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Continuously Increasing Subsequences of Random Multiset Permutations

Alexander Clifton111Dept. of Mathematics, Emory University [email protected]    Bishal Deb222Dept. of Mathematics, University College London [email protected]    Yifeng Huang333Dept. of Mathematics, University of Michigan [email protected]    Sam Spiro444Dept. of Mathematics, UCSD [email protected]. This material is based upon work supported by the National Science Foundation Graduate Research Fellowship under Grant No. DGE-1650112.    Semin Yoo555School of Computational Sciences, KIAS [email protected]. The author is supported by the KIAS Individual Grant (CG082701) at Korea Institute for Advanced Study.
Abstract

For a word π\pi and integer ii, we define Li(π)L^{i}(\pi) to be the length of the longest subsequence of the form i(i+1)ji(i+1)\cdots j, and we let L(π):=maxiLi(π)L(\pi):=\max_{i}L^{i}(\pi). In this paper we estimate the expected values of L1(π)L^{1}(\pi) and L(π)L(\pi) when π\pi is chosen uniformly at random from all words which use each of the first nn integers exactly mm times. We show that 𝔼[L1(π)]m\mathbb{E}[L^{1}(\pi)]\sim m if nn is sufficiently larger in terms of mm as mm tends towards infinity, confirming a conjecture of Diaconis, Graham, He, and Spiro. We also show that 𝔼[L(π)]\mathbb{E}[L(\pi)] is asymptotic to the inverse gamma function Γ1(n)\Gamma^{-1}(n) if nn is sufficiently large in terms of mm as mm tends towards infinity.

1 Introduction

1.1 Main Results

Given integers mm and nn, let 𝔖m,n\mathfrak{S}_{m,n} denote the set of words π\pi which use each integer in [n]:={1,2,,n}[n]:=\{1,2,\ldots,n\} exactly mm times, and we will refer to π𝔖m,n\pi\in\mathfrak{S}_{m,n} as a multiset permutation. For example, π=211323𝔖2,3\pi=211323\in\mathfrak{S}_{2,3}. For π𝔖m,n\pi\in\mathfrak{S}_{m,n} and ii an integer, we define Lm,ni(π)L^{i}_{m,n}(\pi) to be the length of the longest subsequence of π\pi of the form i(i+1)(i+2)ji(i+1)(i+2)\cdots j, which we call an ii-continuously increasing subsequence. We say that a subsequence is a continuously increasing subsequence if it is an ii-continuously increasing subsequence for some ii, and we define Lm,n(π)=maxiLm,ni(π)L_{m,n}(\pi)=\max_{i}L^{i}_{m,n}(\pi) to be the length of a longest continuously increasing subsesquence of π\pi. For example, if π=2341524315\pi=2341524315 then L2,5(π)=L2,52(π)=4L_{2,5}(\pi)=L^{2}_{2,5}(\pi)=4 due to the subsequence 23452345, and L2,51(π)=3L^{1}_{2,5}(\pi)=3 due to the subsequence 123123.

The focus of this paper is to study Lm,n1(π)L^{1}_{m,n}(\pi) and Lm,n(π)L_{m,n}(\pi) when π\pi is chosen uniformly at random from 𝔖m,n\mathfrak{S}_{m,n}. We focus on the regime where nn is much larger than mm, as in the opposite regime Lm,ni(π)L^{i}_{m,n}(\pi) is very likely to be its maximum possible value ni+1n-i+1 for all ii.

We first consider 𝔼[Lm,n1(π)]\mathbb{E}[L^{1}_{m,n}(\pi)]. This quantity was briefly studied by Diaconis, Graham, He, and Spiro [DGHS21] due to its relationship with a certain card game that we describe later in this paper. They proved 𝔼[Lm,n1(π)]m+Cm3/4logm\mathbb{E}[L^{1}_{m,n}(\pi)]\leq m+Cm^{3/4}\log m for some absolute constant CC provided nn is sufficiently large in terms of mm. It was conjectured in [DGHS21] that this upper bound for 𝔼[Lm,n1(π)]\mathbb{E}[L^{1}_{m,n}(\pi)] is asymptotically tight for nn sufficiently large in terms of mm. We verify this conjecture in a strong form, obtaining an exact formula for limn𝔼[Lm,n1(π)]\lim_{n\to\infty}\mathbb{E}[L^{1}_{m,n}(\pi)] for any fixed mm and precise estimates of this value as mm tends towards infinity.

Theorem 1.1.
  • (a)

    For any integer m1m\geq 1, let α1,,αm\alpha_{1},\ldots,\alpha_{m} be the zeroes of Em(x):=k=0mxkk!E_{m}(x):=\sum_{k=0}^{m}\frac{x^{k}}{k!}. If π𝔖m,n\pi\in\mathfrak{S}_{m,n} is chosen uniformly at random, then

    m1:=limn𝔼[Lm,n1(π)]=1αi1eαi.\mathcal{L}_{m}^{1}:=\lim_{n\to\infty}\mathbb{E}[L_{m,n}^{1}(\pi)]=-1-\sum\alpha_{i}^{-1}e^{-\alpha_{i}}. (1)
  • (b)

    There exists an absolute constant β>0\beta>0 such that

    |m1(m+11m+2)|O(eβm).\left|\mathcal{L}_{m}^{1}-\left(m+1-\frac{1}{m+2}\right)\right|\leq O(e^{-\beta m}).

For example, when m=1m=1 we have E1(x)=1+xE_{1}(x)=1+x and α1=1\alpha_{1}=-1, implying 11=1+e\mathcal{L}_{1}^{1}=-1+e, which can also be proven by elementary means. For m=2m=2 we have E2(x)=1+x+x2/2E_{2}(x)=1+x+x^{2}/2 and α1=1i,α2=1+i\alpha_{1}=-1-i,\alpha_{2}=-1+i. From this Theorem 1.1(a) gives the following closed form expression for 21\mathcal{L}_{2}^{1}.

Corollary 1.2.
21=e(cos(1)+sin(1))1.\mathcal{L}_{2}^{1}=e\left(\cos(1)+\sin(1)\right)-1.

Our next result gives precise bounds for the length of a longest continuously increasing subsequence in a random permutation of 𝔖m,n\mathfrak{S}_{m,n}. We recall that the gamma function Γ(x)\Gamma(x) is a function which, in particular, gives a bijection from x1x\geq 1 to y1y\geq 1 and which satisfies Γ(n)=(n1)!\Gamma(n)=(n-1)! for non-negative integers nn.

Theorem 1.3.

If nn is sufficiently large in terms of mm, then

𝔼[Lm,n(π)]=Γ1(n)+Θ(1+logmlog(Γ1(n))Γ1(n)),\mathbb{E}[L_{m,n}(\pi)]=\Gamma^{-1}(n)+\Theta\left(1+\frac{\log m}{\log(\Gamma^{-1}(n))}\Gamma^{-1}(n)\right),

where Γ1(n)\Gamma^{-1}(n) is the inverse of the gamma function when restricted to x1x\geq 1.

Note when m=1m=1 the error term of Theorem 1.3 is Θ(1)\Theta(1), but for m2m\geq 2 it is Θ(logmlogΓ1(n)Γ1(n))\Theta(\frac{\log m}{\log\Gamma^{-1}(n)}\Gamma^{-1}(n)), which is fairly close to the main term of Γ1(n)\Gamma^{-1}(n). Thus the behavior of 𝔼[Lm,n(π)]\mathbb{E}[L_{m,n}(\pi)] changes somewhat dramatically as soon as one starts to consider multiset permutations as opposed to just permutations.

1.2 History and Related Work

Determining Lm,ni(π)L^{i}_{m,n}(\pi) and Lm,n(π)L_{m,n}(\pi) can be viewed as variants of the well-studied problem of determining the length of the longest increasing subsequence in a random permutation of length nn, and we denote this quantity by L~n\widetilde{L}_{n}. It was shown by Logan and Shepp [LS77] and Vershick and Kerov [VK77] that 𝔼[L~n]2n\mathbb{E}[\widetilde{L}_{n}]\sim 2\sqrt{n}, answering a famous problem of Ulam. Later Baik, Deift, and Johansson [BDJ99] showed that the limiting distribution of L~n\widetilde{L}_{n} is the Tracy-Widom distribution. Some work with the analogous problem for multiset permutations has been considered recently by Almeanazel and Johnson [AMJ20]. Much more can be said about this topic, and we refer the reader to the excellent book by Romik [Rom15] for more information.

The initial motivation for studying L1(π)L^{1}(\pi) was due to its relationship to a card guessing experiment introduced by Diaconis and Graham [DG81]. To start the experiment, one shuffles a deck of mnmn cards which consists of nn distinct card types each appearing with multiplicity mm. In each round, a subject iteratively guesses what the top card of the deck is according to some strategy GG. After each guess, the subject is told whether their guess was correct or not, the top card is discarded, and then the experiment continues with the next card. This experiment is known as the partial feedback model. For more on the history of the partial feedback model we refer the reader to [DGS21].

If GG is a strategy for the subject in the partial feedback model and π𝔖m,n\pi\in\mathfrak{S}_{m,n}, we let P(G,π)P(G,\pi) denote the number of correct guesses made by the subject if they follow strategy GG and the deck is shuffled according to π\pi. We say that GG is an optimal strategy if 𝔼[P(G,π)]=maxG𝔼[P(G,π)]\mathbb{E}[P(G,\pi)]=\max_{G^{\prime}}\mathbb{E}[P(G^{\prime},\pi)], where GG^{\prime} ranges over all strategies and π𝔖m,n\pi\in\mathfrak{S}_{m,n} is chosen uniformly at random. Optimal strategies are unknown in general, and even if they were known they would likely be too complex for a human subject to implement in practice. As such there is interest in coming up with (simple) strategies GG such that 𝔼[P(G,π)]\mathbb{E}[P(G,\pi)] is relatively large.

