Highly Composite Numbers and the Riemann hypothesis

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Highly Composite Numbers and the Riemann hypothesis

Jean-Louis Nicolas

Key words: Divisor function, Highly Composite Numbers, Riemann hypothesis.

2018 Mathematics Subject Classification:11N56, 11N37.

Abstract Let us denote by d(n) the number of divisors of n, by li(t) the logarithmic integral of t, by β₂ the number ^{log 3/2}_{log 2} = 0.584. . . and by R(t) the function t 7→ ²

√t+P

ρt^ρ/ρ²

log²t , where ρ runs over the non-trivial zeros of the Riemannζfunction. In his PHD thesis about highly composite numbers, Ramanujan proved, under the Riemann hypothesis, that

logd(n)

log 2 6li(logn) +β2li(log^β²n)− log^β²n

log logn−R(logn) +O √

logn (log logn)³

holds whenn tends to infinity. The aim of this paper is to give an effective form to the above asymptotic result of Ramanujan.

1 Introduction

Let us denote byd(n)the number of divisors ofnand byli(t)the logarithmic integral oft(see, below, §2.2). In [24, (235)], under the Riemann hypothesis,

Jean-Louis Nicolas

Univ Lyon, Université Claude Bernard Lyon 1, CNRS UMR 5208, Institut Camille Jordan, 43 blvd du 11 Novembre 1918, F-69622 Villeurbanne cedex, France

e-mail:nicolas@math.univ-lyon1.fr

Research partially supported by CNRS, UMR 5208.

1

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2 Jean-Louis Nicolas

Ramanujan proved that whenN → ∞,D(N) = max₁₆_n₆_Nd(n)satisfies logD(N)

log 2 = li(logN) +β₂li(log^β²N)− log^β²N

log logN −R(logN) + O(√ logN) (log logN)³,

(1.1) with, fork>1,

β_k= log(1 + 1/k)

log 2 (1.2)

and, fort >1,

R(t) = 2√

t+S(t)

log²t with S(t) =X

ρ

t^ρ

ρ², (1.3)

whereρruns over the non-trivial zeros of the Riemannζfunction. Moreover, in [24, (226)], Ramanujan writes under the Riemann hypothesis

|S(t)|=

X

ρ

t^ρ ρ²

6X

ρ

t^ρ ρ²

=√ tX

ρ

1 ρ(1−ρ)

=√ tX

ρ

1 ρ+ 1

1−ρ

= 2√ tX

ρ

1 ρ =τ√

t (1.4) with

τ= 2 +γ₀−log(4π) = 0.046 191 417 932 242 0. . . (1.5) where γ0 is the Euler constant. The value of τ = 2P

ρ1/ρ can be found in several books, for instance [8, p. 67] or [5, p. 272]. (1.4) implies that R(t) defined in (1.3) satisfies

(2−τ)

√t

log²x6R(t)6(2 +τ)

√t

log²t. (1.6)

It is convenient to use the following notation fort >1 F(t) = li(t) +β₂li(t^β²)− t^β²

logt with β₂=log 3/2

log 2 = 0.584. . . (1.7) The aim of this paper is to give an effective form to the result (1.1) of Ra- manujan and, more precisely, to prove

Theorem 1.1.(i) Under the Riemann hypothesis, forn >183783600, logd(n)

log 2 6F(logn)−R(logn)−5.12

√logn

(log logn)³ +1.52 log^β³n

log logn (1.8) with β3= (log(4/3))/log 2 = 0.415. . .

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(ii) If the Riemann hypothesis is not true, there exists infinitely manyn’s for which (1.8)does not hold. In other words, (i) is equivalent to the Riemann hypothesis.

(iii) Independently of the Riemann hypothesis, there exists infinitely many n’s such that

logd(n)

log 2 >F(logn)−R(logn)−25.3

√logn

(log logn)³ −1.45 log^β³n

log logn . (1.9) Corollary 1.2.Under the Riemann hypothesis, we have, forn >122522400,

logd(n)

log 2 6F(logn)−(2−τ)√ logn (log logn)² −5.12

√logn

(log logn)³+1.52 log^β³n log logn (1.10) forn /∈ {1,2,3,4,5,6,8,9,10,12,18,24},

logd(n)

log 2 6li(logn) +β2li(log^β²n), (1.11) for4324320< n6exp(10⁸) or forn >exp(1.56×10¹⁷),

logd(n)

log 2 6F(logn) (1.12)

and, for N⁽¹⁾< n <exp(10⁸) or forn >exp(1.11×10⁴⁰) logd(n)

log 2 6F(logn)−R(logn), (1.13) where

N⁽¹⁾= 2¹⁰3⁶5⁴7³

19

Y

p=11

p²

157

Y

p=23

p= 1.245143. . .×10⁷⁵. (1.14) Corollary 1.2 is easy to prove from Theorem 1.1 and some computation (see below Section 4.3). Under the Riemann hypothesis, (1.12) probably holds for all n >4324320 and also (1.13) for all n > N⁽¹⁾ but we have not been able to prove it.

Let us recall some effective upper bounds for ^log_{log 2}^d(n), obtained without any hypothesis:

logd(n)

log 2 61.5379398606. . . logn

log logn, n>3 (1.15) with equality forn= 6983776800 = 2⁵3³5²

Q19 p=7p

,

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logd(n)

log 2 6 logn

log logn+ 1.9348509679. . . logn

(log logn)², n>2, (1.16) with equality forn= 2⁸3⁵5³7²11²13²

Q83 p=17p

and logd(n)

log 2 6

2

X

i=1

(i−1)! logn

(log logn)ⁱ + 4.7623501211. . . logn

(log logn)³, n>2 (1.17) with equality forn= 2¹¹3⁶5⁴7³

Q23

p=11p² Q293 p=29p

.

