On strong and almost sure local limit theorems for a probabilistic model of the Dickman distribution

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On strong and almost sure local limit theorems for a

probabilistic model of the Dickman distribution

Régis de La Bretèche, Gérald Tenenbaum

To cite this version:

Régis de La Bretèche, Gérald Tenenbaum. On strong and almost sure local limit theorems for a probabilistic model of the Dickman distribution. Lithuanian Mathematical Journal, Springer Verlag, In press. �hal-03167177�

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On strong and almost sure local

limit theorems for a probabilistic

model of the Dickman distribution

Régis de la Bretèche & Gérald Tenenbaum

To the memory of Jonas Kubilius,

who stood on the bridge and invited us all.

Abstract.

Let {Zk}k>1 denote a sequence of independent Bernoulli random variables defined by

P(Zk= 1) = 1/k = 1 − P(Zk= 0) (k > 1) and put Tn :=P_16k6nkZk. It is then known that

Tn/n converges weakly to a real random variable D with density proportional to the Dickman

function, defined by the delay-differential equation u%0(u) + %(u − 1) = 0 (u > 1) with initial condition %(u) = 1 (0 6 u 6 1). Improving on earlier work, we propose asymptotic formulae with remainders for the corresponding local and almost sure limit theorems, namely

X m>0 P(T n= m) − e−γ n % m n =2 log n π2_n 1 + O ₁ log2n (n → ∞), and (∀u > 0) X n6N Tn=bunc

1 = e−γ%(u) log N + O(log N )2/3+o(1) a.s. (N → ∞),

where γ denotes Euler’s constant.

Keywords: almost sure limit theorems, almost sure local limit theorems, Dickman function, Dickman distribution, quickselect algorithm, friable integers, random permutations.

2020 Mathematics Subject Classification: primary 60F15; secondary: 11N25, 60C05.

1. Introduction and statement of results

Dickman’s function is defined on [0, ∞[ as the continuous solution to the delay-differential equation u%0(u) + %(u − 1) = 0 (u > 1) with initial condition %(u) = 1 (0 6 u 6 1). It is known (see, e.g., [24; th. III.5.10]) that R∞

0 %(u) du = e

γ_{, where γ denotes Euler’s constant.}

The Dickman distribution is defined as the law of a random variable D on [0, ∞[ with density %0(u) := e−γ%(u) (u > 0).

This law appears in a large variety of mathematical topics, such as (the following list being non limiting):

Number theory, in the context of friable integers,(1) _{after the seminal paper of}

Dickman [10] : see [24] for a expositary account;

Random polynomials over finite fields : see, e.g., Car [6], Manstaviˇcius [18], Arratia, Barbour & Tavar´e [1], Knopfmacher & Manstaviˇcius [17];

Random permutations: see in particular, Shepp & Lloyd [21], Kingman [16], Arratia, Barbour & Tavar´e [3], Manstaviˇcius & Petuchovas [19].

In number theory, the Dickman function also appears in Billingsley’s model [5] for the vector distribution of large prime factors of integers (see [23] for an effective version) and in Kubilius’ model(2): see Elliott [11], Arratia, Barbour & Tavar´e [2], Tenenbaum [22], and [24; § III.6.5] for an expositary account.

1. That is integers free of large prime factors

2. A probabilistic model for the uniform probability defined on the set of the first N integers with σ-algebra comprising those events that can be defined by divisibility conditions involving solely small primes

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2 R. de la Bret`eche & G. Tenenbaum

A simple probabilistic description of D is provided by the almost surely convergent series X

n>1

Y

16j6n

Xj,

where the Xjare independent and uniform on [0, 1]: see Goldie & Gr¨ubel [14], Fill & Huber [12],

Devroye [8].

There is a vast bibliography on the various probabilistic models of the Dickman distribution: see, e.g., Chen & Hwang [7], Devroye & Fawzi [9], Pinsky [20].

In 2002, Hwang & Tsai [15] used a simple model to show that, suitably normalized, the cost of Hoare’s quickselect algorithm converges weakly to D. This model may be described as follows: if {Zk}k>1 denotes a sequence of independent Bernoulli random variables such that

P(Zk = 1) = 1/k = 1 − P(Zk = 0) (k > 1) and if Tn := P 16k6nkZk, then Tn/n converges weakly to D, viz. lim n→∞P(Tn6 nu) = e −γ Z u 0 %(v) dv (u > 0).

