation  −1 t=0 exp(rω t ) = 0 for r. This can be simplified by using the Taylor series for e z to expand exp(rω t ) and then collecting terms according to powers of r:  ∞  k=0 r k (k)! =0, (3.2) since the sum of ω tn over t vanishes when n is not a multiple of . Thisistheresult of Leeming and MacLeod [5] mentioned after Theorem 2. In their notation, r = p  ,the smallest magnitude zero of (3.2). By (2.8), we can rewrite (3.2) as 0= ∞  k=0 (−1/λ) k (k)! , (3.3) which can be solved numerically for the largest λ>0. By (2.9) and Taylor series expansion of the exponentials, φ(x, µ)=D 0 −1  t=0 exp(rω t x)=D 0 ∞  k=0 (rx) k (k)! = D 0 ∞  k=0 (−1/λ) k x k (k)! . Integrating over [0, 1] gives 1 = D 0  ∞ k=0 (−1/λ) k (k+1)! and so φ(x, ud −1 )= ∞  k=0 (−1/λ) k x k (k)!  ∞  k=0 (−1/λ) k (k +1)! . (3.4) For µ R = u −1 d, the conditions (8.11) for 0 ≤ k ≤  − 2 become 0= −1  t=0 ω kt D t exp(rω t )= −1  t=0 ω (k+1)t  ω −t D t exp(rω t )  . the electronic journal of combinatorics 10 (2003), #R40 7 With E t = ω −t exp(rω t )D t , these become  −1 t=0 ω jt E t = 0 for 1 ≤ j ≤  − 1andso,as before, E 0 = E 1 = ···= E −1 .Fork =  − 1wehave 0= −1  t=0 ω t(−1) D t = −1  t=0 ω −t (ω t exp(−rω t )E t )=E 0 −1  t=0 exp(−rω t ). This is the same as (3.1) with −r replacing r.Thusr = −p  and φ(x, u −1 d)= −1  t=0 D t exp(−p  xω t )=E 0 −1  t=0 ω t exp(p  ω t )exp(−p  xω t ) = E 0 −1  t=0 ω t exp(p  (1 − x)ω t )= E 0 p  ∞  k=1 (−1/λ) k (1 − x) k−1 (k − 1)! , by expanding the exponentials in Taylor series as before. Integrating over [0, 1] gives 1= E 0 p  ∞  k=1 (−1/λ) k (k)! = − E 0 p  by (3.3). Thus φ(x, u −1 d)= − ∞  k=1 (−1/λ) k (1 − x) k−1 (k − 1)! . (3.5) Combining (3.4) and (3.5) with the (2.10), we have C(ud −1 )= − ∞  k=0 (−1/λ) k (k +1)! ∞  s=0 ∞  t=1  1 0 (−1/λ) s x s (s)! (−1/λ) t (1 − x) t−1 (t − 1)! dx = − ∞  k=0 (−1/λ) k (k +1)! ∞  s=0 ∞  t=1 (−1/λ) s+t ((s + t))! = − ∞  k=0 (−1/λ) k (k +1)! ∞  k=1 k(−1/λ) k (k)! . (3.6) The following table contains some values of λ(ud −1 )andC(ud −1 )andwellasthede- nominator of (3.4) and λ 1/ . The denominator is needed in computing φ and λ 1/ is used the electronic journal of combinatorics 10 (2003), #R40 8 in (1.1). λ(ud −1 ) C(ud −1 ) (3.4) denom. λ 1/ 2 0.405285 0.810569 0.636620 0.636620 3 0.157985 0.786835 0.744142 0.540595 4 4.1064e−2 0.810569 0.798696 0.450158 5 8.3001e−3 0.836374 0.833030 0.383546 6 1.3874e−3 0.858002 0.857071 0.333964 7 1.9835e−4 0.875238 0.874983 0.295844 8 2.4800e−5 0.888954 0.888885 0.265647 9 2.7557e−6 0.900018 0.899999 0.241128 To further illustrate the calculation procedure, we compute asymptotics in Theorem 1 when the permutation alternates up/down, except for some internal cases of uu.Todo this, we take all a i to be even except possibly a k , M i = 1 for all i, L i = 1 for 1 ≤ i ≤ k−1, and L 0 and L k empty. We need to compute B i . Ehrenborg’s Theorem 4.1 [3] gives the value. When a is even and b is odd, what he calls β(1 a , 2, 1 b ) is the number of permutations with pattern (ud) a/2 u(ud) (b+1)/2 .Hecompares this with E n , the number of alternating permutations of the same length n = 1 a 21 b . On the other hand, B i compares it with E n−1 Since the fraction of n-long permutations that alternate is asymptotic to C(1)λ(1) n , we obtain an extra factor of λ(1) = 2/π: B i ∼ λ(1) β(1 a , 2, 1 b ) E n ∼ 4 π 2 , where min(a, b) →∞. Thus B i =4/π 2 . Ehrenborg also discusses computing β(1 a ,L,1 b ). To illustrate the use of our formulas, we now compute B i without using