One strategy is the trivial strategy which guesses card type 1 every single round, guaranteeing a score of exactly mm at the end of the experiment. A slightly better strategy is the safe strategy GsafeG_{safe} which guesses card type 1 every round until all mm are guessed correctly, then 2’s until all mm are guessed correctly, and so on. It can be deduced from arguments given by Diaconis, Graham, and Spiro [DGS21] that 𝔼[P(Gsafe,π)]\mathbb{E}[P(G_{safe},\pi)] is m+11m+1m+1-\frac{1}{m+1} plus an exponential error term, so the safe strategy does just a little better than the trivial strategy.

Another natural strategy is the shifting strategy GshiftG_{shift}, defined by guessing 1 until you are correct, then 2 until you are correct, and so on; with the strategy being defined arbitrarily in the (very rare) event that one correctly guesses a copy of each card type. It is not difficult to see that P(Gshift,π)Lm,n1(π)P(G_{shift},\pi)\geq L^{1}_{m,n}(\pi), with equality holding provided the player does not correctly guess nn. Thus Theorem 1.1(b) shows that the expected number of correct guesses under the shifting strategy is close to m+11m+2m+1-\frac{1}{m+2}, which is slightly better than the trivial strategy, and very slightly better than the safe strategy.

1.3 Preliminaries

We let [n]:={1,2,,n}[n]:=\{1,2,\ldots,n\} and let [m]n[m]^{n} be the set of tuples of length nn with entries in [m][m]. Whenever we write, for example, Pr[Lm,n(π)k]\Pr[L_{m,n}(\pi)\geq k], we will assume π\pi is chosen uniformly at random from 𝔖m,n\mathfrak{S}_{m,n} unless stated otherwise.

Throughout this paper we use several basic results from probability theory. One such result is that if XX is a non-negative integer-valued random variable, then

𝔼[X]=k=1Pr[Xk].\mathbb{E}[X]=\sum_{k=1}^{\infty}\Pr[X\geq k].

A crucial observation that we use throughout the text is the following.

Observation 1.4.

For knk\leq n, if π𝔖m,n\pi\in\mathfrak{S}_{m,n} and τ𝔖m,k\tau\in\mathfrak{S}_{m,k} are drawn uniformly at random, then

Pr[Lm,n1(π)k]=Pr[Lm,k1(τ)=k].\Pr[L^{1}_{m,n}(\pi)\geq k]=\Pr[L^{1}_{m,k}(\tau)=k].
Proof.

For π𝔖m,n\pi\in\mathfrak{S}_{m,n}, let ϕ(π)𝔖m,k\phi(\pi)\in\mathfrak{S}_{m,k} be the word obtained by deleting every letter from π\pi which is larger than kk. Note that Lm,n1(π)kL^{1}_{m,n}(\pi)\geq k if and only if Lm,k1(ϕ(π))=kL^{1}_{m,k}(\phi(\pi))=k. Moreover, it is not difficult to see that ϕ(π)\phi(\pi) is distributed uniformly at random in 𝔖m,k\mathfrak{S}_{m,k} provided π\pi is distributed uniformly at random in 𝔖m,n\mathfrak{S}_{m,n}, proving the result. ∎

2 Proof of Theorem 1.1

2.1 Theorem 1.1(a): Generating Functions

We say that a word π𝔖m,n\pi\in\mathfrak{S}_{m,n} has a complete increasing subsequence if Lm,n1(π)=nL^{1}_{m,n}(\pi)=n. Let hm(n)h_{m}(n) be the number of words π𝔖m,n\pi\in\mathfrak{S}_{m,n} which have a complete increasing subsequence. Horton and Kurn [HK81, Corollary (c)] give the following formula for hm(n)h_{m}(n).

Theorem 2.1 ([HK81]).

The number of words π𝔖m,n\pi\in\mathfrak{S}_{m,n} which have a complete increasing subsequence, hm(n)h_{m}(n), is given by

hm(n)=(i1,,im)𝒩(n,m)(ni1,,im)(mn)!l!(1)lnj=1m(mj)!ij,h_{m}(n)=\sum_{(i_{1},\ldots,i_{m})\in\mathcal{N}(n,m)}\binom{n}{i_{1},\ldots,i_{m}}\dfrac{(mn)!}{l!}\dfrac{(-1)^{l-n}}{\prod_{j=1}^{m}(m-j)!^{i_{j}}},

where

l=j=1mjij,l=\sum_{j=1}^{m}ji_{j},

𝒩(n,m)\mathcal{N}(n,m) is the set of weak compositions of nn into mm parts, i.e.,

𝒩(n,m){(i1,,im)0m|j=1mij=n},\mathcal{N}(n,m)\coloneqq\left\{(i_{1},\ldots,i_{m})\in\mathbb{Z}_{\geq{0}}^{m}\middle|\sum_{j=1}^{m}i_{j}=n\right\},

and

(ni1,,im)=n!j=1mij!\binom{n}{i_{1},\ldots,i_{m}}=\dfrac{n!}{\prod_{j=1}^{m}i_{j}!}

is a multinomial coefficient.

Notice that m1\mathcal{L}_{m}^{1} can be expressed in terms of hm(n)h_{m}(n) as follows:

m1=limk𝔼[Lm,k1(π)]=limkn=1kPr[Lm,k1(π)n]=limkn=1khm(n)|𝔖m,n|=n=1hm(n)|𝔖m,n|,\mathcal{L}_{m}^{1}=\lim_{k\to\infty}\mathbb{E}[L_{m,k}^{1}(\pi)]=\lim_{k\to\infty}\sum_{n=1}^{k}\mathrm{Pr}[L^{1}_{m,k}(\pi)\geq n]=\lim_{k\to\infty}\sum_{n=1}^{k}\dfrac{h_{m}(n)}{|\mathfrak{S}_{m,n}|}=\sum_{n=1}^{\infty}\dfrac{h_{m}(n)}{|\mathfrak{S}_{m,n}|}, (2)

where the third equality is due to Observation 1.4. Note that |𝔖m,n|=(mn)!/(m!)n|\mathfrak{S}_{m,n}|=(mn)!/(m!)^{n}. Thus, as a consequence of Theorem 2.1, we have

hm(n)|𝔖m,n|=(m!)n(i1,,im)𝒩(n,m)(ni1,,im)1j=1m(mj)!ij(1)ll!,\dfrac{h_{m}(n)}{|\mathfrak{S}_{m,n}|}=(-m!)^{n}\sum_{(i_{1},\ldots,i_{m})\in\mathcal{N}(n,m)}\binom{n}{i_{1},\ldots,i_{m}}\dfrac{1}{\prod_{j=1}^{m}(m-j)!^{i_{j}}}\dfrac{(-1)^{l}}{l!}, (3a)
=(m!)n(i1,,im)𝒩(n,m)(ni1,,im)j=1m((1)j(mj)!)ij1l!.\hskip 25.00003pt=(-m!)^{n}\sum_{(i_{1},\ldots,i_{m})\in\mathcal{N}(n,m)}\binom{n}{i_{1},\ldots,i_{m}}\prod_{j=1}^{m}\left(\dfrac{(-1)^{j}}{(m-j)!}\right)^{i_{j}}\dfrac{1}{l!}. (3b)

Intuitively, if the 1/l!1/l! were removed from the right-hand-side expression in (3b), then by using the multinomial theorem we could write this expression as an nthn^{\text{th}} power, turning (2) into a geometric series. The next few paragraphs formalise this idea.

We begin by replacing (1)l(-1)^{l} by xlx^{l} in the right-hand-side of (3a) to obtain the polynomial

pm,n(x)\displaystyle p_{m,n}(x)\coloneqq (m!)n(i1,,im)𝒩(n,m)(ni1,,im)1j=1m(mj)!ijxll!\displaystyle(-m!)^{n}\sum_{(i_{1},\ldots,i_{m})\in\mathcal{N}(n,m)}\binom{n}{i_{1},\ldots,i_{m}}\dfrac{1}{\prod_{j=1}^{m}(m-j)!^{i_{j}}}\dfrac{x^{l}}{l!}
=\displaystyle= (m!)n(i1,,im)𝒩(n,m)(ni1,,im)j=1m(xj(mj)!)ij1l!.\displaystyle(-m!)^{n}\sum_{(i_{1},\ldots,i_{m})\in\mathcal{N}(n,m)}\binom{n}{i_{1},\ldots,i_{m}}\prod_{j=1}^{m}\left(\dfrac{x^{j}}{(m-j)!}\right)^{i_{j}}\dfrac{1}{l!}.

Thus,

pm,n(1)=hm(n)|𝔖m,n|.p_{m,n}(-1)=\dfrac{h_{m}(n)}{|\mathfrak{S}_{m,n}|}.

Next, we define an operator in order to remove the l!l! from the denominator. Let RR be a commutative ring containing \mathbb{Q} and let Φ:R[x]R[x]\Phi:R[x]\to R[x] be an RR-linear map defined on the monomials by

Φ(xn)=xnn!.\Phi(x^{n})=\dfrac{x^{n}}{n!}.