A strong local limit theorem was then obtained by Giuliano, Szewczak & Weber [13], in the form (1·1) vn:= X m>0 P(Tn = m) − e −γ n % m n = o(1) (n → ∞).

We propose a sharp estimate of the speed of convergence. Here and in the sequel, we let log_kdenote the k-fold iterated logarithm.

Theorem 1.1. We have (1·2) vn= 2 log n π2_n n 1 + O 1 log₂n o (n → ∞).

This estimate may be put in perspective with the following result of Manstaviˇcius [18]. Let {Xk}k>1denote a sequence of independent Poisson variables such that E(Xk) = 1/k, and put

Yn:=P_16k6nkXk. Then [18; cor. 2] readily yields the strong local limit theorem

(1·3) X m>0 P(Y n= m) − e−γ n % m n 1 n (n > 1).

Thus, as may be expected, Poissonian approximations to the Bernoulli random variables Zk

provide a closer model of the Dickman distribution. As a byproduct of (1·2) and (1·3), we get an estimate of the total variation distance between Tn and Yn, viz.

dT V(Tn, Yn) := X m>0 |P(Tn= m) − P(Yn= m)| = vn+ O 1 n = 2 log n π2_n n 1 + O 1 log₂n o .

We also point, without details, to a recent estimate of Bhattacharjee & Goldstein [4; th.1.1], which provides a bound 6 3/(4n) for a smooth Wasserstein-type distance between Tn/n and D.

For u > 0, let εn denote a non-negative sequence tending to 0 at infinity, and let {mn}n>1

denote a non-decreasing integer sequence such that mn = un + O(εnn) as n → ∞. We may

then define a sequence of random variables {LN(u)}∞N =1 by the formula

LN(u) :=

X

n6N, Tn=mn

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By a complicated proof resting on a general correlation inequality, an almost sure local limit theorem is established in [13] assuming furthermore that {mn}n>1 is strictly increasing: for

any u > 1, the asymptotic formula LN(u) ∼ e−γ%(u) log N holds almost surely as N → ∞.(3)

The following result, proved by a simple, direct method, provides an effective version. Theorem 1.2. Let u > 1, εn= o(1) as n → ∞, and let {mn}n>1denote a strictly increasing

sequence of integers such that mn= un + O(εnn) (n > 1). We have, almost surely,

(1·4) LN(u) = 1 + OηN+ (log₂N )1/2+o(1) (log N )1/3 e−γ%(u) log N, where ηN := (1/ log N ) P 16Nεn/n = o(1).

Furthermore, for any u > 0, the formula LN(u) ∼ e−γ%(u) log N holds almost surely provided

ϑm := |{n > 1 : mn= m}| = o(log m) (m → ∞),

and assuming only that the sequence {mn}n>1 is non-decreasing. If ϑm (log m)α with

0 6 α < 1, the estimate (1·4) holds with remainder ηN+ 1/(log N )(1−α)/3+o(1).

We note that, for all u > 0, the case mn := bunc is covered by the second part of the

statement with α = 0.

2. Proof of Theorem 1.1

Let c be a large constant and put M (n) := cn(log n)/(log₂3n). We first show that the contribution to vn of those m > M (n) is negligible. Indeed, since %(v) v−v, we first have

1 n X m>M (n) %m n 1 n X m>M (n) e−m(log2n)/2n 1 n·

Then, we have, for all y > 0

(2·1) X

m>M (n)

P(Tn= m) 6 e−yM (n)E(eyTn) = e−yM (n) Y 16k6n 1 +e ky_{− 1} k .

Selecting y = (log₂n)/n, we see that the last product is

expn Z n 0 eyv− 1 v dv o n2/ log2n_,

hence the left-hand side of (2·1) is also 1/n, and we infer that

(2·2) vn= X 16m6M (n) P(Tn = m) − e −γ n % m n + O1 n .

Recall the definition %0(u) := e−γ%(u) (u ∈ R) and let I(s) :=

R1

0(e

vs_{− 1) dv/v (s ∈ C).}

From [24; th. III.5.10], we know that

(2·3) %_b0(τ ) :=

Z

R

eiτ u%0(u) du = eI(iτ ) (τ ∈ R).