Ehrenborg’s result. Note that Run(M n i M n i+1 ) is just counting alternating permutations. To evaluate Run(M n i L i M n i+1 ), we apply (2.5) twice to compute f(x, y, (ud) m u(ud) m ) and integrate this over x and y.Since f(x, y, (ud) m )=C(1)  φ(x, u)φ(y,u)+o(1)  λ(1) 2m , where C and λ are given by (1.2), we need to know φ(x, u). One can use (3.5) with  =2 or [4] to conclude that φ(x, u)=(π/2) sin(πx/2). In computing B i in Theorem 1, the formulas we are using are probabilities and so we will be estimating Run(P )/P ! for patterns P . Remembering that  1 0 φ(x, α) dx =1, B i = C(1) 2 λ(1) 2n  1 0  1 0 sin(πs/2) sin(πt/2)f(s, t, u) ds dt C(1) 2 λ(1) 2n  1 0 sin(πs/2) 2 ds , where f (s, t, u) is given by (2.3). The integral in the denominator is 1/2 and the integral in the numerator is the electronic journal of combinatorics 10 (2003), #R40 9  0<s<t<1 sin(πs/2) sin(πt/2) ds dt =  1 0 sin(πs/2)(2/π)cos(πs/2) ds =(1/π)  1 0 sin(πs) ds =2/π 2 . Hence B i =4/π 2 . 4 Proof of Theorem 3 Lemma 3 Let α and β be descent words. (a) If |α| > 1, then f(x, y, α) is a monotonic uniformly continuous function of x and y on the unit square. In fact, it is increasing in x if and only if α begins with d and is increasing in y if and only if α ends with u. (b) f(x, y, αduβ) ≥ f(x, ∗,αd) f(∗,y,uβ). (c) If α contains both u and d, there is are functions U(k) and L(x, y, k) such that, for each k, L(x, y, k) is strictly positive for (x, y) in the interior of [0, 1] 2 and such that U(k) ≥ f(x, y | α) ≥ L(x, y, k) where k =max{A(α),Z(α)}. Similarly, U 1 (A(α)) ≥ f (x, ∗|α) ≥ L 1 (x, A(α)) and U 1 (Z(α)) ≥ f(∗,y | α) ≥ L 1 (y, Z(α)) for functions U 1 and L 1 where L 1 (x, k) is strictly positive for 0 <x<1. Proof It is easily seen that (2.3) is monotonic. It follows by induction that f (x, y, α)is continuous if |α| > 1. Suppose α = uβ where β is not the empty word. By (2.5) f(x, y, α)=  1 x f(t, y, β) dt, which is clearly a decreasing function of x. We now prove (b). By (2.5), f(x, y, αduβ)=  1 0 f(x, t, αd) f(t, y, uβ) dt. By (a), both f(x, t, αd)andf(t, y, uβ) are monotonic decreasing functions of t.Bythe integral form of Chebyshev’s integral inequality [7],  1 0 f(x, t, αd) f(t, y, uβ) dt ≥  1 0 f(x, t, αd) dt  1 0 f(t, y, uβ) dt = f(x, ∗,αd) f(∗,y,uβ). the electronic journal of combinatorics 10 (2003), #R40 10 [...]... whose kernel is K If we were dealing with matrices, we would simply be taking powers of a matrix K with strictly positive entries and so K would have a unique eigenvalue λ1 of maximum modulus It would be positive real and have left and right eigenspaces of dimension 1 Thus we would have K n ∼ Cλn uv for 1 left and right eigenvectors u and v This is the discrete form of Theorem 4 We prove analogous results... = g(y) dy x 22 correspond to prepending d and u to α when g(y) = f (y, ∗, α) To study the behavior of f (x, ∗, ab z) one studies the eigenfunctions of T = Na Nb · · · Nz Equivalently, we −1 can work with T −1 and thus deal with eigenfunctions of the differential operators Nd −1 and Nu , which are d/dx and −d/dx, respectively, with boundary conditions at 0 and 1, respectively Using this approach,... the proof of Theorem 1 We now turn our attention to Theorem 2 If |L| is even and corresponds to the descent word µ, the fraction of permutations with run pattern Ln is f (∗, ∗, µn ) and so we are done by Theorem 4 Suppose |L| is odd and L corresponds to descent word α Let M = LL Then M corresponds to µ = αα Us the previous paragraph and