We can extend Φ\Phi to an RR-linear map on R[[x]]R[[x]]R[[x]]\to R[[x]], which we also refer to as Φ\Phi by abuse of notation. Throughout this article, RR is either \mathbb{C} or [[y]]\mathbb{C}[[y]] for an indeterminate yy, and we shall refer to this RR-linear map as Φ\Phi in both cases. Notice that Φ\Phi is invertible for any such ring RR. A key property that we use about Φ\Phi is

Φ(11ax)=Φ(i=0(ax)i)=i=0(ax)ii!=eax.\Phi\left(\frac{1}{1-ax}\right)=\Phi\left(\sum_{i=0}^{\infty}(ax)^{i}\right)=\sum_{i=0}^{\infty}\frac{(ax)^{i}}{i!}=e^{ax}. (4)

Consider the polynomial

qm,n(x)Φ1(pm,n(x))=(m!)n(i1,,im)𝒩(n,m)(ni1,,im)xlj=1m(mj)!ijq_{m,n}(x)\coloneqq\Phi^{-1}\left(p_{m,n}(x)\right)=(-m!)^{n}\sum_{(i_{1},\ldots,i_{m})\in\mathcal{N}(n,m)}\binom{n}{i_{1},\ldots,i_{m}}\dfrac{x^{l}}{\prod_{j=1}^{m}(m-j)!^{i_{j}}} (5)
=(m!)n(i1,,im)𝒩(n,m)(ni1,,im)j=1m(xj(mj)!)ij.=(-m!)^{n}\sum_{(i_{1},\ldots,i_{m})\in\mathcal{N}(n,m)}\binom{n}{i_{1},\ldots,i_{m}}\prod_{j=1}^{m}\left(\dfrac{x^{j}}{(m-j)!}\right)^{i_{j}}.

Notice that,

qm,n(x)=(m!j=1mxj(mj)!)n=(qm,1(x))n.q_{m,n}(x)=\left(-m!\sum_{j=1}^{m}\dfrac{x^{j}}{(m-j)!}\right)^{n}=(q_{m,1}(x))^{n}.

Let Pm(x,y)P_{m}(x,y) and Qm(x,y)Q_{m}(x,y) be the ordinary generating functions of pm,n(x)p_{m,n}(x) and qm,n(x)q_{m,n}(x) respectively, i.e.

Pm(x,y)n=0pm,n(x)yn,P_{m}(x,y)\coloneqq\sum_{n=0}^{\infty}p_{m,n}(x)y^{n}, (6)
Qm(x,y)n=0qm,n(x)yn=Φ1(Pm(x,y)).Q_{m}(x,y)\coloneqq\sum_{n=0}^{\infty}q_{m,n}(x)y^{n}=\Phi^{-1}\left(P_{m}(x,y)\right).

Putting everything together, we see that

m1=Pm(1,1)1.\mathcal{L}_{m}^{1}=P_{m}(-1,1)-1. (7)

and thus it suffices to find a nice closed form expression for Pm(x,y)P_{m}(x,y). Note that

qm,1(x)=m!xmEm1(1/x),q_{m,1}(x)=-m!x^{m}E_{m-1}(1/x),

where we recall the polynomial Em1(x)E_{m-1}(x) is defined in Theorem 1.1 by Em1(x)=k=0m1xk/k!E_{m-1}(x)=\sum_{k=0}^{m-1}x^{k}/k!. As qm,n(x)=(qm,1(x))nq_{m,n}(x)=\left(q_{m,1}(x)\right)^{n}, we have

Qm(x,y)=11yqm,1(x)=11+m!xmyEm1(1/x).Q_{m}(x,y)=\dfrac{1}{1-yq_{m,1}(x)}=\dfrac{1}{1+m!x^{m}yE_{m-1}(1/x)}. (8)

Hence,

Pm(x,y)=Φ(Qm(x,y))=Φ(11+m!xmyEm1(1/x)),P_{m}(x,y)=\Phi\left(Q_{m}(x,y)\right)=\Phi\left(\dfrac{1}{1+m!x^{m}yE_{m-1}(1/x)}\right),

and thus

Pm(x,1)=Φ(11+m!xmEm1(1/x))=Φ(1m!xmEm(1/x)).P_{m}(x,1)=\Phi\left(\dfrac{1}{1+m!x^{m}E_{m-1}(1/x)}\right)=\Phi\left(\dfrac{1}{m!x^{m}E_{m}(1/x)}\right).

We now prove the main result of this subsection.

Proposition 2.2.

Let α1,,αm\alpha_{1},\ldots,\alpha_{m} be the zeroes of the polynomial Em(x)E_{m}(x). The formal power series Pm(x,1)P_{m}(x,1) satisfies

Pm(x,1)=i=1mαi1eαix.P_{m}(x,1)=-\sum_{i=1}^{m}\alpha_{i}^{-1}e^{\alpha_{i}x}.
Proof.

Let g(x)m!xmEm(1/x)g(x)\coloneqq m!x^{m}E_{m}(1/x). Since α11,,αm1\alpha_{1}^{-1},\ldots,\alpha_{m}^{-1} are the zeroes of g(x)g(x), we have

g(x)=m!(xα11)(xαm1).g(x)=m!(x-\alpha_{1}^{-1})\cdots(x-\alpha_{m}^{-1}).

Notice that Em(x)E_{m}(x) has no repeated zeroes. This is true because, if α\alpha is a repeated zero of Em(x)E_{m}(x), it is also a zero of its derivative Em(x)=Em1(x)E_{m}^{\prime}(x)=E_{m-1}(x). But then α\alpha has to be a zero of Em(x)Em1(x)=xm/m!E_{m}(x)-E_{m-1}(x)=x^{m}/m!, which is only possible if α=0\alpha=0, a contradiction as 0 is not a zero of Em(x)E_{m}(x).

Thus α1,,αm\alpha_{1},\ldots,\alpha_{m} are pairwise distinct, and hence the zeroes of g(x)g(x) being α11,,αm1\alpha_{1}^{-1},\ldots,\alpha_{m}^{-1}, are also pairwise distinct. This and (8) means Qm(x,1)Q_{m}(x,1) has the partial fraction decomposition

Qm(x,1)=1g(x)=i=1m1g(αi1)1xαi1.Q_{m}(x,1)=\dfrac{1}{g(x)}=\sum_{i=1}^{m}\dfrac{1}{g^{\prime}\left(\alpha_{i}^{-1}\right)}\cdot\dfrac{1}{x-\alpha_{i}^{-1}}.

The derivative of gg is

g(x)=m!(mxmEm(1/x)xxmEm(1/x)x2)=m!(mxmEm(1/x)xxm(Em(1/x)xm/m!)x2)g^{\prime}(x)=m!\left(\dfrac{mx^{m}E_{m}(1/x)}{x}-\dfrac{x^{m}E_{m}^{\prime}(1/x)}{x^{2}}\right)=m!\left(\dfrac{mx^{m}E_{m}(1/x)}{x}-\dfrac{x^{m}(E_{m}(1/x)-x^{-m}/m!)}{x^{2}}\right)

Hence for any ii,

g(αi1)=αi2g^{\prime}\left(\alpha_{i}^{-1}\right)=\alpha_{i}^{2}

which gives

Pm(x,1)=Φ(Qm(x,1))=i=1mΦ(1αi(1αix))=i=1mαi1eαix,P_{m}(x,1)=\Phi\left(Q_{m}(x,1)\right)=-\sum_{i=1}^{m}\Phi\left(\dfrac{1}{\alpha_{i}(1-\alpha_{i}x)}\right)=-\sum_{i=1}^{m}\alpha_{i}^{-1}e^{\alpha_{i}x},

where this last step used (4). ∎

This proposition together with (7) gives Theorem 1.1(a).

2.2 Theorem 1.1(b): Exponential Sums of Zeroes

We remind the reader that Theorem 1.1(b) claims

|m1(m+11m+2)|O(eβm)\left|\mathcal{L}_{m}^{1}-\left(m+1-\frac{1}{m+2}\right)\right|\leq O(e^{-\beta m})

for some constant β>0\beta>0. Given Theorem 1.1(a), proving this claim is equivalent to showing that iαi1eαi=m2+1m+2+O(eβm)\displaystyle\sum_{i}\alpha_{i}^{-1}e^{-\alpha_{i}}=-m-2+\frac{1}{m+2}+O(e^{-\beta m}) for some positive constant β\beta. We do this by following the approach used by Conrey and Ghosh [CG88] to evaluate a similar exponential sum.

For 1im1\leq i\leq m, let sis_{i} and tit_{i} be the real and imaginary parts, respectively, of αi\alpha_{i}. Let γ,γ,γ+\gamma^{-},\gamma,\gamma^{+} be arbitrary positive numbers such that 0<γ<γ<γ+<1log20<\gamma^{-}<\gamma<\gamma^{+}<1-\log{2}. We partition [m][m] into disjoint sets SS and LL where iSi\in S when siγms_{i}\leq{\gamma m} and iLi\in L when si>γms_{i}>\gamma m, allowing us to rewrite our desired sum as

i=1mαi1eαi=iSαi1eαi+iLαi1eαi.\displaystyle\sum_{i=1}^{m}\alpha_{i}^{-1}e^{-\alpha_{i}}=\displaystyle\sum_{i\in S}\alpha_{i}^{-1}e^{-\alpha_{i}}+\displaystyle\sum_{i\in L}\alpha_{i}^{-1}e^{-\alpha_{i}}.

Define

Rm(x)=exEm(x)=k=m+1xkk!.R_{m}(x)=e^{x}-E_{m}(x)=\displaystyle\sum_{k=m+1}^{\infty}\frac{x^{k}}{k!}.

The following results are proven by Conrey and Ghosh [CG88] and Zemyan [Zem05]:

Proposition 2.3.
  • (a)

    [CG88, Equations (6) and (7)] For sufficiently large mm, we have |αi|meγ1|\alpha_{i}|\geq{me^{\gamma^{-}-1}} for iLi\in L and |αi|meγ+1|\alpha_{i}|\leq{me^{\gamma^{+}-1}} for iSi\in S. Consequently, we have |αi|<m/2|\alpha_{i}|<m/2 for iSi\in S.

