Number of Singletons in Involutions of large size: a central range and a large deviation analysis

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Number of Singletons in Involutions of large size: a

central range and a large deviation analysis

Guy Louchard

To cite this version:

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Number of Singletons in Involutions of large size: a

central range and a large deviation analysis

Guy Louchard* December 14, 2016

Abstract

In this paper, we analyze the asymptotic number I(m, n) of involutions of large size n with m singletons. We consider a central region and a non-central region. In the range m = n−nα_{, 0 < α <}

1, we analyze the dependence of I(m, n) on α. This paper fits within the framework of Analytic Combinatorics.

Keywords: Involutions, Singletons, Asymptotics, Saddle point method, Multiseries expansions, Analytic Combinatorics.

2010 Mathematics Subject Classification: 05A16 60C05 60F05.

1 Introduction

During the last few years, we have been interested in asymptotic properties of some permutations parameters. For instance, in [9] (in cooperation with H.Prodinger), [7], [8], we analyzed the number of inversions, of cycles (related to the Stirling numbers of the first kind), of rises (related to Eulerian numbers). We extended the Gaussian approximation with more terms and we also considered some large deviation expansions. In this paper, we turn to another property: the number of singletons in involutions: an involution is a permutation σ such that σ2is the identity permutation. this corresponds to cycles of size 1 and 2. See Bona [1] for details. We will use the Saddle point method: see Flajolet and Sedgewick [2, ch. VIII] for a nice introduction.

We denote by In the total number of involutions of size n and by I(m, n) the number of all

involutions of size n with m singletons . We define a random variable Jn by the relation

P(Jn= m) =

I(m, n) In

.

This the number of singletons in an involution chosen (uniformly) at random among all involutions of size n. In [2, ch. VIII, p.558], using the Saddle point technique, Flajolet and Sedgewick give the first terms of the asymptotic expansion of In, also obtained by Knuth [6]. See also Moser and Wyman

[10]. In Section 2, we provide a more detailed expansion of In. In [2, ch. VIII, p.691], the authors

provide the first terms of the mean and variance of Jn. In Section 2, we consider a detailed analysis of

all moments of Jn. In [2, ch. VIII, p.692], the authors prove the dominant Gaussian asymptotic of Jn

by using together the Saddle point technique and a generalized quasi-powers technique (see Sachkov [11], Hwang [4], [5]). In Section 3, we give a detailed analyzis of the asymptotic distribution of Jn. In

Section 4, we consider a large deviation range: m = n − nα, 0 < α < 1. In this section, we will use multiseries expansions: multiseries are in effect power series (in which the powers may be non-integral but must tend to infinity) and the variables are elements of a scale. The scale is a set of variables of increasing order. The series is computed in terms of the variable of maximum order, the coefficients

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of which are given in terms of the next-to-maximum order, etc. This is more precise than mixing different terms.

An appendix provides a justification of some integration procedures.

Note finally that our approach can be used in generalizations of the involution: we can deal with cycles of any chosen sizes and deal with singletons or other specific cycle.

2 The moments

We have the classical (exponential) generating functions

f1(z) = ∞ X n=0 In zn n! = e z+z2_/2 , f2(z, y) = ∞ X m=0 ∞ X n=0 I(m, n)z n n!y m_{= e}zy+z2_/2 , f3(z, m) = ∞ X n=0 I(m, n)z n n! = e z2_/2zm m!. We note that m and n do have the same parity: n − m is even.

We define m`:=Q`−1

j=0(m − j) as the `th falling factorial of m.

We have E(Jn`) = ∞ X m=0 m`P(Jn= m) = Sl In , Sl= ∞ X m=0 m`I(m, n) = n! [zn] ∂ ` ∂y`f2(z, y) y=1 = n! [zn]z`ez+z2/2= n![zn−`]ez+z2/2, E(Jn`) = n! In In−` (n − `)!.

Now we turn to an asymptotic expansion of In.