3. The authors of [13] state that this almost sure asymptotic formula holds for all u > 0. However, the requirement that {mn}∞n=1should be strictly increasing is incompatible with the assumption mn∼ un

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Next, for |τ | < π, we have

(2·4) E eiτ Tn = Y 16k6n 1 +e iτ k_{− 1} k = exp Sn(τ ) + U (τ ) + Wn(τ ) + O τ n(1 + n|τ |) , with Sn(τ ) := X 16k6n eiτ k− 1 k , Wn(τ ) := X k>n (eiτ k− 1)2 2k2 U (τ ) :=X k>1 log 1 +e iτ k_{− 1} k − e iτ k_{− 1} k .

For |τ | < 2π, we may write

Sn(τ ) = X 16k6n Z iτ 0 ekvdv = Z iτ 0 env_{− 1} 1 − e−v dv = Z n 0 eiτ v_{− 1} v dv + Vn(τ ) with Vn(τ ) := Z iτ 0 (env− 1) 1 1 − e−v − 1 v dv = Z 1 0 (einτ v− 1)gτ(v) dv, gτ(v) := iτ 1 − e−iτ v − 1 v (0 6 v 6 1).

Since gτ(v) is twice continuously differentiable on [0, 1], partial integration yields

Vn(τ ) = V (τ ) + a(τ )einτ₋ 1 2 n + O 1 n2 (|τ | 6 π), with V (τ ) := − Z 1 0 gτ(v) dv, a(τ ) := gτ(1) iτ = 1 1 − e−iτ − 1 iτ, We have Wn(τ ) τ if |τ | 6 1/n. When 1/n 6 |τ | 6 π, we have

(2·5) Wn(τ ) = X k>n e2ikτ − 2eikτ 2k2 + 1 2n + O 1 n2 = 1 2n 1 + O 1 1 + n min(|τ |, π − |τ |) .

by Abel’s summation. This estimate is hence also valid for |τ | 6 1/n, and so we deduce that

Vn(τ ) + Wn(τ ) = V (τ ) + a(τ )einτ n + O τ n(1 + n|τ |) + 1 n + n2_{min(|τ |, π − |τ |)} .

Put F (τ ) := eU (τ )+V (τ )_{− 1, so that F (0) = 0 and F may be analytically continued to the}

disc {z ∈ C : |z| < 2π}. We finally get

(2·6) _E(eiτ Tn_{) =} b %0(nτ ) 1 + F (τ ) + W_n∗(τ ) + O τ 1 + n2_τ2 (n > 1, |τ | 6 π), with (2·7) W_n∗(τ ) := a(τ ){1 + F (τ )}e inτ n + O 1 n + n2_{min(|τ |, π − |τ |)} ·

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It follows that, for m > 1, (2·8) P(Tn= m) = 1 2π Z π −πE(e iτ Tn_)e−imτ_dτ = 1 2πn Z nπ −nπ b %0(τ )e−iτ m/n 1 + Fτ n + W_n∗τ n + O τ n(1 + τ2₎ dτ.

By (2·3) and, say, [24; lemma III.5.9], we have

(2·9) %_b0(τ ) = −1 iτ + O 1 τ (1 + |τ |) (τ 6= 0), %_b0(τ ) 1 1 + |τ | (τ ∈ R).

Therefore, the error term of (2·8) contributes 1/n2 to the right-hand side. Summing over m 6 M (n), we obtain that the corresponding contribution to the right-hand side of (2·2) is (log n)/(n log₂n), in accordance with (1·2).

We first evaluate (2·10) 1 2πn Z nπ −nπ b %0(τ )e−iτ m/ndτ n > 1, 1 6 m 6 M (n)

by extending the integration range to R and inserting the first estimate (2·9) to bound the integral over R r [−πn, πn]. This yields

(2·11) 1 2πn Z nπ −nπ b %0(τ )e−iτ m/ndτ − %0(m/n) n = −1 πn Z ∞ nπ sin(τ m/n) τ dτ + O 1 n2 = (−1) m+1 π2_mn + O 1 m2_n+ 1 n2 .

In order to estimate the contributions from F and W_n∗ to the main term of (2·8), we use the more precise formula

(2·12) %_b0(τ ) = i τ − eiτ τ2 + O 1 τ3 (|τ | > 1). Writing F (τ ) = τ G(τ ), we indeed deduce from by (2·12) that

Z nπ −nπ b %0(τ )e−iτ m/nF τ n dτ = Z nπ −nπ τ%_b0(τ ) e−iτ m/n n G τ n dτ = i Z In e−iτ mG(τ )1 + ie iτ n nτ dτ + O1 n ,

with In:= [−π, π] r [−1/n, 1/n]. A standard computation furnishes G(π) − G(−π) = −2/π.(4)

