(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15)(16)(17)(18)(19)
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
idenotes 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
kis of the form n
k= p
r11p
r22· · · 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
iµ
for X → ∞ and log p
ilog 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
0is 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
εmaximizes d ( n )/ n
ε. The problem of finding h.c. numbers is in fact an optimization problem
max
n≤xd ( 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
jlarge 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
1a 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−τlog 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 )
0.465log 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−εlog ( 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 3log 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
jsuch that the number of l.c. numbers between n
j−1and n
j(which is exactly F ( n
j− 1 , 1 )) satisfies F ( n
j− 1 , 1 ) ≥ exp (( log n
j)
b1) for n
jlarge
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 2log log X)(log X), and thus log ( D ( X )/ D ( √
X )) ∼ log 2 2
log X log log X .
Observing that X < n
2j, 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−1log log X
¶
≥ exp µ 1
3
1
( 2 log n
j)
µ−1log ( 2 log n
j)
¶
≥ exp µ 1
( log n
j)
a¶
≥ 1 + 1 ( log n
j)
afor 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
jand following the proof of Theorem 8 in [7, p. 174], which yields a =
loglog 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
jsatisfying (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
ithe 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
iκ¢ . (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
j2 ≤ β ≤ 3 β
j2 . (26)
Indeed, as α
jand β
jare coprime, one can write B, the nearest integer to (β
jlog zlog 2
+ 2 ) , as B = u
1α
j− u
2β
jwith | u
1| ≤ β
j/ 2, and then α = α
j+ u
2and β = β
j+ u
1satisfy (25).
With the notation of Section 1, we write ˆ
n = n
kp
m+1p
m+2· · · p
m+βp
`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 )
β2
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
mτ+i, i
A log p
m+i¶
(31) and consequently,
Y
β i=1p
m+ip
m= exp Ã
βX
i=1
log p
m+ip
m!
≤ exp Ã
βX
i=1
p
m+i− p
mp
m!
≤ exp µ β
p
mmax µ
p
mτ+β, β
A log p
m+ⶶ
≤ exp ¡ O ¡
max ¡
( log X )
δ+θ(τ−1), ( log X )
2δ−θlog log X ¢¢¢
(32) by (26), (24), (3) and (1). Similarly, we get
α−
Y
1 i=0p
`p
`−i≤ exp µ α
p
`−α+1max µ
p
lτ, α A log p
`¶¶
≤ exp µ
O µ
max
µ ( log X )
δ− log z
( log X )
1−τ, (( log X )
δ− log z )
2log X log log X
¶¶¶
(33) by (28). Further, it follows from (3) and (25) that
p
βmp
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)θ¾
. (34)
It follows from (23) and (24) that
β
jÀ ( 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
12 δ − θ < −λ
1κλ
1< δ < 1 . It is possible to find such a δ if λ
1satisfies
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
r1ˆ1p
r2ˆ2· · · p
trˆt(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+1p
t+2· · · p
t+vp
t−v+1p
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+ip
t−v+i= 1 + p
t+i− p
t−v+ip
t−v+i≤ 1 + ( 1 + o ( 1 )) 1 p
tmax µ
p
tτ+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
vi=1
p
t+ip
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 )
τlog 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−τlog 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−τlog 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
rˆ1p
r2ˆ2· · · p
rtˆ−vt−vp
i1· · · p
ivwhere 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
i1p
i2· · · p
ivp
t−v+1p
t−v+2· · · p
tˆ
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
vin (46) can be chosen in (
2vv) 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
1/6and
f(xx)& 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 loglog 4(1/z))
1/2and 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≥
εlog4 log 2(1/z)and α, β by (25) and (26); we set ` = t + α and choose X = n
ka 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
vi=1
p
t+ip
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
¶
2log 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
0such that
N
0≤ NP and d ( N
0) ≤ 2d ( N ) .
Definition. Let ε, 0 < ε < 1, and N
εsatisfy (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+1α
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
pα
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
αpp
¶
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
pα
p+ 1
¶ .
We have ϕ
2≥ 0 , ψ
2≥ 0 , ψ
2(α
2, α
p− 1 ) = 0. Moreover, observe that ψ
1is an increasing function of β
p− α
p, and ψ
2is an increasing function of α
p− β
p, for α
pfixed.
We will prove:
Theorem 6. Let x → +∞, ε be defined by ( 60 ) and N
εby ( 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
kbe 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
kbe the kth prime. The number of solutions of the inequality
p
1x
1+ p
2x
2+ · · · + p
kx
k+ · · · ≤ x (69) in integers x
1, x
2, . . . , is exp (( 1 + o ( 1 ))
2√π3q
x log x
) .