define C(L) = C(M) and λ(L) = λ(M) This completes the proof for... we can replace (4.2) with (4.3) plus f (ti−1 , si−1 , βi−1 ) o(1) The effect of this latter is to add a term of products of Cj ’s with Ci−1 Ci replaced by o(1) Since the Ci will be shown to be nonzero, the asymptotics are unchanged Now suppose A(αi,n ) → ∞ and a(αi,n ) = u the cases of Z and d are handled by (2.4) For simplicity, we drop the i subscripts Write αn = um γ where m → ∞ and a(γ) = d By assumption,... Theorem 5 and Lemma 6 to T and obtain, like before, m−1 (T )m − λ1 Q L(B ) ≤ (|λ2 | + ε)n (6.5) As before Q has the form Q (η) = Lα(η) for all η ∈ B Here L ∈ P and α ∈ (B ) The definition of adjoint and (6.3) imply L(b)α(η) = Q (η)(b) = η(Q(b)) = η(φ) (b) for all η and b Thus L = (6.6) and α(η) = η(φ) for all η ∈ B Thus we have proved Lemma 7 In the notation of Lemma 6, there exist φ ∈ P and L ∈ P... and (8.5) This completes the proof of the first part of Theorem 4 Equation (2.7) and the remark after the theorem follow from (8.1), provided we obtain formulas for C, λ and φ (There is no C in (8.1), but it is needed now because we normalize 1 φ to have 0 φ = 1 and we incorporate a factor of λ in C.) One can interpret finding the eigenfunction φ in terms of the inverse of the integral operator Equivalently,... absolute value is λ1 and the remaining points in the spectrum of T on the complexification of Me have absolute value at most |λ2 | There is a unique (up to scalar multiple) eigenfunction, φdx in Pe , of T Moreover, spectrum(T on Mc ) = spectrum(T on Mc ) e Proof It remains to prove the last assertion of the lemma An eigenvector ν of T on M has the form g(x) dx where g ∈ C[0, 1] and |g(x)| ≤ τ e(x) for... the electronic journal of combinatorics 10 (2003), #R40 12 The case of unbounded runs at the end of αi,n can be handled as in the derivation of (4.5) Otherwise, stability guarantees that f (∗, s | αi,n ) and f (t, ∗ | αi+1,n ) approach a limit and Lemma 3(c) guarantees that the limits are bounded Since f (s, t, βi) is well behaved, Ci exists Furthermore, it is positive because of the lower bound in... Q(b) = φL(b), 7 T φ = λ1 φ and T (L) = λ1 L Asymptotics for Integral Operators We now take K(x, y) of Section 5 to be the kernel of an integral operator acting on a space of measures as follows The space C[0, 1] of continuous functions on [0, 1] with norm given by g ∞ = sup |g(x)| is a real Banach space whose dual M is the Banach space of finite 0≤x≤1 total mass Borel measures with ν by M T (νy ) = = 1... are divided by factorials and so can be thought of as functions of the form f (∗, ∗, γ) according to Lemma 1 Thus one could apply Theorems 3 and 4 and deduce Theorems 1 and 2 except for two minor complications which we now discuss The first complication is the fact that the forms are not quite the same: A direct application would give k i=1 Run(Miai ) Miai ! and Ci = instead of a a Run(M1 1 Mk k ) . Asymptotics of Permutations with Nearly Periodic Patterns of Rises and Falls Edward A. Bender Department of Mathematics University of California, San Diego La Jolla, CA. results for nearly alternating permutations and conjectured an asymptotic formula for the number of permutations that have a nearly periodic run pattern. We prove a generalization of this conjecture,. dx , (2.10) and D =(D 0 , ,D |µ|−1 ) t is the solution of MD = 0 such that  1 0 φ(x, µ) dx =1.The value of φ(y, µ R ) is found by replacing x with y and µ with µ R . The values of λ and |r| are