Integrating by parts, we get

i Z In e−iτ mG(τ ) dτ = 2(−1) m+1 πm + 1 m Z π −π e−iτ mG0(τ ) dτ + O1 n = 2(−1) m+1 πm + O 1 m2 + 1 n , and, similarly, −1 n Z In e−iτ mG(τ )e iτ n τ dτ = −1 n Z In G(τ ) − G(0) τ e iτ (n−m) dτ + O 1 n 1 n· 4. V (±π) = log(π/2) ∓1 2iπ, e U (±π)_{= −2e}γ_{, F (±π) = ∓iπe}γ_{− 1.}

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We can thus state that, for n > 1, 1 6 m 6 M (n), we have

(2·13) 1 2πn Z nπ −nπ b %0(τ )e−iτ m/nF τ n dτ = (−1) m+1 π2_mn + O 1 m2_n+ 1 n2 .

It remains to estimate the contribution to (2·8) involving W_n∗(τ /n). The error term of (2·7) clearly contributes 1/n2_{. Arguing as for the proof of (2·11), we finally show that the}

contribution to (2·8) arising from a(τ /n)(1 + F (τ /n))eiτ_{/n is also 1/n}2_.

The above estimates and (2·11) furnish together

(2·14) _P(Tn= m) − 1 n%0 m n = (−1)m+1 2 π2_mn + O 1 m2_n + 1 n2 1 6 m 6 M (n).

Summing over m 6 M (n) provides the required estimate (1·2).

3. Proof of Theorem 1.2

We first note that (2·14) yields

(3·1) _P(Tn= m) = 1 n%0 m n + O 1 n2 (m n > 1).

Hence, writing LN for LN(u) here and throughout,

(3·2) _E(LN) = X n6N 1 n%0 m_n n

+ O(1) = %0(u) log N + O(ηNlog N + 1) (N > 1).

The stated result will follow from an estimate of the variance V(LN). We have

(3·3) E(L2N) = E(LN) + 2 X 16ν<n6N P(Tν = mν)ανn, with ανn:= P Tn− Tν = mn− mν (1 6 ν < n 6 N ).

Let us initially assume that {mn}n>1 is strictly increasing and hence that u > 1. We

note right away that, for large n and ν > (u + 1)n/(u + 2), we have ανn = 0, since the

corresponding event is then impossible: either Tn− Tν > ν > (u + 1)(n − ν) > mn− mν or

Tn− Tν = 0 6= mn− mν. Therefore, we may assume ν 6 n(u + 1)/(u + 2) in the sequel.

By (2·4), we have, writing ϕj(τ ) := E eiτ Tj (j > 1),

(3·4) ανn= 1 2π Z π −π ϕn(τ ) ϕν(τ ) e−i(mn−mν)τ_{dτ = β} νn+ ∆νn,

where βνn denotes the contribution of the interval |τ | 6 1/2ν, and ∆νn that of the

complementary range 1/2ν < |τ | 6 π. Now we may derive from (2·6) that

βνn= 1 2πn Z n/2ν −n/2ν b %0(τ ) ϕν(τ /n) n 1 + Fτ n + O1 n o e−i(mn−mν)τ /n_dτ.

Invoking (2·9) and (2·6)-(2·7) with ν in place of n to estimate the contribution of the error term, we get

(3·5) βνn= βνn∗ + O

log n n2

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with (3·6) β_νn∗ := 1 2πn Z n/2ν −n/2ν b %0(τ ) ϕν(τ /n) 1 + Fτ n e−i(mn−mν)τ /n_dτ.

Note that the remainder term of (3·5) in turn contributes 1 to (3·3). Still assuming that ν 6 (u + 1)n/(u + 2) and observing that

ψν(z) := Y k6ν {1 + (ekz− 1)/k}−1 1 (z ∈ C, |z| 6 2/3ν), we may write 1 + F (τ /n) ϕν(τ /n) = 1 +X j>1 µνj τ n j (|τ | 6 n/2ν)

with µνj (3ν/2)j (j > 1). Hence, in view of (2·9), the contribution of the series to (3·6) is

1 2πn X j>1 µνj nj Z n/2ν −n/2ν τj−1eiτ (mn−mν)/n_{+ O} τj−1 1 + |τ | dτ 1 n X j>1 µνj Z 1/2ν −1/2ν vj−1eiv(mn−mν)_{+ O} vj−1 1 + n|v| dv ν n(n − ν)+ ν n2 + ν log(2n/ν) n2 ν log(2n/ν) n2 ,

since the assumption that {mn}∞n=1is strictly increasing implies mn− mν > n − ν.

