denotes the i th prime number. The number of the positive divisors of n ∈ N is denoted by d ( n ) , and we write

On Large Values of the Divisor Function

P. ERD ˝OS

J.-L. NICOLAS jlnicola@in2p3.fr

Institut Girard Desargues, UPRES-A-5028, Math´ematiques, Bat. 101, Universit´e Claude Bernard (LYON 1), F-69622 Villeurbanne c´edex, France

A. S ´ARK ¨OZY^∗ sarkozy@cs.elte.hu

E¨otv¨os Lor´and University, Dept. of Algebra and Number Theory, H-1088 Budapest, M´uzeum krt. 6-8, Hungary

Jean-Louis Nicolas and Andr`as S´ark¨ozy dedicate this paper to the memory of Paul Erd˝os Received June 10, 1997; Accepted January 7, 1998

Abstract. Let d(n)denote the divisor function, and let D(X)denote the maximal value of d(n)for n≤X . For 0<z≤1, both lower and upper bounds are given for the number of integers n with n≤X,z D(X)≤d(n).

Key words: division function, highly composite numbers, maximal order 1991 Mathematics Subject Classification: Primary 11N56

1. Introduction

Throughout this paper, we shall use the following notations: N denotes the set of the positive integers, π( x ) denotes the number of the prime numbers not exceeding x, and p

i

denotes the i th prime number. The number of the positive divisors of n ∈ N is denoted by d ( n ) , and we write

D ( X ) = max

n≤X

d ( n ).

Following Ramanujan we say that a number n ∈ N is highly composite, briefly h.c., if d ( m ) < d ( n ) for all m ∈ N , m < n. For information about h.c. numbers, see [13, 15] and the survey paper [11].

The sequence of h.c. numbers will be denoted by n

1

, n

2

, . . . : n

1

= 1 , n

2

= 2 , n

3

= 4 , n

4

= 6 , n

5

= 12 , . . . (for a table of h.c. numbers, see [13, Section 7, or 17]. For X > 1, let n

k

= n

k(X)

denote the greatest h.c. number not exceeding X , so that

D ( X ) = d ¡ n

k(X)

¢

.

∗Research partially supported by Hungarian National Foundation for Scientific Research, Grant No. T017433 and by C.N.R.S, Institut Girard Desargues, UPRES-A-5028.

It is known (cf. [13, 8]) that n

k

is of the form n

k

= p

^r₁¹

p

^r₂²

· · · p

^r<sub>`</sub><sup>`</sup>

, where r

1

≥ r

2

≥ · · ·

≥ r

_`

,

` = ( 1 + o ( 1 )) log X

log log X , (1)

r

i

= ( 1 + o ( 1 )) log p

_`

log 2 log p

_i

µ

for X → ∞ and log p

_i

log p

_`

→ 0

¶

(2) and, if m is the greatest integer such that r

m

≥ 2,

p

_m

= p

^θ_`

+ O ¡ p

^τ_`^oθ

¢

(3) where

θ = log ( 3 / 2 )

log 2 = 0 . 585 . . . (4)

and τ

0

is a constant < 1 which will be given later in (8).

For 0 < z ≤ 1 , X > 1 , let S ( X , z ) denote the set of the integers n with n ≤ X , d ( n ) ≥ z D ( X ) . In this paper, our goal is to study the function F ( X , z ) = Card ( S ( X , z )) .

In Section 4, we will study F ( X , 1 ) , further we will prove (Corollary 1) that for some c > 0 and infinitely many X ’s with X → +∞ , we have F ( X , z ) = 1 for all z and X satisfying

1 − 1

( log X )

^c

< z ≤ 1 .

Thus, to have a non trivial lower bound for F ( X , z ) for all X , one needs an assumption of the type z < 1 − f ( X ), cf. (6).

In Section 2, we shall give lower bounds for F ( X , z ) . Under a strong, but classical, assumption on the distribution of primes, the lower bound given in Theorem 1 is similar to the upper bound given in Section 3. The proofs of the lower bounds will be given in Section 5: in the first step we construct an integer n ˆ ∈ S ( X , z ) such that d (ˆ n ) is as close to z D ( X ) as possible. This will be done by using diophantine approximation of θ (defined by (4)), following the ideas of [2, 8]. Further, we observe that slightly changing large prime factors of n will yield many numbers n not much greater than ˆ n, and so belonging to ˆ S ( X , z ) . The proof of the upper bound will be given in Section 7. It will use the superior h.c. numbers, introduced by Ramanujan (cf. [13]). Such a number N

_ε

maximizes d ( n )/ n

^ε

. The problem of finding h.c. numbers is in fact an optimization problem

max

n≤x

d ( n )

and, in this optimization problem, the parameter ε plays the role of a Lagrange mul-

tiplier. The properties of the superior h.c. numbers that we shall need will be given in

Section 6.

In [10, p. 411], it was asked whether there exists a positive constant c such that, for n

j

large enough,

d ( n

j+1

)

d ( n

_j

) ≤ 1 + 1 ( log n

_j

)

^c

.

In Section 8, we shall answer this question positively, while in Section 4 we shall prove that for infinitely many n

j

, one has d ( n

j+1

)/ d ( n

j

) ≥ 1 + ( log n

j

)

^−0.71

.

We are pleased to thank J. Rivat for communicating us reference [1].