  • (b)

    [CG88, Lemma 1] For |x|<12(m+2)|x|<\frac{1}{2}(m+2), we have

    1Rm(x)=(m+1)!xm+1(1+k=1ckxk)\frac{1}{R_{m}(x)}=\frac{(m+1)!}{x^{m+1}}\left(1+\displaystyle\sum_{k=1}^{\infty}c_{k}x^{k}\right)

    with |ck|12(2m+2)k|c_{k}|\leq{\frac{1}{2}(\frac{2}{m+2})^{k}}.

  • (c)

    [Zem05, Theorem 7]

    i=1mαit={1t=1,02tm,1/m!t=m+1,1/m!t=m+2.\displaystyle\sum_{i=1}^{m}\alpha_{i}^{-t}=\left\{\begin{array}[]{ll}-1&t=1,\\ 0&2\leq{t}\leq{m},\\ 1/m!&t=m+1,\\ -1/m!&t=m+2.\end{array}\right.

We begin our argument by restricting our attention to the indices in LL:

Lemma 2.4.
|iLαi1eαi|γ1eγm.\left|\displaystyle\sum_{i\in L}\alpha_{i}^{-1}e^{-\alpha_{i}}\right|\leq{\gamma^{-1}e^{-\gamma m}}.
Proof.

By the triangle inequality,

|iLαi1eαi|iL|αi1eαi|.\left|\displaystyle\sum_{i\in L}\alpha_{i}^{-1}e^{-\alpha_{i}}\right|\leq{\displaystyle\sum_{i\in L}|\alpha_{i}^{-1}e^{-\alpha_{i}}|}.

Note that |αi1eαi|=|αi1||esi|<|αi1|eγm|\alpha_{i}^{-1}e^{-\alpha_{i}}|=|\alpha_{i}^{-1}||e^{-s_{i}}|<|\alpha_{i}^{-1}|e^{-\gamma m}. Since si>γms_{i}>\gamma m, we know that |αi|>γm|\alpha_{i}|>\gamma m, so |αi1|<(γm)1|\alpha_{i}^{-1}|<(\gamma m)^{-1}. Thus for iLi\in L, we have that

|αi1eαi|<(γm)1eγm.|\alpha_{i}^{-1}e^{-\alpha_{i}}|<(\gamma m)^{-1}e^{-\gamma m}.

Adding over the elements of LL, of which there are at most mm, we have |iLαi1eαi|γ1eγm|\displaystyle\sum_{i\in L}\alpha_{i}^{-1}e^{-\alpha_{i}}|\leq{\gamma^{-1}e^{-\gamma m}}. ∎

To evaluate the sum for the indices in SS, we utilize Rm(x)R_{m}(x). Since the αi\alpha_{i}’s are the roots of Em(x)E_{m}(x), we have eαi=Rm(αi)e^{\alpha_{i}}=R_{m}(\alpha_{i}) for i=1,,mi=1,\cdots,m, so eαi=1Rm(αi)e^{-\alpha_{i}}=\frac{1}{R_{m}(\alpha_{i})}.

Lemma 2.5.

For |x|<12(m+2)|x|<\frac{1}{2}(m+2), we have

1Rm(x)=(m+1)!xm+1(1xm+2+k=2ckxk)\frac{1}{R_{m}(x)}=\frac{(m+1)!}{x^{m+1}}\left(1-\frac{x}{m+2}+\displaystyle\sum_{k=2}^{\infty}c_{k}x^{k}\right)

with |ck|12(2m+2)k|c_{k}|\leq{\frac{1}{2}(\frac{2}{m+2})^{k}}.

Proof.

All of this follows from Proposition 2.3(b) except for showing that c1=1/(m+2)c_{1}=-1/(m+2). To show this, we observe by definition of Rm(x)R_{m}(x) that

xm+1(m+1)!Rm(x)=(1+(m+1)!k=1xk(m+1+k)!)1.\frac{x^{m+1}}{(m+1)!R_{m}(x)}=\left(1+(m+1)!\displaystyle\sum_{k=1}^{\infty}\frac{x^{k}}{(m+1+k)!}\right)^{-1}.

Conrey and Ghosh [CG88] note that |(m+1)!k=1xk(m+1+k)!|<1\left|(m+1)!\displaystyle\sum_{k=1}^{\infty}\frac{x^{k}}{(m+1+k)!}\right|<1 when |x|<12(m+2)|x|<\frac{1}{2}(m+2), so this can be expanded as a convergent geometric series in such cases. Thus,

1Rm(x)=(m+1)!xm+1(1+j=1((m+1)!k=1xk(m+1+k)!)j).\frac{1}{R_{m}(x)}=\frac{(m+1)!}{x^{m+1}}\left(1+\displaystyle\sum_{j=1}^{\infty}\left(-(m+1)!\displaystyle\sum_{k=1}^{\infty}\frac{x^{k}}{(m+1+k)!}\right)^{j}\right).

The only time an xx term can appear in the double infinite sum is when k,j=1k,j=1, so this term has coefficient (m+1)!1(m+1+1)!=1m+2-(m+1)!\frac{1}{(m+1+1)!}=\frac{-1}{m+2} as desired.

Note that for iSi\in S we have |αi|<m/2|\alpha_{i}|<m/2 from Proposition 2.3(a). Thus we can use Lemma 2.5 to conclude that

iSαi1eαi=iSαi1Rm(αi)=(m+1)!iSk=0ckαikm2,\displaystyle\sum_{i\in S}\alpha_{i}^{-1}e^{-\alpha_{i}}=\displaystyle\sum_{i\in S}\frac{\alpha_{i}^{-1}}{R_{m}(\alpha_{i})}=(m+1)!\displaystyle\sum_{i\in S}\displaystyle\sum_{k=0}^{\infty}c_{k}\alpha_{i}^{k-m-2}, (9)

where c0=1c_{0}=1 and c1=1m+2c_{1}=\frac{-1}{m+2}.

Using the values of i=1mαit\displaystyle\sum_{i=1}^{m}\alpha_{i}^{-t} for t=1,,m+2t=1,\cdots,m+2 from Proposition 2.3(c) and that c0=1c_{0}=1 and c1=1m+2c_{1}=\frac{-1}{m+2}, we see that

k=0m+1iSckαikm2=k=0m+1(i=1mckαikm2iLckαikm2)=1m!1m!(m+2)cm+1k=0m+1iLckαikm2.\displaystyle\sum_{k=0}^{m+1}\displaystyle\sum_{i\in S}c_{k}\alpha_{i}^{k-m-2}=\displaystyle\sum_{k=0}^{m+1}\left(\displaystyle\sum_{i=1}^{m}c_{k}\alpha_{i}^{k-m-2}-\displaystyle\sum_{i\in L}c_{k}\alpha_{i}^{k-m-2}\right)=\frac{-1}{m!}-\frac{1}{m!(m+2)}-c_{m+1}-\displaystyle\sum_{k=0}^{m+1}\displaystyle\sum_{i\in L}c_{k}\alpha_{i}^{k-m-2}.

Plugging this into (9) gives

iSαi1eαi\displaystyle\displaystyle\sum_{i\in S}\alpha_{i}^{-1}e^{-\alpha_{i}} =(m+1)!(1m!1m!(m+2)cm+1k=0m+1iLckαikm2+iSk=m+2ckαikm2)\displaystyle=(m+1)!\left(\frac{-1}{m!}-\frac{1}{m!(m+2)}-c_{m+1}-\displaystyle\sum_{k=0}^{m+1}\displaystyle\sum_{i\in L}c_{k}\alpha_{i}^{k-m-2}+\displaystyle\sum_{i\in S}\displaystyle\sum_{k=m+2}^{\infty}c_{k}\alpha_{i}^{k-m-2}\right)
=(m+1)m+1m+2(m+1)!(cm+1+k=0m+1iLckαikm2iSk=m+2ckαikm2)\displaystyle=-(m+1)-\frac{m+1}{m+2}-(m+1)!\left(c_{m+1}+\displaystyle\sum_{k=0}^{m+1}\displaystyle\sum_{i\in L}c_{k}\alpha_{i}^{k-m-2}-\displaystyle\sum_{i\in S}\displaystyle\sum_{k=m+2}^{\infty}c_{k}\alpha_{i}^{k-m-2}\right)
=m2+1m+2(m+1)!(cm+1+k=0m+1iLckαikm2iSk=m+2ckαikm2).\displaystyle=-m-2+\frac{1}{m+2}-(m+1)!\left(c_{m+1}+\displaystyle\sum_{k=0}^{m+1}\displaystyle\sum_{i\in L}c_{k}\alpha_{i}^{k-m-2}-\displaystyle\sum_{i\in S}\displaystyle\sum_{k=m+2}^{\infty}c_{k}\alpha_{i}^{k-m-2}\right).

By Lemma 2.4, we have i=1mαi1eαi=iSαi1eαi+O(eβm)\displaystyle\sum_{i=1}^{m}\alpha_{i}^{-1}e^{-\alpha_{i}}=\displaystyle\sum_{i\in S}\alpha_{i}^{-1}e^{-\alpha_{i}}+O(e^{-\beta m}) for some β>0\beta>0, so we are left to consider the sum over SS. The first three terms above match our claimed expression, so it suffices to show that the leftover terms (m+1)!(cm+1+k=0m+1iLckαikm2iSk=m+2ckαikm2)-(m+1)!\left(c_{m+1}+\displaystyle\sum_{k=0}^{m+1}\displaystyle\sum_{i\in L}c_{k}\alpha_{i}^{k-m-2}-\displaystyle\sum_{i\in S}\displaystyle\sum_{k=m+2}^{\infty}c_{k}\alpha_{i}^{k-m-2}\right) are O(eβm)O(e^{-\beta m}).