Proof: The number T ( n ) of partitions of n into primes satisfies (cf. [5]) log T ( n ) ∼
2π
√3
q
nlog 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
ris ≤ ( 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
ydivides 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
1A
2· · · A
6with ( A
i, A
j) = 1 and p | A
1H⇒ p > 2x
p | A
2H⇒ x < p ≤ 2x p | A
3H⇒ 2x
2< p ≤ x p | A
4H⇒ x
2< p ≤ 2x
2p | A
5H⇒ 2x
3< p ≤ x
2p | A
6H⇒ p ≤ 2x
3,
where x
2and x
3are defined by (62). Similarly, we write D = D
1D
2. . . D
5, with ( D
i, D
j) = 1 and
p | D
1H⇒ x / 2 < p ≤ x p | D
2H⇒ x
2< p ≤ x / 2 p | D
3H⇒ x
2/ 2 < p ≤ x
2p | D
4H⇒ 2x
3< p ≤ x
2/ 2 p | D
5H⇒ p ≤ 2x
3. We have
ben M = X
6 i=1ben ( A
iN
ε) + X
5i=1
ben ( N
ε/ D
i), and denoting by ν
i(resp. ν
i0) the number of solutions of
ben ( A
iN
ε) ≤ B ( resp. ben ( N
ε/ D
i) ≤ B ),
we have
ν ≤ Y
6 i=1ν
iY
5 i=1ν
i0. (72)
In (72), we shall see that the main factors are ν
2and ν
10and 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
1y1P
2y2· · · P
ryr, 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
2N
ε) ≥ X
ri=2
ϕ
1(ε, P
i, 0 , y
i) = X
ri=2
ε y
ilog ( P
i/ x )
≥ X
ri=2
ε y
iP
i− x P
i≥ X
ri=2
ε y
i2x ( P
i− x ) ≥ X
r i=20 . 115 ε y
ix p
i.
By (71), the number of possible choices for y
1is less than ( x + 1 ) , so that ν
2is certainly less than ( x + 1 ) times the number of solutions of:
X
∞ i=2p
iy
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 xlog 2) . If λ <
13, there is just one interval in the subdivision. Further, we write A
1= a
1a
2. . . a
swith p | a
iH⇒ T
i−1< p ≤ T
i, and if we denote the number of solutions of ben ( a
iN
ε) ≤ B by ν
1(i)clearly we have
ν
1≤ Y
s i=1ν
(1i).
To estimate ν
1(i)let us denote the primes between T
i−1and T
iby T
i−1< P
1< · · · < P
r≤ T
i, and let a
i= P
1y1· · · P
ryr. We have
B ≥ ben ( a
iN
ε) ≥ X
ri=1
ϕ
1(ε, P
i, 0 , y
i) = X
ri=1
ε y
ilog P
ix
≥ X
ri=1
ε y
ilog T
i−1x . If i = 1 , T
0= 2x, this implies P
ri=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
ri=1
y
i≤
εxBλi−1, and by Lemma 3, ν
1(i)≤ exp
µ B ε x
λi−1log T
i¶
≤ exp ©
( 1 + o ( 1 )) B x
λi−λi−1ª ,
and from the choice of the λ
i’s, one can easily see that, for B ≤ x
λ, ν
1= Q
si=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
2and x by 2x
2< P
r< P
r−1<
· · · < P
1≤ x. By (62) and (4), x
2= x
θ, and by (63), α
Pi= 1. Let us write A
3= P
1y1· · · P
ryr. We have
B ≥ ben ( A
3M ) ≥ X
ri=1
ϕ
1(ε, P
i, 1 , 1 + y
i) = X
ri=1
ε y
ilog P
ix
2≥
X
r i=1( log 2 )
2log x y
i. So, P
ri=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
2the upper bound obtained for ν
2becomes:
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
3becomes:
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
3be the first primes and write A
6= p
1y1p
2y2· · · p
ryr. By (71), y
i≤ x, and thus by (62),
ν
6≤ ( x + 1 )
r≤ ( x + 1 )
x3= exp ( x
1−θlog ( x + 1 )) and for B ≥ x
−µand µ < 0 . 16, this is negligible compared with ν
2.
Estimation of ν
10. Let us denote the primes between
x2and x by
x2< P
r< P
r−1< · · · <
P
1≤ x, and let D
1= P
1y1· · · P
ryr. We have α
Pi= 1 and since D
1divides N
ε, y
i= 0 or 1.
By a computation similar to that of ν
2, we obtain B ≥ ben N
εD
1≥ X
ri=2
ϕ
2(ε, P
i, 1 , y
i) = X
ri=2
ε y
ilog x P
i≥ X
ri=2