2. Lower bounds We will show that

Theorem 1. Assume that τ is a positive number less than 1 and such that π( x ) − π( x − y ) > A y

log x for x

^τ

< y < x (5) for some A > 0 and x large enough. Then for all ε > 0 , there is a number X

0

= X

0

(ε) such that , if X > X

0

(ε) and

exp (−( log X )

^λ

) < z < 1 − log X )

^−λ1

(6) where λ is any fixed positive real number < 1 and λ

1

a positive real number ≤ 0 . 03 , then we have:

F ( X , z ) > exp (( 1 − ε) min { 2 ( A log 2 log X log ( 1 / z ))

^1/2

, 2 ( log X )

^1−τ

log log X log ( 1 / z )}).

(7) Note that (5) is known to be true with

τ = τ

0

= 0 . 535 and A = 1 / 20 (8)

(cf. [1]) so that we have

F ( X , z ) > exp (( 1 − ε) 2 ( log X )

^0.465

log log X log ( 1 / z ))

for all z satisfying (6), and assuming the Riemann hypothesis, (5) holds for all τ > 1 / 2 so that

F ( X , z ) > exp (( log X )

^1/2−ε

log ( 1 / z ))

for all ε > 0 , X large enough and z satisfying (6). Moreover, if (5) holds with some τ < 1 / 2 and A > 1 − ε/ 2 (as it is very probable), then for a fixed z we have

F ( X , z ) > exp (( 2 − ε)(( log 2 )( log X ) log ( 1 / z ))

^1/2

). (9)

In particular,

F ( X , 1 / 2 ) > exp (( 1 − ε)( log 2 )( log X )

^1/2

). (10) While we need a very strong hypothesis to prove (9) for all X , we will show without any unproved hypothesis that, for fixed z and with another constant in the exponent, it holds for infinitely many X ∈ N:

Theorem 2. If z is a fixed real number with 0 < z < 1 , and ε > 0 , then for infinitely many X ∈ N we have

F ( X , z ) > exp (( 1 − ε)( log 4 log X log ( 1 / z )

^1/2

) (11) so that , in particular

F ( X , 1 / 2 ) > exp (( 1 − ε) √

2 log 2 ( log X )

^1/2

). (12) We remark that the constant factor √

2 log 2 on the right hand side could be improved by the method used in [12] but here we will not work out the details of this. It would also be possible to extend Theorem 2 to all z depending on X and satisfying (6).

3. Upper bounds We will show that:

Theorem 3. There exists a positive real number γ such that , for z ≥ 1 − ( log X )

^−γ

, as X → +∞ we have

log F ( X , z ) = O ¡

( log X )

^{(1−γ )/2}

¢

, (13)

and if λ, η are two real numbers , 0 < λ < 1 , 0 < η < γ, we have for

1 − ( log X )

^−γ+η

≥ z ≥ exp (−( log X )

^λ

), (14) and X large enough:

F ( X , z ) ≤ exp µ 24

√ 1 − γ ( log ( 1 / z ) log X )

^1/2

¶

. (15)

The constant γ will be defined in Lemma 5 below. One may take γ = 0 . 03. Then for z = 1 / 2, (15) yields

log F ( X , 1 / 2 ) ≤ 21 ( p log X )

which, together with the results of Section 2, shows that the right order of magnitude of log F ( X , 1 / 2 ) is, probably, √

log X .

4. The cases z = 1 and z close to 1

Let us first define an integer n to be largely composite (l.c.) if m ≤ n ⇒ d ( m ) ≤ d ( n ) . S. Ramanujan has built a table of l.c. numbers (see [14, p. 280 and 15, p. 150]). The distribution of l.c. numbers has been studied in [9], where one can find the following results:

Proposition 1. Let Q

_`

( X ) be the number of l.c. numbers up to X . There exist two real numbers 0 . 2 < b

1

< b

2

< 0 . 5 such that for X large enough the following inequality holds:

exp (( log X )

^b1

) ≤ Q

_`

( X ) ≤ exp (( log X )

^b2

).

We may take any number <( 1 −

^{log 3}_{log 2}^/2

)/ 2 = 0 . 20752 for b

1

, and any number >( 1 − γ )/ 2 with γ > 0 . 03 defined in Lemma 5 , for b

₂

.

From Proposition 1, it is easy to deduce:

Theorem 4. There exists a constant b

2

< 0 . 485 such that for all X large enough we have F ( X , 1 ) ≤ exp (( log X )

^b2

). (16) There exists a constant b

1

> 0 . 2 such that , for a sequence of X tending to infinity , we have F ( X , 1 ) ≥ exp (( log X )

^b1

). (17) Proof: F ( X , 1 ) is exactly the number of l.c. numbers n such that n

k

≤ n ≤ X . Thus F ( X , 1 ) ≤ Q

_`

( X ) and (16) follows from Proposition 1.

The proof of Proposition 1 in [9, Section 3] shows that for any b

1

< 0 . 207, there exists an infinite number of h.c. numbers n

_j

such that the number of l.c. numbers between n

_j−1

and n

_j

(which is exactly F ( n

_j

− 1 , 1 )) satisfies F ( n

_j

− 1 , 1 ) ≥ exp (( log n

_j

)

^b1

) for n

_j

large

enough, which proves (17). 2

We shall now prove:

Theorem 5. Let ( n

j

) be the sequence of h.c. numbers. There exists a positive real number a , such that for infinitely many n

j

’s , the following inequality holds:

d ( n

j

)

d ( n

_j−1

) ≥ 1 + 1

( log n

_j

)

^a

. (18)

One may take any a > 0 . 71 in ( 18 ) .