Using the Triangle Inequality, and recalling that |ck|(2m)k|c_{k}|\leq{(\frac{2}{m})^{k}} for all kk by Proposition 2.3(b), we obtain

|cm+1+k=0m+1iLckαikm2iSk=m+2ckαikm2|\displaystyle\left|c_{m+1}+\displaystyle\sum_{k=0}^{m+1}\displaystyle\sum_{i\in L}c_{k}\alpha_{i}^{k-m-2}-\displaystyle\sum_{i\in S}\displaystyle\sum_{k=m+2}^{\infty}c_{k}\alpha_{i}^{k-m-2}\right|
\displaystyle\leq (2/m)m+1+k=0m+1iL(2/m)k|αi|km2+iSk=m+2(2/m)k|αi|km2.\displaystyle{(2/m)^{m+1}+\displaystyle\sum_{k=0}^{m+1}\displaystyle\sum_{i\in L}(2/m)^{k}|\alpha_{i}|^{k-m-2}+\displaystyle\sum_{i\in S}\displaystyle\sum_{k=m+2}^{\infty}(2/m)^{k}|\alpha_{i}|^{k-m-2}}.

Since |L|,|S|m|L|,|S|\leq{m}, this is at most (2/m)m+1+mk=0m+1(2/m)k|αi|km2+mk=m+2(2/m)k|αi|km2(2/m)^{m+1}+m\displaystyle\sum_{k=0}^{m+1}(2/m)^{k}|\alpha_{i}|^{k-m-2}+m\displaystyle\sum_{k=m+2}^{\infty}(2/m)^{k}|\alpha_{i}|^{k-m-2}.

Now we make use of Proposition 2.3(a). In the first summation, the quantity km2k-m-2 is negative, so |αi|km2(meγ1)km2|\alpha_{i}|^{k-m-2}\leq{(me^{\gamma^{-}-1})^{k-m-2}}. In the second summation, km2k-m-2 is nonnegative, so |αi|km2(meγ+1)km2|\alpha_{i}|^{k-m-2}\leq{(me^{\gamma^{+}-1})^{k-m-2}}. Putting this altogether, we have

|cm+1+k=0m+1iLckαikm2iSk=m+2ckαikm2|\displaystyle\left|c_{m+1}+\displaystyle\sum_{k=0}^{m+1}\displaystyle\sum_{i\in L}c_{k}\alpha_{i}^{k-m-2}-\displaystyle\sum_{i\in S}\displaystyle\sum_{k=m+2}^{\infty}c_{k}\alpha_{i}^{k-m-2}\right|
(2/m)m+1+mk=0m+1(2/m)k(meγ1)km2+mk=m+2(2/m)k(meγ+1)km2\displaystyle\leq{(2/m)^{m+1}+m\displaystyle\sum_{k=0}^{m+1}(2/m)^{k}(me^{\gamma^{-}-1})^{k-m-2}+m\displaystyle\sum_{k=m+2}^{\infty}(2/m)^{k}(me^{\gamma^{+}-1})^{k-m-2}}
=(2/m)m+1+mm1(eγ1)m+2k=0m+1(2eγ1)k+mm1(eγ+1)m+2k=m+2(2eγ+1)k.\displaystyle=(2/m)^{m+1}+\frac{m^{-m-1}}{(e^{\gamma^{-}-1})^{m+2}}\displaystyle\sum_{k=0}^{m+1}(2e^{\gamma^{-}-1})^{k}+\frac{m^{-m-1}}{(e^{\gamma^{+}-1})^{m+2}}\displaystyle\sum_{k=m+2}^{\infty}(2e^{\gamma^{+}-1})^{k}.

Note that the first summation is finite and it is bounded above by the convergent infinite sum k=0(2eγ1)k\sum_{k=0}^{\infty}(2e^{\gamma^{-}-1})^{k}, which is a constant. Since 0<γ+<1log20<\gamma^{+}<1-\log{2}, we have that 2eγ+1<12e^{\gamma^{+}-1}<1, so the second summation also converges. Let CC be some constant which serves as an upper bound for both of these summations. The total expression is then at most

(2/m)m+1+2Cmm1(eγ1)m+2.(2/m)^{m+1}+2C\frac{m^{-m-1}}{(e^{\gamma^{-}-1})^{m+2}}.

We need to show that this expression will still be O(eβm)O(e^{-\beta m}) after we multiply it by (m+1)!(m+1)!. By the Stirling approximation, m!m! is asymptotically 2πm(me)m\sqrt{2\pi m}(\frac{m}{e})^{m}, so for sufficiently large mm, we have

(m+1)!=(m+1)m!(m+1)(m1)(m/e)mm2(m/e)m.(m+1)!=(m+1)m!\leq{(m+1)(m-1)(m/e)^{m}}\leq{m^{2}(m/e)^{m}}.

Now we examine

m2(m/e)m((2/m)m+1+2Cmm1(eγ1)m+2)\displaystyle m^{2}(m/e)^{m}\left((2/m)^{m+1}+2C\frac{m^{-m-1}}{(e^{\gamma^{-}-1})^{m+2}}\right) =2m+1emm+2Cmem(eγ1)m+2.\displaystyle=\frac{2^{m+1}}{e^{m}}m+2C\frac{m}{e^{m}(e^{\gamma^{-}-1})^{m+2}}.

Let D:=(eγ1)1D:=(e^{\gamma^{-}-1})^{-1}. Since γ<1log2\gamma^{-}<1-\log{2}, we have that eγ1<12e^{\gamma^{-}-1}<\frac{1}{2} and hence that D>2D>2. However, γ\gamma^{-} can be chosen arbitrarily close to 1log21-\log{2} to ensure D<eD<e. In this case

m2(m/e)m((2/m)m+1+2Cmm1(eγ1)m+2)\displaystyle m^{2}(m/e)^{m}\left((2/m)^{m+1}+2C\frac{m^{-m-1}}{(e^{\gamma^{-}-1})^{m+2}}\right) =mem(2m+1+2CDm+2)\displaystyle=me^{-m}\left(2^{m+1}+2CD^{m+2}\right)
=O(mem(2m+CD2Dm))\displaystyle=O\left(me^{-m}(2^{m}+CD^{2}D^{m})\right)
=O(memDm)\displaystyle=O\left(me^{-m}D^{m}\right)
=O(em(logmm1+logD)).\displaystyle=O\left(e^{m(\frac{\log{m}}{m}-1+\log{D})}\right).

Note that 1+logD<0-1+\log{D}<0 and that the logmm\frac{\log{m}}{m} is negligible for sufficiently large mm, so indeed the sum we are considering is O(eβm)O(e^{-\beta m}) for some positive β\beta, completing our proof of Theorem 1.1(b).

3 Proof of Theorem 1.3

At a high level, the proof of Theorem 1.3 revolves around showing that Pr[Lm,n(π)k]\Pr[L_{m,n}(\pi)\geq k] tends to 0 or 1 depending on if nmk/k!nm^{k}/k! tends to 0 or infinity as nn tends to infinity. The following lemma (which we will apply for t!nt!\approx n) will be used to determine the threshold when nmk/k!nm^{k}/k! shifts from being very small to very large. Here and throughout the text, log\log denotes the natural logarithm.

Lemma 3.1.

Given integers m1m\geq 1 and t2t\geq 2, let C>0C>0 be a real number such that k=t+Clogmlogttk=t+\frac{C\log m}{\log t}t is an integer. We have

k!t!mCt,k!\geq t!\cdot m^{Ct},

and if tm10Ct\geq m^{10C}, we have

k!t!(1.1)kmCtk!\leq t!\cdot(1.1)^{k}m^{Ct}

Here and throughout the text we define the falling factorial (N)a:=N(N1)(Na+1)(N)_{a}:=N(N-1)\cdots(N-a+1).

Proof.

Note that k!=t!(k)ktk!=t!(k)_{k-t}, so it suffices to show

mCt(k)kt(1.1)kmCt,m^{Ct}\leq(k)_{k-t}\leq(1.1)^{k}m^{Ct},

with the upper bound holding when tm10Ct\geq m^{10C}. When tm10Ct\geq m^{10C}, we have logt10Clogm\log t\geq 10C\log m, so k=t+Clogmlogtt1.1tk=t+\frac{C\log m}{\log t}t\leq 1.1t. This implies

(k)ktkkt(1.1t)kt(1.1)ktkt=(1.1)kmCt.(k)_{k-t}\leq k^{k-t}\leq(1.1t)^{k-t}\leq(1.1)^{k}\cdot t^{k-t}=(1.1)^{k}\cdot m^{Ct}.

Similarly (k)kttkt=mCt(k)_{k-t}\geq t^{k-t}=m^{Ct} for all tt, proving the result. ∎

Before delving into the details of the proof, we introduce some auxiliary definitions that will make our arguments somewhat cleaner. The main idea is that we wish to reduce multiset permutations to set permutations by labeling each of the mm copies of i[n]i\in[n].

