**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

**N is denoted by d**

*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*

^{r}_{1}

^{1}

*p*

^{r}_{2}

^{2}

### · · · *p*

^{r}_{`}

^{`}*, 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}_{`}

^{o}

^{θ}### ¢

### (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* *)*

^{−λ}

^{1}

### (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**

**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**

**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* *)*

^{b}^{1}

*)* ≤ *Q*

_{`}*(* *X* *)* ≤ exp *((* *log X* *)*

^{b}^{2}

*).*

*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* *)*

^{b}^{2}

*).* (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* *)*

^{b}^{1}

*).* (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}*)*

^{b}^{1}

*) 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<n**j*≤*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*

*1*

_{`−α+}### (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*

*1*

_{`−α+}### = 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* *)*

^{2}

^{δ−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*

*1*

_{`−α+}### 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*

^{β}m*p*

^{α}_{`}### = *p*

_{`}^{βθ−α}### ¡ 1 + *O* ¡

*p*

_{`}^{(τ}^{−}

^{1}

^{)θ}### ¢¢

_{β}### ≤ *p*

*log z*
log 2+_{βj}^{4}

*`*

### 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*

^{r}_{1}

^{ˆ}

^{1}

*p*

^{r}_{2}

^{ˆ}

^{2}

### · · · *p*

_{t}^{r}^{ˆ}

^{t}### (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

^{r}^{ˆ}

^{1}

*p*

^{r}_{2}

^{ˆ}

^{2}

### · · · *p*

^{r}_{t}^{ˆ}

_{−v}

^{t}^{−v}

*p*

*i*

_{1}

### · · · *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*

*i*1

*p*

*i*2

### · · · *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 −

^{ε}_{3}

*)* *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*

### ≥

_{ε}_{log}

^{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*

^{α}

^{p}### (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*

^{β}

^{p}*,*

*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*

_{α}

_{p}_{+}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*

_{α}

_{p}*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}

^{√}

^{π}_{3}

### q

*x*
*log x*

*)* *.*

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

2*π*

√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*

_{i}^{0}

### ) 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*

*i*

^{0}

*.* (72)

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

2### and *ν*

1^{0}

### 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}

^{y}^{1}

*P*

_{2}

^{y}^{2}

### · · · *P*

_{r}^{y}

^{r}*,* *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*

*P*

*i*

### = 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}

^{−λ}

_{2}

*. We set T*

_{0}

### = *2x* *,* *T*

_{i}### = *x exp* *(* *x*

^{λ}

^{i}*)* for 1 ≤ *i* ≤ *s* − 1, and *T*

_{s}### = *x exp* *(*

^{B log x}_{log 2}

*)* . If *λ <*

^{1}

_{3}

### , 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*

*1*

^{(}

^{i}^{)}*.*

### 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}

^{y}^{1}

### · · · *P*

*r*

^{y}

^{r}### . 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}

*2*

^{(}^{log x}_{)}

^{)}### ≤ *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*

^{λ}^{1}

*).*

*If i* *>* 1 *,* we have P

*r*

*i*=1

*y*

*i*

### ≤

_{ε}_{x}^{B}*λ*

*i−1*

### , and by Lemma 3, *ν*

1

^{(}^{i}^{)}### ≤ exp

### µ *B* *ε* *x*

^{λ}

^{i−1}*log T*

*i*

### ¶

### ≤ exp ©

*(* 1 + *o* *(* 1 *))* *B x*

^{λ}

^{i}^{−λ}

^{i−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}

^{y}^{1}

### · · · *P*

*r*

^{y}

^{r}### . 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}

^{y}^{1}

*p*

_{2}

^{y}^{2}

### · · · *p*

*r*

^{y}

^{r}*. By (71), y*

*i*

### ≤ *x, and thus by (62),*

*ν*

6### ≤ *(* *x* + 1 *)*

^{r}### ≤ *(* *x* + 1 *)*

^{x}^{3}

### = 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*

_{1}

^{0}

*.* Let us denote the primes between

^{x}_{2}

*and x by*

^{x}_{2}

*<* *P*

_{r}*<* *P*

_{r}_{−}

_{1}

*<* · · · *<*

*P*

_{1}

### ≤ *x, and let D*

_{1}

### = *P*

_{1}

^{y}^{1}

### · · · *P*

*r*

^{y}

^{r}### . We have *α*

*P*

*i*

### = *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*

1^{0}

### ≤ 2 exp *((* 1 + *o* *(* 1 *))* 2 *π*

### √ 3 s

### 6 *.* *3Bx log x*

### log *(* *B x* *)* ≤ exp *(* 9 *.* 2 √ *Bx* *).*

*Estimation of* *ν*

2^{0}

*.* By an estimation similar to that of *ν*

3### , replacing *ϕ*

1 ### by *ϕ*

2 ### and using Lemma 3, we get

*ν*

2^{0}

### ≤ exp *(* *3B log*

^{2}

*x* *).*

*Estimation of* *ν*

3^{0}

*.* *Replacing x by x*

2### , it is similar to that of *ν*

1^{0}

### and we get *ν*

3^{0}

### = exp *(* *O* *(* p

*B x*

2*)).*

*Estimation of* *ν*

4^{0}

*.* *Replacing x by x*

2### , we get, as for *ν*

2^{0}

### ,

*ν*

4^{0}

### ≤ exp *(* *3B log x log x*

2*)* = exp *(* 3 *θ* *B log*

^{2}

*x* *).*

*Estimation of* *ν*

_{5}

^{0}

*.* As we have seen for *ν*

6### , we have *D*

_{5}

### = *p*

_{1}

^{y}^{1}

### · · · *p*

_{r}^{y}

^{r}*with y*

*i*

### ≤ *α*

*p*

*i*

### ≤ *3 log x and r* ≤ *π(* *2x*

3*)* ≤ *x*

3### . Thus

*ν*

5^{0}

### ≤ *(* 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*

*i*

^{0}

### , 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