Proof: Let X tend to infinity, and define k = k ( X ) by n

_k

≤ X < n

_k+1

. By [8], the number k ( X ) of h.c. numbers up to X satisfies

k ( X ) ≤ ( log X )

^µ

(19)

for X large enough, and one may choose for µ the value µ = 1 . 71, cf. [10, p. 411 or 11, p. 224]. From (19), the proof of Theorem 5 follows by an averaging process: one has

Y

√X<nj≤X

d ( n

j

)

d ( n

j−1

) = D ( X ) D ( √

X ) . The number of factors in the above product is k ( X )− k ( √

X ) ≤ k ( X ) so that there exists j , k ( √

X ) + 1 ≤ j ≤ k ( X ) , with d ( n

j

) d ( n

j−1

) ≥

µ D ( X ) D ( √

X )

¶

1/k(X)

. (20)

But it is well known that log D ( X ) ∼

^{(log 2}_{log log X}^{)(log X)}

, and thus log ( D ( X )/ D ( √

X )) ∼ log 2 2

log X log log X .

Observing that X < n

²_j

, it follows from (19) and (20) for X large enough:

d ( n

j

) d ( n

j−1

) ≥ exp

µ 1 3

1

( log X )

^µ−1

log log X

¶

≥ exp µ 1

3

1

( 2 log n

j

)

^µ−1

log ( 2 log n

j

)

¶

≥ exp µ 1

( log n

j

)

^a

¶

≥ 1 + 1 ( log n

j

)

^a

for any a > µ − 1, which completes the proof of Theorem 5. 2 A completely different proof can be obtained by choosing a superior h.c. number for n

_j

and following the proof of Theorem 8 in [7, p. 174], which yields a =

^log_{log 2}^(3/2)

= 0 . 585 . . . See also [10, Proposition 4].

Corollary 1. For c > 0 . 71 , there exists a sequence of values of X tending to infinity such that F ( X , z ) = 1 for all z, 1 − 1 /( log X )

^c

< z ≤ 1.

Proof: Let us choose X = n

j

, with n

j

satisfying (18), and c > a. For all n < X , we have

d ( n ) ≤ d ( n

j−1

) ≤ d ( n

j

)

1 + ( log n

j

)

^−a

= D ( X )

1 + ( log X )

^−a

< z D ( X ).

Thus S ( X , z ) = { n

j

} , and F ( X , z ) = 1. 2

5. Proofs of the lower estimates

Proof of Theorem 1: Let us denote by α

i

/β

i

the convergents of θ , defined by (4). It is known that θ cannot be too well approximated by rational numbers and, more precisely, there exists a constant κ such that

| q θ − p | À q

^−κ

(21) for all integers p , q 6= 0 ( cf. [4]). The best value of κ

κ = 7 . 616 (22)

is due to G. Rhin (cf. [16]). It follows from (21) that β

i+1

= O ¡

β

i^κ

¢ . (23)

Let us introduce a positive real number δ which will be fixed later, and define j = j ( X , δ) so that

β

j

≤ ( log X )

^δ

< β

j+1

. (24) By Kronecker’s theorem (cf. [6], Theorem 440), there exist two integers α and β such that

¯¯ ¯¯ βθ − α − log z log 2 − 2

β

j

¯¯ ¯¯ < 2 β

j

(25) and

β

j

2 ≤ β ≤ 3 β

j

2 . (26)

Indeed, as α

j

and β

j

are coprime, one can write B, the nearest integer to (β

jlog z

log 2

+ 2 ) , as B = u

1

α

j

− u

2

β

j

with | u

1

| ≤ β

j

/ 2, and then α = α

j

+ u

2

and β = β

j

+ u

1

satisfy (25).

With the notation of Section 1, we write ˆ

n = n

_k

p

m+1

p

m+2

· · · p

m+β

p

_`

p

_`−1

· · · p

_`−α+1

(27) for X large enough. By (26), (24), and (6), (25) yields

α ≤ βθ + log ( 1 / z )

log 2 ¿ max (( log X )

^δ

, ( log X )

^λ

) (28) and

α ≥ βθ − log z log 2 − 4

β

j

> βθ − 6

β + log ( 1 / z )

log 2 > 0

for X large enough. Thus, if we choose δ < 1, from (3) and (1) we have r

_`

= r

_`−1

= · · · = r

_`−α+1

= 1. By (1) and the prime number theorem, we also have

p

_`

∼ log X (29)

and by (3), we have r

m+1

= r

m+2

= · · · = r

m+β

= 1 so that, by (25), d (ˆ n ) = d ( n

k

) ( 3 / 2 )

^β

2

^α

= d ( n

k

) exp ( log 2 (βθ − α)) ≥ zd ( n

k

) = z D ( X ). (30) Now we need an upper bound for n ˆ / n

k

. First, it follows from (5) that for i = o ( m ) we have

p

m+i

− p

m

≤ max µ

p

_m^τ_+i

, i

A log p

m+i

¶

(31) and consequently,

Y

β i=1

p

m+i

p

m

= exp Ã

_β

X

i=1

log p

m+i

p

m

!