To this end, let 𝔖m,n\mathfrak{S}_{m,n}^{*} denote the set of permutations of the set {ih:i[n],h[m]}\{i_{h}:i\in[n],\ h\in[m]\}. For example, τ:=312132122211𝔖2,3\tau^{\prime}:=3_{1}2_{1}3_{2}1_{2}2_{2}1_{1}\in\mathfrak{S}_{2,3}^{*}. If τ𝔖m,n\tau\in\mathfrak{S}_{m,n}^{*} contains a subsequence of the form (w1)x1(wk)xk(w_{1})_{x_{1}}\cdots(w_{k})_{x_{k}}, then we will say that τ\tau has a subsequence of type (w,x)(w,x) where w=w1wkw=w_{1}\cdots w_{k} and x=x1xkx=x_{1}\cdots x_{k}. We say that τ𝔖m,n\tau\in\mathfrak{S}_{m,n}^{*} has a subsequence of type ww if it has a subsequence of type (w,x)(w,x) for some xx. For example, τ\tau^{\prime} defined above has a subsequence of type (12,22)(12,22) and hence of type 1212, but it contains no subsequence of type 123123.

Observation 3.2.

If π𝔖m,n\pi\in\mathfrak{S}_{m,n} and τ𝔖m,n\tau\in\mathfrak{S}_{m,n}^{*} are chosen uniformly at random, then for any word ww with letters in [n][n] we have

Pr[π contains w as a subsequence]=Pr[τ contains a subsequence of type w].\Pr[\pi\textrm{ contains }w\textrm{ as a subsequence}]=\Pr[\tau\textrm{ contains a subsequence of type }w].

The intuition for this observation is as follows. We can view {ih:i[n],h[m]}\{i_{h}:i\in[n],\ h\in[m]\} as a deck of cards with nn card types each having mm suits, and we view τ𝔖m,n\tau\in\mathfrak{S}_{m,n}^{*} as a way of shuffling this deck. The property that τ\tau contains a subsequence of type ww is independent of the suits of the cards. Thus if we let π𝔖m,n\pi\in\mathfrak{S}_{m,n} denote the shuffling τ\tau after ignoring suits, then π\pi contains ww as a subsequence if and only if τ\tau contains a subsequence of type ww. More formally, one can prove this result by considering the map ϕ:𝔖m,n𝔖m,n\phi:\mathfrak{S}_{m,n}^{*}\to\mathfrak{S}_{m,n} which deletes the subscripts in the letters of τ𝔖m,n\tau\in\mathfrak{S}_{m,n}^{*}. We omit the details.

3.1 The Upper Bound

To prove the upper bound of Theorem 1.3, essentially the only fact we need is that there are at most nn continuously increasing subsequences of a given length kk, and as such our proof easily generalizes to a wider set of subsequence problems.

To this end, let 𝒲\mathcal{W} be a set of words with letters in [n][n]. For π𝔖m,n\pi\in\mathfrak{S}_{m,n}, we define Lm,n(π;𝒲)L_{m,n}(\pi;\mathcal{W}) to be the maximum length of a word w𝒲w\in\mathcal{W} which appears as a subsequence in π\pi. For example, if 𝒲\mathcal{W} consists of every word of the form i(i+1)ji(i+1)\cdots j for some iji\leq j, then Lm,n(π;𝒲)=Lm,n(π)L_{m,n}(\pi;\mathcal{W})=L_{m,n}(\pi). We will say that a set of words 𝒲\mathcal{W} is prefix closed if for every w1wk𝒲w_{1}\cdots w_{k}\in\mathcal{W} we have w1w𝒲w_{1}\cdots w_{\ell}\in\mathcal{W} for all k\ell\leq k.

Lemma 3.3.

Let 𝒲\mathcal{W} be a prefix closed set of words with letters in [n][n] and let 𝒲k𝒲\mathcal{W}_{k}\subseteq\mathcal{W} be the set of words of length kk in 𝒲\mathcal{W}. If π𝔖m,n\pi\in\mathfrak{S}_{m,n} is chosen uniformly at random, then

Pr[Lm,n(π;𝒲)k]|𝒲k|mkk!.\Pr[L_{m,n}(\pi;\mathcal{W})\geq k]\leq\frac{|\mathcal{W}_{k}|m^{k}}{k!}.
Proof.

For τ𝔖m,n\tau\in\mathfrak{S}_{m,n}^{*} we define Lm,n(τ;𝒲)L_{m,n}^{*}(\tau;\mathcal{W}) to be the length of a longest w𝒲w\in\mathcal{W} such that τ\tau contains a subsequence of type ww. By Observation 3.2, it suffices to bound Pr[Lm,n(τ;𝒲)k]\Pr[L_{m,n}^{*}(\tau;\mathcal{W})\geq k] with τ\tau chosen uniformly at random from 𝔖m,n\mathfrak{S}_{m,n}^{*}.

Because 𝒲\mathcal{W} is prefix closed, we have Lm,n(τ;𝒲)kL_{m,n}^{*}(\tau;\mathcal{W})\geq k if and only if τ\tau contains some subsequence of type w𝒲kw\in\mathcal{W}_{k}, and by definition this happens if and only if τ\tau contains some subsequence of type (w,x)(w,x) with w𝒲kw\in\mathcal{W}_{k} and x[m]kx\in[m]^{k}. For w𝒲kw\in\mathcal{W}_{k} and x[m]kx\in[m]^{k}, let 1w,x(τ)1_{w,x}(\tau) be the indicator variable which is 1 if τ\tau contains a subsequence of type (w,x)(w,x) and which is 0 otherwise. Let X(τ)=w𝒲k,x[m]k1w,x(τ)X(\tau)=\sum_{w\in\mathcal{W}_{k},x\in[m]^{k}}1_{w,x}(\tau). By our observations above and Markov’s inequality, we find

Pr[Lm,n(τ;𝒲)k]=Pr[X(τ)1]𝔼[X(τ)]=w𝒲k,x[m]kPr[1w,x(τ)=1]=|𝒲k|mk1k!,\Pr[L_{m,n}^{*}(\tau;\mathcal{W})\geq k]=\Pr[X(\tau)\geq 1]\leq\mathbb{E}[X(\tau)]=\sum_{w\in\mathcal{W}_{k},x\in[m]^{k}}\Pr[1_{w,x}(\tau)=1]=|\mathcal{W}_{k}|m^{k}\cdot\frac{1}{k!},

where the last step used that 1w,x(τ)=11_{w,x}(\tau)=1 if and only if the distinct letters (w1)x1,,(wk)xk(w_{1})_{x_{1}},\ldots,(w_{k})_{x_{k}} appear in the correct relative order in τ\tau, and this happens with probability 1/k!1/k!. This proves the result. ∎

Proposition 3.4.

Let 𝒲\mathcal{W} be a prefix closed set of words with letters in [n][n] and let 𝒲k𝒲\mathcal{W}_{k}\subseteq\mathcal{W} be the words of length kk in 𝒲\mathcal{W}. Assume there exists an NN such that |𝒲k|N|\mathcal{W}_{k}|\leq N for all kk. If π𝔖m,n\pi\in\mathfrak{S}_{m,n} is chosen uniformly at random and NN is sufficiently large in terms of mm, then

𝔼[Lm,n(π;𝒲)]Γ1(N)+O(1+logmlog(Γ1(N))Γ1(N)).\mathbb{E}[L_{m,n}(\pi;\mathcal{W})]\leq\Gamma^{-1}(N)+O\left(1+\frac{\log m}{\log(\Gamma^{-1}(N))}\Gamma^{-1}(N)\right).

In particular, for 𝒲\mathcal{W} the set of continuously increasing words i(i+1)ji(i+1)\cdots j, we have |𝒲k|n|\mathcal{W}_{k}|\leq n for all kk, so taking N=nN=n gives the upper bound of Theorem 1.3. As another example, if 𝒲\mathcal{W} is the set of arithmetic progressions, then one can take N=n2N=n^{2} to give an upper bound of roughly Γ1(n2)\Gamma^{-1}(n^{2}) for 𝔼[Lm,n(π;𝒲)]\mathbb{E}[L_{m,n}(\pi;\mathcal{W})]. Recent work of Goh and Zhao [GZ20] shows that this bound for arithmetic progressions is tight.

Proof.

By using Lemma 3.3 and the trivial bound Pr[Lm,n(π;𝒲)K]1\Pr[L_{m,n}(\pi;\mathcal{W})\geq K]\leq 1, we find for all integers kk that

𝔼[Lm,n(π;𝒲)]\displaystyle\mathbb{E}[L_{m,n}(\pi;\mathcal{W})] =K1Pr[Lm,n(π;𝒲)K]k+K>kNmKK!.\displaystyle=\sum_{K\geq 1}\Pr[L_{m,n}(\pi;\mathcal{W})\geq K]\leq k+\sum_{K>k}\frac{Nm^{K}}{K!}. (10)

Let tt be the integer such that (t1)!<Nt!(t-1)!<N\leq t! and let k=t+2logmlogttk=\left\lceil t+\frac{2\log m}{\log t}t\right\rceil. Note that this implies k=t+Clogmlogttk=t+\frac{C\log m}{\log t}t for some C2C\geq 2. Assume NN is sufficiently large so that 2mk2t2m\leq k\leq 2t. By Lemma 3.1, we have for K>ktK>k\geq t that

NmKK!Nmkk!(mk)KkNmkt!mCt2kK2kK,\frac{Nm^{K}}{K!}\leq\frac{Nm^{k}}{k!}\cdot\left(\frac{m}{k}\right)^{K-k}\leq\frac{Nm^{k}}{t!m^{Ct}}\cdot 2^{k-K}\leq 2^{k-K},

with this last step using k2tCtk\leq 2t\leq Ct and Nt!N\leq t!. Plugging this and our choice of kk into (10) gives, after setting =Kk\ell=K-k,

𝔼[Lm,n(π;𝒲)]t+2logmlogtt+>02t+2logmlogtt+2.\mathbb{E}[L_{m,n}(\pi;\mathcal{W})]\leq\left\lceil t+\frac{2\log m}{\log t}t\right\rceil+\sum_{\ell>0}2^{-\ell}\leq t+\frac{2\log m}{\log t}t+2.