Let Ω denote the circle ρeiθ. By Cauchy’s theorem, it follows that

In/n! = 1 2πi Z Ω f1(z) zn+1 dz = 1 ρn 1 2π Z π −π

f1(ρeiθ)e−niθdθ using z = ρeiθ

= 1 ρn 1 2π Z π −π

exp ln(f1(ρeiθ)) − niθ dθ

= 1 ρn f1(ρ) 2π Z π −π exp iκ1θ − niθ − 1 2κ2θ 2₋ i 6κ3θ 3_{+ · · ·} dθ, (1) where κi(ρ) := ∂ ∂u i ln(f1(ρeu))|u=0.

Now we have κ1 = ρ + ρ2 and we set κ1− n = 0 such that the saddle point is the root (of smallest

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ρ = −1 2+ 1 2 √ 4n + 1 =√n − 1/2 + 1/8√1 n− 1 128n −3/2 + 1 1024n −5/2₋ 5 32768n −7/2 + O 1 n9/2 , ln(ρ) = 1/2 ln (n) − 1/2√1 n+ 1/48 n −3/2₋ 3 1280n −5/2₊ 5 14336n −7/2₋ 35 589824n −9/2_{+ O} 1 n5 . See Good [3] for a neat description of this technique.

The dominant part of (1) gives f1(ρ)

ρn = exp(E1),

E1= ρ + ρ2/2 − n ln(ρ) = n/2 + ρ/2 − n ln(ρ),

with the substitution ρ2= n − ρ. (This substitution will be frequently used in the sequel.) Now we turn to the integral. We have

κ2 = −ρ + 2n,

and more generally

κj = −(2j−1− 1)ρ + 2j−1n.

We choose a splitting value θ0 such that κ2θ20 → ∞, and κ3θ30 → 0, as n → ∞. If we choose θ0 = nβ,

we must have n2β+1→ ∞, n3β+1_{→ 0. For instance, we can use θ}

0 = n−5/12. We must prove that the

integral

Kn=

Z 2π−θ0

θ0

exp ln(f1(ρeiθ)) − niθ dθ

is such that |Kn| is exponentially small (tail pruning). This is done in [2, ch. VIII, p.559]. Now we

use the classical trick of setting

∞ X j=2 κj(iθ)j/j! = 1 2 h

(n − ρ)(e2iθ− 1 − 2iθ)i+ ρ(eiθ− 1 − iθ) = −u2/2.

Computing θ as a series in u, this gives, by Lagrange’s inversion, θ = ∞ X i=1 ai ui ni/2, (2)

with, for instance, a1= 1/2 √ 2 + 1/8 √ 2 √ n − 1 64 √ 2 n − 3 256 √ 2 n3/2 + 11 4096 √ 2 n2 + O 1 n5/2 , a2= −1/6 i − 1/24 i √ n + 1/48 i n + 1 192i n3/2 − 1 192i n2 + O 1 n5/2 . This expansion is valid in the dominant integration domain

|u| ≤ √

n θ0

a1

= n1/12.

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integration gives 1 2√π√nF1, F1 = 1 + 1/4 1 √ n− 19 96n −1₋ 13 384n −3/2 + O 1 n2 , In n! ∼ 1 2√π√nF1exp(E1). Now we turn to In−` (n−`)!. We successively have ρ`+ ρ2` − n + ` = 0,

ρ2_` = n − ` − ρ` is used as a next substitution,

ρ` = √ n − 1/2 +1/8 − 1/2 `√ n + − 1 128+ 1/16 ` − 1/8 ` 2 n−3/2+ 1 1024− 3 256` + 3 64` 2_{− 1/16 `}3 n−5/2 + − 5 32768+ 5 2048` − 15 1024` 2₊ 5 128` 3₋ 5 128` 4 n−7/2+ O 1 n9/2 , ln(ρ`) = 1/2 ln (n) − 1/2 1 √ n− 1/2 ` n + 1/48 − 1/4 ` n3/2 − 1/4 `2 n2 + − 3 1280+ 1/32 ` − 3/16 ` 2 n−5/2− 1/6 ` 3 n3 + 5 14336− 3 512` + 5 128` 2₋ 5 32` 3 n−7/2− 1/8 ` 4 n4 + − 35 589824+ 5 4096` − 21 2048` 2₊ 35 768` 3₋ 35 256` 4 n−9/2+ O 1 n5 , f1(ρ`) ρn_` = exp(E1,`), E1,` = n/2 + ρ`/2 − `/2 − (n − `) ln(ρ`), κj,`= −(2j−1− 1)ρ`+ 2j−1(n − `), θ is again given by (2), a1,` = 1/2 √ 2 + 1/8 √ 2 √ n+ − 1 64 √ 2 + 1/4 `√2 n−1+ 1/8 `√2 − 3 256 √ 2 n−3/2+ O 1 n2 , F1,` = 1 + 1/4 1 √ n+ −19 96 + 1/2 ` n−1+ 1/4 ` − 13 384 n−3/2+ O 1 n2 , In−` (n − `)! ∼ 1 2√π√nF1,`exp(E1,`).