≤ exp Ã

_β

X

i=1

p

m+i

− p

m

p

m

!

≤ exp µ β

p

m

max µ

p

_m^τ_+β

, β

A log p

m+β

¶¶

≤ exp ¡ O ¡

max ¡

( log X )

^{δ+θ(τ−1)}

, ( log X )

^2δ−θ

log log X ¢¢¢

(32) by (26), (24), (3) and (1). Similarly, we get

α−

Y

1 i=0

p

_`

p

_`−i

≤ exp µ α

p

_`−α+1

max µ

p

_l^τ

, α A log p

_`

¶¶

≤ exp µ

O µ

max

µ ( log X )

^δ

− log z

( log X )

^1−τ

, (( log X )

^δ

− log z )

²

log X log log X

¶¶¶

(33) by (28). Further, it follows from (3) and (25) that

p

^βm

p

^α_`

= p

_`^βθ−α

¡ 1 + O ¡

p

_`^(τ−1)θ

¢¢

_β

≤ p

log z log 2+_βj⁴

`

exp ¡

O ¡

β p

_`^(τ−1)θ

¢¢

≤ exp

½µ log z log 2 log p

_`

¶

+ 4 log p

_`

β

j

+ β p

_`^(1−τ)θ

¾

. (34)

It follows from (23) and (24) that

β

j

À ( log X )

^δ/κ

. (35)

Multiplying (32), (33) and (34), we get from (27) and (29):

ˆ

n / n

k

≤ exp

½

( 1 + o ( 1 )) log z log log X log 2

¾

(36)

if we choose δ in such a way that the error terms in (32), (33) and (34) can be neglected.

More precisely, from (6) and (36), δ should satisfy:

δ + θ(τ − 1 ) < −λ

1

2 δ − θ < −λ

1

κλ

1

< δ < 1 . It is possible to find such a δ if λ

1

satisfies

λ

1

< min

µ ( 1 − τ)θ 1 + κ , θ

1 + 2 κ

¶ .

(4), (8) and (22) yield λ

1

< 0 . 03157.

For convenience, let us write ˆ

n = p

^r₁^ˆ1

p

^r₂^ˆ2

· · · p

_t^rˆt

(37) with, by (27), t = ` − α . It follows from (1) and (28) that

t = ( 1 + o ( 1 )) log X

log log X ; p

_t

∼ log X (38)

and from (24) and (26) that ˆ

r

_i

= 1 for i ≥ t − t

^9/10

. (39)

Now, consider the integers v satisfying P ( t , v)

^def

= p

t+1

p

t+2

· · · p

t+v

p

t−v+1

p

t−v+2

· · · p

t

≤ exp µ

( 1 − ε) log ( 1 / z ) log X log 2

¶

(40) and

v ≤ t

^9/10

. (41)

By a calculation similar to that of (32) and (33), by (5) and the prime number theorem, for all v satisfying (41) and for all 1 ≤ i ≤ v we have:

p

t+i

p

t−v+i

= 1 + p

t+i

− p

t−v+i

p

t−v+i

≤ 1 + ( 1 + o ( 1 )) 1 p

t

max µ

p

_t^τ_+v

, v

A log p

t+v

¶

= 1 + ( 1 + o ( 1 )) 1 t max

µ

t

^τ

( log t )

^τ−1

, v A

¶

so that, by (38), the left hand side of (40) is P ( t , v) = Y

^v

i=1

p

t+i

p

t−v+i

≤ exp µ

v( 1 + o ( 1 )) 1 t max

µ

t

^τ

( log t )

^τ−1

, v A

¶¶

= exp µ

( 1 + o ( 1 ))v log log X log X max

µ ( log X )

^τ

log log X , v

A

¶¶

= exp µ

( 1 + o ( 1 ))v max µ

( log X )

^τ−1

, v A

log log X log X

¶¶

. (42)

By (42), (40) follows from exp

µ

( 1 + o ( 1 ))v max µ

( log X )

^τ−1

, v A

log log X log X

¶¶

< exp µµ

1 − ε 2

¶ log ( 1 / z ) log X log 2

¶ . (43) An easy computation shows that with

µ 1 − 5 ε

6

¶ min

µµ A log X

log 2 log ( 1 / z )

^1/2

¶

, ( log X )

^1−τ

log log X

log 2 log ( 1 / z )

¶

in place of v both (41) and (43) hold. Thus fixing v now as the greatest integer v satisfying (41) and (43), we have

v >

µ 1 − 3 ε

4

¶ min

µµ A log X

log 2 log ( 1 / z )

^1/2

¶

, ( log X )

^1−τ

log log X

log 2 log ( 1 / z )

¶ . (44) Then it follows from (39) and (41) that

ˆ

r

_t−v+i

= 1 for i = 1 , 2 , . . . , v. (45) Let now A denote the set of the integers a of the form

a = 2

^rˆ1

p

^r₂^ˆ2

· · · p

^r_t^ˆ_−v^t−v

p

i₁

· · · p

i_v

where t − v + 1 ≤ i

1

< i

2

< · · · < i

_v

≤ t + v. (46) Then, by (37), (46) and (30) we have

d ( a ) = d (ˆ n ) ≥ z D ( X ). (47) Moreover, by (40) and (36) such an a satisfies

a = p

i1

p

i2

· · · p

i_v

p

t−v+1

p

t−v+2

· · · p

t

ˆ

n ≤ P ( t , v)ˆ n ≤ n

k

. (48)