This gives the desired result since t<Γ1(N)t<\Gamma^{-1}(N). ∎

3.2 The Lower Bound

For x,y[m]nx,y\in[m]^{n}, we define their Hamming distance dH(x,y):=|{i[n]:xiyi}|d_{H}(x,y):=|\{i\in[n]:x_{i}\neq y_{i}\}|. Our key lemma for proving the lower bound of Theorem 1.3 is the following:

Lemma 3.5.

Let T[m]nT\subseteq[m]^{n} be such that any distinct x,yTx,y\in T have dH(x,y)δd_{H}(x,y)\geq\delta for some integer δ\delta. Then

Pr[Lm,n(π)=n]|T|n!(1|T|δ!).\Pr[L_{m,n}(\pi)=n]\geq\frac{|T|}{n!}\left(1-\frac{|T|}{\delta!}\right).
Proof.

For τ𝔖m,n\tau\in\mathfrak{S}_{m,n}^{*}, let Lm,n(τ)L^{*}_{m,n}(\tau) denote the length of the longest subsequence of τ\tau of type i(i+1)ji(i+1)\cdots j. By Observation 3.2, it suffices to prove this lower bound for Pr[Lm,n(τ)=n]\Pr[L_{m,n}^{*}(\tau)=n] where τ𝔖m,n\tau\in\mathfrak{S}_{m,n}^{*} is chosen uniformly at random. For x[m]nx\in[m]^{n}, let Ax(τ)A_{x}(\tau) be the event that τ\tau contains a subsequence of type (12n,x)(12\cdots n,x). Observe that

Pr[Lm,n(τ)=n]\displaystyle\Pr[L_{m,n}^{*}(\tau)=n] =Pr[x[m]nAx(τ)]Pr[xTAx(τ)]\displaystyle=\Pr[\bigcup_{x\in[m]^{n}}A_{x}(\tau)]\geq\Pr[\bigcup_{x\in T}A_{x}(\tau)]
xTPr[Ax(τ)]x,yT,xyPr[Ax(τ)Ay(τ)],\displaystyle\geq\sum_{x\in T}\Pr[A_{x}(\tau)]-\sum_{x,y\in T,\ x\neq y}\Pr[A_{x}(\tau)\cap A_{y}(\tau)], (11)

where the last inequality used the Bonferroni inequality (which is essentially a weakening of the principle of inclusion-exclusion); see e.g. [Spe14] for further details on this inequality. To bound (11), we use the following:

Claim 3.6.

If x,yTx,y\in T with xyx\neq y, then Pr[Ax(τ)]=1/n!\Pr[A_{x}(\tau)]=1/n! and

Pr[Ax(τ)Ay(τ)]1δ!n!.\Pr[A_{x}(\tau)\cap A_{y}(\tau)]\leq\frac{1}{\delta!n!}.
Proof.

Observe that Ax(τ)A_{x}(\tau) occurs if and only if 1x1,,nxn1_{x_{1}},\ldots,n_{x_{n}} occur in the correct relative order in τ\tau, so Pr[Ax(τ)]=1/n!\Pr[A_{x}(\tau)]=1/n!. Let S={i1<i2<<iδ}S=\{i_{1}<i_{2}<\cdots<i_{\delta}\} be any set of δ\delta indices ii such that yixiy_{i}\neq x_{i}, and note that such a set exists by assumption of TT. Let AyS(τ)A_{y}^{S}(\tau) be the event that (i1)yi1(iδ)yiδ(i_{1})_{y_{i_{1}}}\cdots(i_{\delta})_{y_{i_{\delta}}} is a subsequence of τ\tau. Observe that Pr[AyS(τ)]=1/δ!\Pr[A_{y}^{S}(\tau)]=1/\delta! and that this event is independent of Ax(τ)A_{x}(\tau) since these two events concern disjoint sets of letters. Because Ay(τ)A_{y}(\tau) implies AyS(τ)A_{y}^{S}(\tau), we have

Pr[Ax(τ)Ay(τ)]Pr[Ax(τ)AyS(τ)]=1δ!n!,\Pr[A_{x}(\tau)\cap A_{y}(\tau)]\leq\Pr[A_{x}(\tau)\cap A_{y}^{S}(\tau)]=\frac{1}{\delta!n!},

proving the result. ∎

Plugging the results of this claim into (11) and using that the second sum of (11) has at most |T|2|T|^{2} terms gives the desired result. ∎

The problem of finding T[m]nT\subseteq[m]^{n} such that dH(x,y)δd_{H}(x,y)\geq\delta with |T||T| and δ\delta both large is the central problem of coding theory. In particular, a basic greedy argument from coding theory gives the following:

Lemma 3.7.

For any m2m\geq 2 and 1δn/21\leq\delta\leq n/2, there exists T[m]nT\subseteq[m]^{n} such that any two distinct x,yTx,y\in T have dH(x,y)δd_{H}(x,y)\geq\delta and such that

|T|mnδ(nδ)(m1)δ.|T|\geq\frac{m^{n}}{\delta{n\choose\delta}(m-1)^{\delta}}.
Proof.

Let T[m]nT\subseteq[m]^{n} be a set such that dH(x,y)δd_{H}(x,y)\geq\delta for distinct x,yTx,y\in T and such that |T||T| is as large as possible. Let B(x)={y[m]n:dH(x,y)<δ}B(x)=\{y\in[m]^{n}:d_{H}(x,y)<\delta\}, and note that for all xx,

|B(x)|=d=0δ1(nd)(m1)dδ(nδ)(m1)δ,|B(x)|=\sum_{d=0}^{\delta-1}{n\choose d}(m-1)^{d}\leq\delta{n\choose\delta}(m-1)^{\delta},

with this last step using (nd)(nδ){n\choose d}\leq{n\choose\delta} for d<δn/2d<\delta\leq n/2. By the maximality of |T||T|, we must have [m]n=xTB(x)[m]^{n}=\bigcup_{x\in T}B(x), and thus

mn=|xTB(x)||T|δ(nδ)(m1)δ,m^{n}=|\bigcup_{x\in T}B(x)|\leq|T|\cdot\delta{n\choose\delta}(m-1)^{\delta},

giving the desired bound on |T||T|. ∎

Combining Lemmas 3.5 and 3.7 gives the following:

Proposition 3.8.

For nn sufficiently large in terms of m2m\geq 2, we have

Pr[Lm,n(π)=n](m/1.03)n2nn!.\Pr[L_{m,n}(\pi)=n]\geq\frac{(m/1.03)^{n}}{2n\cdot n!}.
Proof.

We start with the following fact.

Claim 3.9.

For all ϵ>0\epsilon>0, there exists a constant 0<cϵ10<c_{\epsilon}\leq 1 such that if δcϵn\delta\leq c_{\epsilon}n, then (nδ)(1+ϵ)n{n\choose\delta}\leq(1+\epsilon)^{n}.

Proof.

In [Cov99] it is noted that (nδ)2H(δ/n)n{n\choose\delta}\leq 2^{H(\delta/n)n} for all n,δn,\delta, where H(p):=plog2(p)(1p)log2(1p)H(p):=-p\log_{2}(p)-(1-p)\log_{2}(1-p) is the binary entropy function. Because H(p)H(p) tends to 0 as pp tends to 0, there exists a constant cc such that 2H(c)1+ϵ2^{H(c)}\leq 1+\epsilon, and the result follows by taking cϵ=cc_{\epsilon}=c. ∎

Let δ=2logmlognn\delta=\frac{2\log m}{\log n}n, and assume nn is sufficiently large in terms of mm so that δc.01n\delta\leq c_{.01}n, i.e. so that (nδ)(1.01)n{n\choose\delta}\leq(1.01)^{n}. We also choose nn sufficiently large so that δlog1.01logmn\delta\leq\frac{\log 1.01}{\log m}n, or equivalently so that mδ(1.01)nm^{\delta}\leq(1.01)^{n}. Let TT be a set as in Lemma 3.7, and by our assumptions above we find

|T|(m/1.03)nn.|T|\geq\frac{(m/1.03)^{n}}{n}.

Possibly by deleting elements from TT we can assume that |T||T| is exactly the quantity stated above666Strictly speaking we should take |T||T| to be the floor of this value to guarantee that it is an integer. This would change our ultimate bound by at most a factor of 2, and this factor of 2 can easily be recovered by sharpening our analysis in various places., so by Lemma 3.5 it suffices to show |T|/δ!12|T|/\delta!\leq\frac{1}{2}. Using the inequality δ!(δ/e)δ\delta!\geq(\delta/e)^{\delta} and that nn is sufficiently large, we have

δ!exp[δ(log(δ)1)]exp[2logmlognn(log(n)log(log(n))1)]exp[log(m)n]=mn.\delta!\geq\exp\left[\delta\cdot(\log(\delta)-1)\right]\geq\exp\left[\frac{2\log m}{\log n}n\cdot(\log(n)-\log(\log(n))-1)\right]\geq\exp\left[\log(m)n\right]=m^{n}.

Thus |T|/δ!(1.03)n/n12|T|/\delta!\leq(1.03)^{-n}/n\leq\frac{1}{2}, proving the result. ∎

With this we can now prove Theorem 1.3.

Proof of Theorem 1.3.

The upper bound follows from Proposition 3.4. To prove the lower bound, fix an integer kk. For 0j<n/k0\leq j<\left\lfloor n/k\right\rfloor, let Aj(π)A_{j}(\pi) be the event that π\pi contains the subsequence (jk+1)(jk+2)((j+1)k)(jk+1)(jk+2)\cdots((j+1)k).