Note that setting ` = 0, we recover of course In. The detailed expansions of ρ` and ln(ρ`) are used in

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We are now ready to compute E(Jn`) = S_I_n`. We derive exp(E1,`− E1) = n`/2T1, T1 = 1 − 1/2 ` √ n− 1/8 `2 n + 1/48 ` − 1/8 `2+ 5 48` 3 n−3/2+ − 1 96` 2_{+ 1/16 `}3₊ 1 384` 4 n−2 + − 1 384` 3₊ 1 64` 4₋ 41 3840` 5 n−5/2+ O 1 n3 , F1,` F1 = T2, T2 = 1 + 1/2 ` n+ 1/8 ` n3/2 + O 1 n2 , T3 = T1T2= 1 − 1/2 ` √ n + −1/8 `2_{+ 1/2 `} n + 7 48` − 3/8 ` 2₊ 5 48` 3 n−3/2+ O 1 n2 .

This leads to the following theorem:

Theorem 2.1 The asymptotic expansion of the factorial moments of Jn are given by

E(Jn`) = n`/2 1 − 1/2√` n+ −1/8 `2_{+ 1/2 `} n + 7 48` − 3/8 ` 2₊ 5 48` 3 n−3/2+ O 1 n2 . The first moments of Jn are now immediate:

M = S1 In =√n − 1/2 + 3/8√1 n− 1/8 n −1_{+ O} 1 n3/2 , E(Jn(Jn− 1)) = S2 In , σ2 = S2 In + M − M2 =√n − 1 + 5/8√1 n+ O 1 n , σ = √4_{n − 1/2} 1 4 √ n+ 3/16 n −3/4_{+ O} 1 n . All moments can be similarly mechanically obtained

3 Distribution of J

n

We consider the central range M − 2σ < m < M + 2σ. We now have κi(ρ) :=

∂ ∂u

i

ln(f3(ρeu))|u=0,

and m = M + xσ, x = Θ(1). This leads to κ1 = ρ2+ m,

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The dominant part of (1) gives f3(ρ) ρn = exp(E2), E2= ρ2/2 + m ln(ρ) − ln(m!) − n ln(ρ), we know that ln(m!) = −m + m ln (m) +1 2ln (2πm) + 1 12m −1_{+ O} 1 m2 , hence E2= (−1/2 ln (n) + 1/2) n + √ n − 1/4 − 1/2 x2− 1/4 ln (n) − 1/2 ln (2) − 1/2 ln (π) +−1/2 x + 1/6 x 3 4 √ n + 5 24 + 1/4 x 2_{− 1/12 x}4 1 √ n+ −1/6 x − 1/4 x3_{+ 1/20 x}5 n3/4 + O 1 n . Now we turn to the integral. Proceeding as previously, we have

1 2(n − m)(e 2iθ_{− 1 − 2iθ) = −}u2 2 , θ is again given by (2), a1 = 1/2 √ 2 + 1/4 √ 2 √ n+ 1/4 √ 2x n3/4 + 1/16 √ 2 n + 1/4 √ 2x n5/4 + −1/16√2 + 1/2√2 −1/8 − 1/8 x2 − 1/4 x2√_{2 − 8} −1/16 x2− 3 128 √ 2 n−3/2+ 7 64 √ 2x n7/4 + O 1 n2 a2 = −1/6 i − 1/6 i √ n − 1/6 ix n3/4 − 1/12 i n − 1/4 ix n5/4 + −1/6 ix2_{− 1/16 i} n3/2 − 17 96ix n7/4 + O 1 n2