It follows from (47) and (48) that a ∈ S ( X , z ) and

F ( X , z ) ≥ |A|. (49) The numbers i

₁

, i

₂

, . . . , i

_v

in (46) can be chosen in (

²_v^v

) ways so that

|A| = µ 2 v

v

¶

> exp µµ

1 − ε 8

¶ ( log 4 )v

¶

. (50)

Now (7) follows from (44), (49) and (50), and this completes the proof of Theorem 1. 2 Proof of Theorem 2: By a theorem of Selberg [19, 9], if the real function f ( x ) is increasing, f ( x ) > x

^1/6

and

^f(_x^x)

& 0, then there are infinitely many integers y such that

π( y + f ( y )) − π( y ) ∼ f ( y )

log y and π( y ) − π( y − f ( y )) ∼ f ( y )

log y . (51) We use this result with f ( y ) = ( 1 −

^ε₃

) log y (

^{y log}_{log 4}^(1/z)

)

^1/2

and for a y value satisfying (51), define t by

p

t

≤ y < P

t+1

. (52)

Further, we define β

j

(instead of (24)) so that β

j

≥

_εlog^{4 log 2}_(1/z)

and α, β by (25) and (26); we set ` = t + α and choose X = n

_k

a h.c. number whose greatest prime factor is p

_`

(such a number exists, see [13] or (59), (60) below). We define n by (27), and (30) and (38) still ˆ hold, while (36) becomes

ˆ n n

k

≤ exp µ

( 1 + o ( 1 )) log log X µ log z

log 2 + 4 β

j

¶¶

≤ exp µ

( 1 + o ( 1 )) log log X

log 2 log z ( 1 − ε)

¶

≤ exp

µ log log X log 2 log z

µ 1 − ε

2

¶¶

(53) for X large enough. Let v denote the greatest integer with

p

t+v

≤ y + f ( y ) and p

t−v

≥ y − f ( y ), (54) so that by the definition of y we have

v ∼ f ( y )

log y . (55)

By (38) and (52), we have

y ∼ log x . (56)

Moreover, by (38), (54) and (55), we have P ( t , v)

^def

= Y

^v

i=1

p

t+i

p

t−v+i

≤

µ y + f ( y ) y − f ( y )

¶

_v

≤ exp µ

( 1 + o ( 1 )) f ( y ) log log X log

µ

1 + 2 f ( y ) y

¶¶

= exp µ

( 2 + o ( 1 )) f

²

( y ) y log log X

¶

= µ 1

log 2 + o ( 1 )

¶µ 1 − ε

3

¶

2

log log X log ( 1 / z ).

(57)

It follows from (53) and (57) that P ( t , v) < n

_k

/ n for X large enough and ˆ ε small enough.

Again, as in the proof of Theorem 1, we consider the set A of the integers a of the form (48). Then as in the proof of Theorem 1, by using (38) and (55) finally we obtain

F ( X , z ) ≥ |A| = µ 2 v

v

¶

> exp µµ

1 − ε 3

¶ ( log 4 )v

¶

> exp (( 1 − ε)( log 4 )

^1/2

( log X )

^1/2

( log ( 1 / z ))

^1/2

)

which completes the proof of Theorem 2. 2

6. Superior highly composite numbers and benefits

Following Ramanujan (cf. [13]) we shall say that an integer N is superior highly composite (s.h.c.) if there exists ε > 0 such that for all positive integer M the following inequality holds:

d ( M )/ M

^ε

≤ d ( N )/ N

^ε

. (58)

Let us recall the properties of s.h.c. numbers (cf. [13], [7, p. 174], [8–11]). To any ε, 0 < ε < 1, one can associate the s.h.c. number:

N

_ε

= Y

p≤x

p

^αp

(59)

where

x = 2

^1/ε

, ε = ( log 2 )/ log x (60)

and

α

p

=

¹ 1 p

^ε

− 1

º

. (61)

For i ≥ 1, we write

x

_i

= x

^{log(1+1/i)/log 2}

(62)

and then (61) yields:

α

p

= i ⇐⇒ x

_i+1

< p ≤ x

_i

. (63)

A s.h.c. number is h.c. thus from (1) we deduce:

x ∼ log N

_ε

. (64)

Let P > x be the smallest prime greater than x. There is a s.h.c. number N

⁰

such that

N

⁰

≤ NP and d ( N

⁰

) ≤ 2d ( N ) .

Definition. Let ε, 0 < ε < 1, and N

_ε

satisfy (58). For a positive integer M, let us define the benefit of M by

ben M = ε log M

N

_ε

− log d ( M )

d ( N

_ε

) . (65)

From (58), we have ben M ≥ 0. Note that ben N depends on ε , but not on N

_ε

: If N

⁽¹⁾

and N

⁽²⁾

satisfy (58), (65) will give the same value for ben M if we set N

_ε

= N

⁽¹⁾

or N

_ε

= N

⁽²⁾

.