Claim 3.10.

We have the following:

  • (a)

    If any Aj(π)A_{j}(\pi) event occurs, then Lm,n(π)kL_{m,n}(\pi)\geq k,

  • (b)

    The Aj(π)A_{j}(\pi) events are mutually independent, and

  • (c)

    For all jj, we have Pr[Aj(π)]=Pr[Lm,k(σ)=k]\Pr[A_{j}(\pi)]=\Pr[L_{m,k}(\sigma)=k] where σ𝔖m,k\sigma\in\mathfrak{S}_{m,k} is chosen uniformly at random.

Proof.

Part (a) is clear, and (b) follows from the fact that the Aj(π)A_{j}(\pi) events involve the relative ordering of disjoint sets of letters. For (c), one can consider the map which sends π𝔖m,n\pi\in\mathfrak{S}_{m,n} to σ𝔖m,k\sigma\in\mathfrak{S}_{m,k} by deleting every letter in π\pi except for (jk+1),,((j+1)k)(jk+1),\ldots,((j+1)k) and then relabeling jk+ijk+i to ii. It is not difficult to see that Aj(π)A_{j}(\pi) occurs if and only if Lm,k(σ)=kL_{m,k}(\sigma)=k occurs, and that π\pi being chosen uniformly at random implies σ\sigma is chosen uniformly at random. ∎

Let pk=Pr[Lm,k(σ)=k]p_{k}=\Pr[L_{m,k}(\sigma)=k] and let tt be the integer such that t!n<(t+1)!t!\leq n<(t+1)!. The claim above implies that for all kk we have

Pr[Lm,n(π)k]Pr[Aj(π)]=1Pr[Ajc(π)]=1(1pk)n/k1exp(t!pk2k),\Pr[L_{m,n}(\pi)\geq k]\geq\Pr\left[\bigcup A_{j}(\pi)\right]=1-\Pr\left[\bigcap A_{j}^{c}(\pi)\right]=1-(1-p_{k})^{\left\lfloor n/k\right\rfloor}\geq 1-\exp\left(-\frac{t!p_{k}}{2k}\right), (12)

with this last step using n/kn/2k\left\lfloor n/k\right\rfloor\geq n/2k for knk\leq n, that 1pkepk1-p_{k}\leq e^{-p_{k}}, and that nt!n\geq t!.

It is easy to see by definition that pk1/k!p_{k}\geq 1/k! for all m,km,k; and for nn sufficiently large, we have et/2t1-e^{-t/2}\geq-t^{-1}. For such nn, by (12) we have for all kt2k\leq t-2 that

Pr[Lm,n(π)k]1exp(t!2kk!)1exp(t2)1t1.\Pr[L_{m,n}(\pi)\geq k]\geq 1-\exp\left(-\frac{t!}{2k\cdot k!}\right)\geq 1-\exp\left(-\frac{t}{2}\right)\geq 1-t^{-1}. (13)

Summing this bound over all kt2k\leq t-2 for m=1m=1 gives

𝔼[L1,n(π)]kt2Pr[L1,n(π)k]t3.\mathbb{E}[L_{1,n}(\pi)]\geq\sum_{k\leq t-2}\Pr[L_{1,n}(\pi)\geq k]\geq t-3.

This gives the desired lower bound of Γ1(n)+Ω(1)\Gamma^{-1}(n)+\Omega(1) for m=1m=1 since tΓ1(n)2t\geq\Gamma^{-1}(n)-2.

We now consider m2m\geq 2. By Proposition 3.8 we have for kk sufficiently large in terms of mm that pk(m/1.03)k2kk!p_{k}\geq\frac{(m/1.03)^{k}}{2k\cdot k!}. Let nn be large enough in terms of mm so that this bound holds for ktk\geq t. Also let nn be large enough so that logmlogt1\frac{\log m}{\log t}\leq 1. By Lemma 3.1, if tkt+logm100logtt1.01tt\leq k\leq t+\frac{\log m}{100\log t}t\leq 1.01t, then k!t!(1.1)kmt/100k!\leq t!(1.1)^{k}m^{t/100}. Thus by (12) we have

Pr[Lm,n(π)k]\displaystyle\Pr[L_{m,n}(\pi)\geq k] 1exp((m/1.03)k4k2(1.1)kmt/100)1exp(m.99t4k2(1.14)1.01t)\displaystyle\geq 1-\exp\left(-\frac{(m/1.03)^{k}}{4k^{2}(1.1)^{k}m^{t/100}}\right)\geq 1-\exp\left(-\frac{m^{.99t}}{4k^{2}\cdot(1.14)^{1.01t}}\right)
1exp((1.7)t8t2),\displaystyle\geq 1-\exp\left(-\frac{(1.7)^{t}}{8t^{2}}\right),

where this last step used m2m\geq 2. This quantity is at least 1/21/2 for nn (and hence tt) sufficiently large. Using this together with (13) for kt2k\leq t-2 gives, for nn sufficiently large in terms of mm,

𝔼[Lm,n(π)]t3+tkt+logm100logttPr[Lm,n(π)k]t3+(logm100logtt)12,\mathbb{E}[L_{m,n}(\pi)]\geq t-3+\sum_{t\leq k\leq t+\frac{\log m}{100\log t}t}\Pr[L_{m,n}(\pi)\geq k]\geq t-3+\left(\frac{\log m}{100\log t}t\right)\cdot\frac{1}{2},

proving the desired result. ∎

4 Concluding Remarks

In this paper we solved a conjecture of Diaconis, Graham, He, and Spiro [DGHS21] by asymptotically determining 𝔼[Lm,n1(π)]\mathbb{E}[L^{1}_{m,n}(\pi)] provided nn is sufficiently large in terms of mm. Using similar ideas, it is possible to compute the asymptotic limit of the rthr^{\text{th}} moment 𝔼[Lm,n1(π)r]\mathbb{E}[L^{1}_{m,n}(\pi)^{r}] for any fixed rr. Based off of computational evidence for these higher moments, we conjecture the following:

Conjecture 4.1.

For all r1r\geq 1, if nn is sufficiently large in terms of mm, then

𝔼[(Lm,n1(π)μ)r]=crmr/2+O(mr/21),\mathbb{E}[(L^{1}_{m,n}(\pi)-\mu)^{r}]=c_{r}m^{\left\lfloor r/2\right\rfloor}+O(m^{\left\lfloor r/2\right\rfloor-1}),

where μ=𝔼[Lm,n1(π)]\mu=\mathbb{E}[L^{1}_{m,n}(\pi)] and

cr={r!2r/2(r/2)!r even,r!32(r1)/2((r3)/2)!r odd.c_{r}=\begin{cases}\frac{r!}{2^{r/2}(r/2)!}&r\textrm{ even},\\ \frac{r!}{3\cdot 2^{(r-1)/2}((r-3)/2)!}&r\textrm{ odd}.\end{cases}

One can show that the standard deviation σ\sigma of Lm,n1(π)L^{1}_{m,n}(\pi) is asymptotic to m1/2m^{1/2}. Thus, this conjecture would imply that the standardized moments (Lm,n1(π)μσ)r(\frac{L^{1}_{m,n}(\pi)-\mu}{\sigma})^{r} converge to 0 for rr odd and to r!2r/2(r/2)!\frac{r!}{2^{r/2}(r/2)!} for rr even. These are exactly the moments of a standard normal distribution, and actually this fact would imply that (Lm,n1(π)μ)/σ(L_{m,n}^{1}(\pi)-\mu)/\sigma converges in distribution to a standardized normal distribution, see for example [FK16, Corollary 21.8].

Perhaps a first step towards proving Conjecture 4.1 would be to get a better understanding of the (non-centralized) moments 𝔼[Lm,n1(π)r]\mathbb{E}[L_{m,n}^{1}(\pi)^{r}], and to this end we conjecture the following:

Conjecture 4.2.

For all r1r\geq 1, if nn is sufficiently large in terms of mm, then

𝔼[Lm,n1(π)r]=mr+(r+12)mr1+O(mr2).\mathbb{E}[L^{1}_{m,n}(\pi)^{r}]=m^{r}+{r+1\choose 2}m^{r-1}+O(m^{r-2}).

We can prove that the rthr^{\text{th}} moment is asymptotic to mrm^{r}, but we do not know how to determine the coefficient of mr1m^{r-1}. We were unable to observe any pattern for the coefficients of lower order terms.

In this paper, we considered continuously increasing subsequences in multiset permutations, and it is natural to consider other types of subsequences in multiset permutations. Perhaps the most natural to consider is the following:

Question 4.3.

For π𝔖m,n\pi\in\mathfrak{S}_{m,n}, let L~m,n(π)\widetilde{L}_{m,n}(\pi) denote the length of a longest increasing subsequence in π\pi. What is 𝔼[L~m,n(π)]\mathbb{E}[\widetilde{L}_{m,n}(\pi)] asymptotic to when mm is fixed?

When m=1m=1 it is well known that 𝔼[L~1,n(π)]2n\mathbb{E}[\widetilde{L}_{1,n}(\pi)]\sim 2\sqrt{n} (see [Rom15]), so Question 4.3 is a natural generalization of this classical problem. See also recent work of Almeanazel and Johnson [AMJ20] for some results concerning the distribution of L~m,n(π)\widetilde{L}_{m,n}(\pi).

Acknowledgments. We thank S. Venkitesh for fruitful conversations and Persi Diaconis for comments regarding an earlier draft. We thank the Graduate Student Combinatorics Conference 2021 for hosting an open problem session, at which the fourth author presented the problem which inspired this work.

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