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This leads to the local limit theorem:

Theorem 3.1 The asymptotic distribution of Jn in the central range is given by the local limit

theo-rem: P(Jn= m) = 2 1 √ 2πn1/4e −x2_/2 1 +−1/2 x + 1/6 x 3 4 √ n + 5 12 + 3/8 x 2_{− 1/6 x}4₊ 1 72x 6 1 √ n + 1/8 x − 47 144x 3₊ 37 240x 5_{− 1/48 x}7₊ 1 1296x 9 n−3/4+ O 1 n . (3)

Of course more terms can be mechanically computed, but the expressions become much more intricate. Note carefully the factor 2 in front of our expression: this is justified in the Appendix. This is also justified by probabilistic reasoning: as only one over 2 values of m leads to a non-zero I(m, n) expres-sion, the probability must be multiplied by 2. The tail pruning is also considered in the Appendix: the choice of θ0 is the same as for In.

To check the quality of our asymptotics, we have chosen n = 2000. this gives M = 44.22968229 . . . , σ = 6.613262555 . . ., a range m ∈ [30, 58].

Figure 1 shows I(m, n)/In (circle) and a first asymptotic 2e

−x2/2 √ 2πσ (line). 0 0.02 0.04 0.06 0.08 0.1 0.12 30 35 40 45 50 55 mm

Figure 1: I(m, n)/In (circle) and a first asymptotic 2e

−x2/2

√

2πσ (line)

Figure 2 shows I(m, n)/In (circle) and the asymptotic Equ. (3) (line). The fit is better.

Figure 3 gives the quotient of I(m, n)/In and the asymptotic 2e

−x2/2

√

2πσ (box) as well as the quotient

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0 0.02 0.04 0.06 0.08 0.1 0.12 30 35 40 45 50 55 mm

Figure 2: I(m, n)/In (circle) and the asymptotic Equ. (3) (line)

0.9 0.92 0.94 0.96 0.98 1 1.02 1.04 1.06 30 35 40 45 50 55 mm

Figure 3: quotient of I(m, n)/Inand the asymptotic 2e

−x2/2

√

2πσ (box) as well as the quotient of I(m, n)/In

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4 Large deviation m = n − n

α

, 0 < α < 1

We use again f3(z, m). The multiseries’s scale is here n nα 1/ε if α > 1/2 and n 1/ε α if

α > 1/2. We set ε = nα−1, m = n(1 − ε), L := ln(n). This leads to κ1= ρ2+ m,

ρ2 = n − m = nα (we use that relation as a substitution in the sequel) , κj = nα2j−1,

ρ = nα/2, ln(ρ) = α

2L,

ln(m) = L − ε − 1/2 ε2− 1/3 ε3− 1/4 ε4+ O(ε5). The dominant part of (1) gives

f3(ρ) ρn = exp(E4), E4 = ρ2/2 + m ln(ρ) − ln(m!) − n ln(ρ) = nα/2 + (n − nα) α 2L − ln(m!) − n α 2L, ln(m!) = −n + nα+ n(1 − ε) ln(m) +1 2ln (2πm) + 1 12m −1₋ 1 360m −2_{+ O} 1 m3 .

Now we use the substitution

nεj = njα−(j−1), this leads to n(1 − ε) ln(m) = nL − nα+ 1/2 n2α−1+ 1/6 n3α−2+ 1/12 n4α−3− nαL + 1/4 n5α−4+ O(n6α−5), exp(E4) = F5exp(E5), F5= 1 p2πn(1 − ε) 1 − 1/12 1 (1 − ε) n+ 1 160 1 (1 − ε)2n2 + O( 1 n3) , E5= (1/2 − 1/2 α L + L) nα+ (−L + 1) n − 1/2 n2α−1− 1/6 n3α−2− 1/12 n4α−3− 1/4 , n5α−4+ O(n6α−5). Now we turn to the integral. Proceeding as previously, we have