Now, let us write a generic integer:

M = Y

p

p

^βp

,

for p > x, let us set α

p

= 0, and define:

ben

_p

( M ) = ε(β

p

− α

p

) log p − log µ β

p+1

α

p+1

¶

. (66)

From the definition (61) of α

p

, we have ben

_p

( M ) ≥ 0, and (65) can be written as ben M = X

p

ben

p

( M ). (67)

If β

p

= α

p

, we have ben

p

( M ) = 0. If β

p

> α

p

, let us set ϕ

1

= ϕ

1

(ε, p , α

p

, β

p

) = (β

p

− α

p

)

µ

ε log p − log α

p

+ 2 α

p

+ 1

¶

= (β

p

− α

p

)ε log µ p

x

_αp+1

¶

ψ

1

= ψ

1

(α

p

, β

p

) = (β

p

− α

p

) log µ

1 + 1

α

p

+ 1

¶

− log µ

1 + β

p

− α

p

α

p

+ 1

¶ . We have

ben

_p

( M ) = ϕ

1

+ ψ

1

,

ϕ

1

≥ 0 , ψ

1

≥ 0 and ψ

1

(α

p

, α

p

+ 1 ) = 0. Similarly, for β

p

< α

p

, let us introduce:

ϕ

2

= ϕ

2

(ε, p , α

p

, β

p

) = (α

p

− β

p

) µ

log α

p

+ 1 α

p

− ε log p

¶

= (α

p

− β

p

) ε log µ x

_αp

p

¶

ψ

2

= ψ

2

(α

p

, β

p

) = (α

p

− β

p

) log µ

1 − 1

α

p

+ 1

¶

− log µ

1 − α

p

− β

p

α

p

+ 1

¶ .

We have ϕ

2

≥ 0 , ψ

2

≥ 0 , ψ

2

(α

2

, α

p

− 1 ) = 0. Moreover, observe that ψ

1

is an increasing function of β

p

− α

p

, and ψ

2

is an increasing function of α

p

− β

p

, for α

p

fixed.

We will prove:

Theorem 6. Let x → +∞, ε be defined by ( 60 ) and N

_ε

by ( 59 ) . Let λ < 1 be a positive

real number , µ a positive real number not too large (µ < 0 . 16 ) and B = B ( x ) such that

x

^−µ

≤ B ( x ) ≤ x

^λ

. Then the number of integers M such that the benefit of M ( defined by ( 65 )) is smaller than B , satisfies

ν ≤ exp µ 23

√ 1 − µ

√ Bx

¶

(68) for x large enough.

In [9], an upper bound for ν was given, with B = x

^−γ

. In order to prove Theorem 6, we shall need the following lemmas:

Lemma 1. Let p

1

= 2 , p

2

= 3 , . . . , p

k

be the kth prime. For k ≥ 2 we have k log k ≥ 0 . 46 p

k

.

Proof: By [18] for k ≥ 6 we have

p

k

≤ k ( log k + log log k ) ≤ 2k log k

and the lemma follows after checking the cases k = 2, 3, 4, 5. 2 Lemma 2. Let p

₁

= 2 , p

₂

= 3 , . . . , p

_k

be the kth prime. The number of solutions of the inequality

p

1

x

1

+ p

2

x

2

+ · · · + p

k

x

k

+ · · · ≤ x (69) in integers x

1

, x

2

, . . . , is exp (( 1 + o ( 1 ))

^2√π₃

q

x log x

) .

Proof: The number T ( n ) of partitions of n into primes satisfies (cf. [5]) log T ( n ) ∼

2π

√3

q

n

log n

, and the number of solutions of (69) is P

n≤x

T ( n ) . 2

Lemma 3. The number of solutions of the inequality

x

₁

+ x

₂

+ · · · + x

_r

≤ A (70) in integers x

1

, . . . , x

r

is ≤ ( 2r )

^A

.

Proof: Let a = b A c . It is well known that the number of solutions of (70) is µ r + a

a

¶

= r + a a

r + a − 1

a − 1 · · · r + 2 2

r + 1

1 ≤ ( r + 1 )

^a

≤ ( 2r )

^a

.

2

Proof of Theorem 6: Any integer M can be written as M = A

D N

_ε

, ( A , D ) = 1 and D divides N

_ε

.

First, we observe that, if p

^y

divides A and ben M ≤ B, we have for x large enough:

y ≤ x . (71)

Indeed, by (61), we have

α

p

≤ 1

p

^ε

− 1 ≤ 1

ε log p = log x

log 2 log p ≤ log x

( log 2 )

²

≤ 3 log x . It follows that

B ≥ ben M ≥ ben

_p

( AN

_ε

) ≥ ψ

1

(α

p

, α

p

+ y )

= y log µ

1 + 1

α

p

+ 1

¶

− log µ

1 + y

α

p

+ 1

¶

≥ y α

p

− log ( 1 + y ) ≥ y

3 log x − log ( 1 + y ), and since B ≤ x

^λ

, this inequality does not hold for y > x and x large enough.