Arguing as in the proof of (2·10) to evaluate the main term, we eventually get,

(3·7) β∗_νn= 1 n%0 m_n− m_ν n + Oν log(2n/ν) n2 (1 6 ν < n).

We now consider ∆νn. Let

Sνn(τ ) :=

X

ν<k6n

eikτ k ,

and note right away that Sνn(τ ) is bounded for 1/2ν < |τ | 6 π. For ν > 1, we have

ϕn(τ ) ϕν(τ ) = ν n Y ν<k6n 1 + e ikτ k − 1 = ν nexp Sνn(τ ) + O 1 ν + |τ |ν2 + 1 ν + (π − |τ |)ν2 ,

where we used the bounds

X ν<k6n eikτ k(k − 1) 1 ν + |τ |ν2, X ν<k6n e2ikτ (k − 1)2 1 ν + |τ |ν2 + 1 ν + (π − |τ |)ν2, X ν<k6n eikjτ (k − 1)j 1 νj−1 (j > 2).

Carrying back into (3·4), we obtain

(3·8) ∆νn= ν πn< Z π 1/2ν eSνn(τ )−i(mn−mν)τ_{dτ + O} log 2ν nν .

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Now observe that |S_νn0 (τ )| 6 π/|τ | 6 12(n − ν) 6 1

2(mn − mν) if, say ν 6 n/15 or

|τ | > 10(u + 1)/ν. Hence, on this assumption, a standard estimate on trigonometric integrals such as [25; Lemma 4.2] furnishes the bound 1/(mn− mν) 1/(n − ν) 1/n for the last

integral. Since, in the case ν > n/15, the contribution of the range 1/2ν < |τ | 6 10(u + 1)/ν to the same integral is trivially 1/ν, we finally get

(3·9) ∆νn

log 2ν nν +

ν

n2 1 6 ν 6 (u + 1)n/(u + 2).

Gathering our estimates and using the fact that % is Lipschitz on [0, ∞[, we obtain (3·10) ανn= 1 n%0 m_n n + Oν log(2n/ν) n2 + log 2ν nν (1 6 ν < n). Carrying back into (3·3) and applying (3·1) for the pair (ν, mν) yields

E(L2N) − E(LN) = 2 X 16ν<n6N %0 m_n n %0 m_ν ν 1 νn+ O log(2n/ν) n2 + log 2ν nν2 , and hence V(LN) log N. Selecting N = Nk:= 2k 3

for k > 1, we deduce from the Borel-Cantelli lemma that, given any ε > 0, the estimate

LNk− E(LNk) k

2_{(log 2k)}1/2+ε

holds almost surely. In view of (3·2), this implies the stated result since LN is a non-decreasing

function of N .

We next consider the case of a non-decreasing sequence {mn}n>1. Accordingly, we fix u > 0.

By hypothesis, for some integer q = qN > 2 such that qN = o log N as N → ∞, we have

mn> mν whenever n − ν > qN. Put LN(u; a) := X n6N n≡a (mod q) Tn=mn 1 (1 6 a 6 q).

By (3·1), we have, for all a ∈ [1, q],

E LN(u; a) = X n6N n≡a (mod q) 1 n%0 m_n n + O 1 a2 , and, by (3·10),

V LN(u; a) − E LN(u; a)

X 16ν<n6N ν,n≡a (mod q) nlog 2n/ν n2 + log 2ν nν2 o X q<n6N n≡a (mod q) n 1 nq + log 2a a2_n o log N q n1 q + log 2a a2 o .

Summing over a ∈ [1, q], we get V LN(u) 6 q

X

16a6q

V LN(u; a) qE(LN(u)) + log N q log N.

This is all needed.

Acknowledgment. The authors express warm thanks to Eugenijus Manstaviˇcius for precise and useful comments on a preliminary version of the manuscript.

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[25] E.C. Titchmarsh, The theory of the Riemann zeta-function, Oxford University Press, 1951.

R´egis de la Bret`eche

Universit´e de Paris, Sorbonne Universit´e, CNRS, Institut de Math. de Jussieu-Paris Rive Gauche, F-75013 Paris,

France

regis.de-la-breteche@imj-prg.fr

Gérald Tenenbaum Institut Élie Cartan Université de Lorraine BP 70239

54506 Vandœuvre-l`es-Nancy Cedex France