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This integration gives 1 2√πnαF6, F6 = 1 − 1/6 1 nα + 1 72 1 n2α + O 1 n3α . Finally I(m, n) n! = exp(E5)F7, F7= 1 2√πnαF6F5 = 1 2√πnα 1 − 1/6 nα−1+ 1 72 1 n2α + O 1 n3α × × 1 p2πn(1 − ε) 1 − 1/12 1 (1 − ε) n + 1 160 1 (1 − ε)2n2 + O( 1 n3) , E5= (1/2 − 1/2 α L + L) nα+ (−L + 1) n − 1/2 n2α−1_{− 1/6 n}3α−2_{− 1/12 n}4α−3_{− 1/4 n}5α−4_{+ O(n}6α−5_).

The choice of θ0 is here n−5α/12, (κ1, κ2 = O(nα)). The tail pruning is the same as for I(m, n).

Finally, we obtain the following asymptotic result:

Theorem 4.1 The asymptotic expression of the I(m, n) for large deviation m = n − nα, 0 < α < 1 is given by I(m, n) n! = 2 exp(E5)F7, = 2 1 2√πnα 1 − 1/6 nα−1+ 1 72 1 n2α + O 1 n3α × × 1 p2πn(1 − ε) 1 − 1/12 1 (1 − ε) n + 1 160 1 (1 − ε)2n2 + O( 1 n3) × × exp [(1/2 − 1/2 α L + L) nα+ (−L + 1) n −1/2 n2α−1− 1/6 n3α−2− 1/12 n4α−3− 1/4 n5α−4+ O(n6α−5) . (4) Note that we prefer to keep two separate factors in (4): one in powers of nα and one in powers of n instead of mixing them.

Let us analyze the importance of the terms in E5. We have two sets: the set A of dominant terms,

which stay in the exponent and the set B of small terms, leading to a coefficient of type (1 + ∆), with ∆ small. The property of each term may depend on α. In E5, each term njα−(j−1) is in A if j > _1−α1

and in B otherwize. We finally mention that our non-central range is not sacred: other types of ranges can be analyzed with similar methods.

To check the quality of our asymptotics, we have first chosen n = 2000 and a range α ∈ (0.125, 0.45). This corresponds to the range m ∈ (1968, 1998).

Figure 4 gives ln(I(n, m)/n!) (circle) and ln(Equ.(4))(line) with the substitutions nα = n − m, ε = (n − m)/n, α = ln(n − m)/ln(n), n2α−1_{= (n − m)}2_{/n, . . ..}

Figure 5 gives the quotient of ln(I(n, m)/n!) and ln(Equ.(4)) (circle).

Another way for checking the quality is to fix α. We choose α = 1/4 and n ∈ (1950, 2000). Of course, we must use an integer value for m: m = bn − nα_{+ 1c. Hence, in (4), we use α as the root of}

n − nα− m = 0.

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–13180 –13160 –13140 –13120 –13100 –13080 –13060 –13040 –13020 1970 1975 1980 1985 1990 1995 mm

Figure 4: n = 2000, ln(I(n, m)/n!) (circle) and ln(Equ.(4)) (line)

0.986 0.988 0.99 0.992 0.994 0.996 0.998 1970 1975 1980 1985 1990 1995

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–13100 –13000 –12900 –12800

1950 1960 1970 1980 1990 2000

Figure 6: α = 1/4, ln(I(n, m)/n!) (circle) and ln(Equ.(4)) (line)

1.000448 1.00045 1.000452 1.000454 1.000456 1.000458 1950 1960 1970 1980 1990 2000

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5 Appendix. Justification of the integration procedure

The tail pruning for the Gaussian asymptotics leads to analyze <[ln(f3(ρeiθ) − niθ] = <[ρ2e2iθ/2 + miθ − niθ] =

ρ2_cos(2θ)

2

which has two dominant peaks at 0 and π. So we must be more precise, we have, with n − m even, <he1/2 ρ2e2 iθ+i(m−n)θi= e1/2 ρ2cos(2 θ)cos 1/2 ρ2sin (2 θ) + θ (m − n) ,

and, indeed, if we set θ = π + δ, we recover the same expression. Hence a factor 2 in (3). The large deviation case leads to the same analysis.

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