Further we write A = A

₁

A

₂

· · · A

₆

with ( A

_i

, A

_j

) = 1 and p | A

1

H⇒ p > 2x

p | A

2

H⇒ x < p ≤ 2x p | A

3

H⇒ 2x

2

< p ≤ x p | A

₄

H⇒ x

₂

< p ≤ 2x

₂

p | A

5

H⇒ 2x

3

< p ≤ x

2

p | A

₆

H⇒ p ≤ 2x

₃

,

where x

₂

and x

₃

are defined by (62). Similarly, we write D = D

₁

D

₂

. . . D

₅

, with ( D

_i

, D

_j

) = 1 and

p | D

1

H⇒ x / 2 < p ≤ x p | D

2

H⇒ x

2

< p ≤ x / 2 p | D

3

H⇒ x

2

/ 2 < p ≤ x

2

p | D

₄

H⇒ 2x

₃

< p ≤ x

₂

/ 2 p | D

5

H⇒ p ≤ 2x

3

. We have

ben M = X

6 i=1

ben ( A

_i

N

_ε

) + X

5

i=1

ben ( N

_ε

/ D

_i

), and denoting by ν

i

(resp. ν

_i⁰

) the number of solutions of

ben ( A

i

N

_ε

) ≤ B ( resp. ben ( N

_ε

/ D

i

) ≤ B ),

we have

ν ≤ Y

6 i=1

ν

i

Y

5 i=1

ν

i⁰

. (72)

In (72), we shall see that the main factors are ν

2

and ν

1⁰

and the other ones are negligible.

Estimation of ν

2

. Let us denote the primes between x and 2x by x < P

1

< P

2

< · · · <

P

r

≤ 2x, and let

A

2

= P

₁^y1

P

₂^y2

· · · P

_r^yr

, y

i

≥ 0 . From the Brun-Titchmarsh inequality, it follows for i ≥ 2 that

i = π( P

i

) − π( x ) ≤ 2 P

_i

− x

log ( P

_i

− x ) ≤ 2 P

_i

− x log 2 ( i − 1 ) and it follows from Lemma 1:

P

i

− x ≥ i

2 log 2 ( i − 1 ) ≥ i log i

2 ≥ 0 . 23 p

i

. By (60) and (61) we have α

Pi

= 0 and

ben ( A

₂

N

_ε

) ≥ X

r

i=2

ϕ

1

(ε, P

_i

, 0 , y

_i

) = X

r

i=2

ε y

_i

log ( P

_i

/ x )

≥ X

r

i=2

ε y

_i

P

i

− x P

i

≥ X

r

i=2

ε y

i

2x ( P

_i

− x ) ≥ X

r i=2

0 . 115 ε y

i

x p

_i

.

By (71), the number of possible choices for y

₁

is less than ( x + 1 ) , so that ν

2

is certainly less than ( x + 1 ) times the number of solutions of:

X

∞ i=2

p

_i

y

_i

≤ B x

ε( 0 . 115 ) ≤ 12 . 6Bx log x , and, by Lemma 2,

ν

2

≤ ( x + 1 ) exp (

( 1 + o ( 1 )) 2 π

√ 3 s

12 . 6B x log x log ( B x )

)

≤ exp µ 13 √

√ Bx 1 − µ

¶ .

Estimation of ν

1

. First we observe that, if a large prime P divides M and ben M ≤ B then we have:

B ≥ ben M ≥ ben

p

( M ) ≥ ϕ

1

(ε, P , 0 , β

p

) ≥ ε log ( P / x ),

so that

P ≤ x exp ( B /ε) = x exp

µ B log x log 2

¶ .

If λ is large, we divide the interval [0 , λ ] into equal subintervals: [ λ

i

, λ

i+1

] , 0 ≤ i ≤ s − 1, such that λ

i+1

− λ

i

<

^1−λ₂

. We set T

₀

= 2x , T

_i

= x exp ( x

^λi

) for 1 ≤ i ≤ s − 1, and T

_s

= x exp (

^{B log x}_{log 2}

) . If λ <

¹₃

, there is just one interval in the subdivision. Further, we write A

₁

= a

₁

a

₂

. . . a

_s

with p | a

_i

H⇒ T

_i−1

< p ≤ T

_i

, and if we denote the number of solutions of ben ( a

i

N

_ε

) ≤ B by ν

₁⁽ⁱ⁾

clearly we have

ν

1

≤ Y

s i=1

ν

⁽1ⁱ⁾

.

To estimate ν

1⁽ⁱ⁾

let us denote the primes between T

i−1

and T

i

by T

i−1

< P

1

< · · · < P

r

≤ T

i

, and let a

i

= P

₁^y1

· · · P

r^yr

. We have

B ≥ ben ( a

_i

N

_ε

) ≥ X

r

i=1

ϕ

1

(ε, P

_i

, 0 , y

_i

) = X

r

i=1

ε y

_i

log P

i

x

≥ X

r

i=1

ε y

_i

log T

i−1

x . If i = 1 , T

₀

= 2x, this implies P

r

i=1

y

_i

≤

^B_{(log 2}^{(log x}₎2⁾

≤ 3B log x, and by Lemma 3, ν

1⁽¹⁾

≤ exp ( 3B log x log ( 2r )) ≤ exp ( 3B log x log T

1

) ≤ exp (( 1 + o ( 1 )) B x

^λ1

).

If i > 1 , we have P

r

i=1

y

i

≤

_εx^Bλi−1

, and by Lemma 3, ν

1⁽ⁱ⁾

≤ exp

µ B ε x

^λi−1

log T

i

¶

≤ exp ©

( 1 + o ( 1 )) B x

^{λi−λi−1}

ª ,

and from the choice of the λ

i

’s, one can easily see that, for B ≤ x

^λ

, ν

1

= Q

s

i=1

ν

1⁽ⁱ⁾

is negligible compared with ν

2

.

The other factors of (72) are easier to estimate:

Estimation of ν

3

. Let us denote the primes between 2x

2

and x by 2x

2

< P

r

< P

r−1

<

· · · < P

1

≤ x. By (62) and (4), x

2

= x

^θ

, and by (63), α

P_i

= 1. Let us write A

3

= P

₁^y1

· · · P

r^yr

. We have

B ≥ ben ( A

3

M ) ≥ X

r

i=1

ϕ

1

(ε, P

i

, 1 , 1 + y

i

) = X

r

i=1

ε y

i

log P

_i

x

₂

≥

X

r i=1

( log 2 )

²

log x y

i

. So, P

r

i=1

y

i

≤ B log x /( log 2 )

²

≤ 3B log x, and by Lemma 3,

ν

3

≤ exp ( 3B log x log ( 2r )) ≤ exp ( 3B ( log x )

²

).

Estimation of ν

4

. Replacing x by x

2

the upper bound obtained for ν

2

becomes:

ν

2

= exp ( O ( p

B x

2

)) = exp ( O ( √ B x

^θ

)).

Estimation of ν

5

. Replacing x by x

2

, the upper bound obtained for ν

3

becomes:

ν

5

≤ exp ( 3B log x log x

2

) = exp ( 3 θ B ( log x )

²

).

Estimation of ν

6

. Let p

1

, p

2

, . . . , p

r

≤ 2x

3

be the first primes and write A

6

= p

₁^y1

p

₂^y2

· · · p

r^yr

. By (71), y

i

≤ x, and thus by (62),

ν

6

≤ ( x + 1 )

^r

≤ ( x + 1 )

^x3

= exp ( x

^1−θ

log ( x + 1 )) and for B ≥ x

^−µ

and µ < 0 . 16, this is negligible compared with ν

2

.

Estimation of ν

₁⁰

. Let us denote the primes between

^x₂

and x by

^x₂

< P

_r

< P

_r−1

< · · · <

P

₁

≤ x, and let D

₁

= P

₁^y1

· · · P

r^yr

. We have α

Pi

= 1 and since D

₁

divides N

_ε

, y

_i

= 0 or 1.

By a computation similar to that of ν

2

, we obtain B ≥ ben N

_ε

D

1

≥ X

r

i=2

ϕ

2

(ε, P

i

, 1 , y

i

) = X

r

i=2

ε y

i

log x P

i

≥ X

r

i=2

ε y

i

x − P

i

x , and by using the Brun-Titchmarsch inequality and Lemma 1, it follows that

X

r i=2

p

i

y

i

≤ Bx

0 . 23 ε ≤ 6 . 3 Bx log x . Thus, as y

1

can only take 2 values, by Lemma 2 we have

ν

1⁰

≤ 2 exp (( 1 + o ( 1 )) 2 π

√ 3 s

6 . 3Bx log x

log ( B x ) ≤ exp ( 9 . 2 √ Bx ).

Estimation of ν

2⁰

. By an estimation similar to that of ν

3

, replacing ϕ

1

by ϕ

2

and using Lemma 3, we get

ν

2⁰

≤ exp ( 3B log

²

x ).

Estimation of ν

3⁰

. Replacing x by x

2

, it is similar to that of ν

1⁰

and we get ν

3⁰

= exp ( O ( p

B x

2

)).

Estimation of ν

4⁰

. Replacing x by x

2

, we get, as for ν

2⁰

,

ν

4⁰

≤ exp ( 3B log x log x

2

) = exp ( 3 θ B log

²

x ).

Estimation of ν

₅⁰

. As we have seen for ν

6

, we have D

₅

= p

₁^y1

· · · p

_r^yr

with y

i

≤ α

pi

≤ 3 log x and r ≤ π( 2x

3

) ≤ x

3

. Thus

ν

5⁰

≤ ( 1 + 3 log x )

^r

≤ exp ( x

^1−θ

log ( 1 + 3 log x )).

By formula (68) and the estimates of ν

i

and ν

i⁰

, the proof of Theorem 6 is completed. 2 By a more careful estimate, it would have been possible to improve the constant in (68).

However, using the Brun-Titchmarsch inequality we loose a factor √

2, and we do not see how to avoid this loss. A similar method was used in [3]. Also, the condition µ < 0 . 16 can be replaced easily by µ < 1.

7. Proof of Theorem 3

We shall need the following lemmas:

Lemma 4. Let n

_j

the sequence of h.c. numbers. There exists a positive real number c such that for j large enough , the following inequality holds:

n

j+1

n

j

≤ 1 + 1 ( log n

j

)

^c

.

Proof: This result was first proved by Erd˝os in [2]. The best constant c is given in [8]:

c = log ( 15 / 8 )

log 8 ( 1 − τ

0

) = 0 . 1405 . . .

with the value of τ

0

given by (8). 2

Lemma 5. Let n

j

be a h.c. number , and N

_ε

the superior h.c. number preceding n

j

. Then the benefit of n

j

( defined by ( 65 )) satisfies:

ben n

_j

= O (( log n

_j

)

^−γ

).

Proof: This is Theorem 1 of [8]. The value of γ is given by γ = θ( 1 − τ

0

)/( 1 + κ) = 0 . 03157 . . .

where θ, τ

0

and κ are defined by (4), (8) and (22). 2

To prove Theorem 3, first recall that n

k

is defined so that

n

k

≤ X < n

k+1

. (73)