Inequality (1.15) is proved in [18]. Inequalities (1.16) and (1.17) are proved in [28, pp. 41–49], cf. also [19, Section VII].

1.1 Notation

d(n) =X

m|n

1is the divisor function.

π(x) =X

p6x

1 is the prime counting function.Π(x) = X

p^k6x

1 k =X

k>1

π(x^1/k)

k .

pi denotes thei-th prime.P ={2,3,5,7,11, . . .}is the set of primes.

θ(x) =X

p6x

logpandψ(x) = X

p^k6x

logpare the Chebychev functions.

li(x)denotes the logarithmic integral ofx(cf. below Section 2.2).

F is defined in (1.7),R(t)andS(t)in (1.3).

Gis defined in (2.32), G1 in (2.33), G2 in (4.10), G3 in Section 4.3.2,H in (2.34) andH₀in (4.9).

τ is defined in (1.5),β_k in (1.2),ξin (3.12) andξ_k=ξ^β^k in (3.13).

A(x) = li(θ(x))−π(x), A₁(x)andA₂(x)are defined in (2.41)–(2.43).

σ₂is defined in Definition 2.4.

ξ⁽⁰⁾ = 10⁸+ 7 is defined in (3.21) andξ_k⁽⁰⁾ in (3.22).

The value of N⁽⁰⁾ is given in (3.23) and the one of logN⁽⁰⁾ in (3.24). The value ofN⁽¹⁾ andN⁽²⁾ are given respectively in (1.14) and in (4.7).

Highly composite (hc) numbers are defined in Section 3.4. Mj denotes the j-th hc number.

AslogNandlog logNoccur many times in the article, they are often replaced byLandλ. Similarly,L0meanslogN⁽⁰⁾ andλ0 meanslog logN⁽⁰⁾.

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We often implicitly use the following results : foruand v positive andw real, the function

t7→ (logt−w)^u

t^v is decreasing for t >exp(w+u/v) (1.18) and

max

t>e^w

(logt−w)^u t^v =u

v ^u

exp(−u−vw). (1.19)

1.2 Plan of the article

The proof of Theorem 1.1 follows the proof of (1.1) in [24]. We have replaced the asymptotic estimates in number theory used by Ramanujan by effective ones.

In Section 2, we recall and prove some results that we use in the sequel, first, in Section 2.1, about effective estimates of classical functions of prime number theory and later, in Section 2.2, about the logarithmic integral. In Section 2.3, the functionS, defined in (1.3), is studied and, finally, in Section 2.4, several lemmas in calculus are proved.

Under the Riemann hypothesis, Ramanujan proved that,A(x) = li(θ(x))−

π(x) is positive forx large enough, and it was an important argument for his proof of (1.1). In [20], it is proved that, under the Riemann hypothesis, A(x)>0 holds forx>11. In Section 2.5, some results of [20] about A(x) are recalled to be used in the proof of Theorem 1.1.

Section 3 is devoted to the study ofsuperior highly composite(shc) numbers. These numbers, introduced by Ramanujan to study the large values of the divisor function, play an important role in the proof of Theorem 1.1. In Section 3.2, the definition of shc numbers is recalled, and some properties and examples are given.

A shc number N is associated to a parameter ε and its largest prime factor is6ξ= 2^1/ε. In Section 3.7 (without any hypothesis) and in Section 3.8 (under the Riemann hypothesis) effective estimates ofN in terms ofξare given.

Let ε be a positive real number. The theorem of the six exponentials implies that there are at most four shc numbers associated toε. However, no ε is known with more than two shc numbers associated to it. This question is discussed in Section 3.1 and in Proposition 3.7.

The definition of highly composite numbers (hc) introduced by Ramanujan is recalled in Section 3.4. These numbers are used to determine the largest number (i.e.183783600) not satisfying (1.8), see Lemma 3.14 and Section 4.2.

The notion of benefit, recalled in Section 3.5, is convenient to find, on some interval, the numbers with a large number of divisors.

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In Section 3.6, an argument of convexity is given, which is used in Lemma 4.3 to shorten the computation in the proof of Theorem 1.1.

The proofs of Theorem 1.1 and Corollary 1.2 are given in Section 4.

The computations, both algebraic and numerical, have been carried out with Maple. On the website [32], one can find the code and a Maple sheet with the results.

2 Preliminary results 2.1 Effective estimates

Without any hypothesis, Büthe [4, Theorem 2] has shown by computation that

θ(x)< x, for 1< x610¹⁹ (2.1) while, Platt and Trudjan, [23, Theorem 1, Corollary 1] proved forx >0,

θ(x)<(1 +η)x with η= 1 + 7.5·10⁻⁷. (2.2) Without any hypothesis, Dusart [7, Théorème 5.2] has proved that

|θ(x)−x]< x

log³x forx>89 967 803. (2.3) From (2.3) and the computation of θ(x)forx <89 967 803, it is possible to show (cf. [6, Table 1, p. 114] or [33]) that we have θ(x)> bx forx>a for each of the following pairs of values ofaandb

a 127 367 1993 47491

b 0.8499 0.9134 0.9629 0.9927 (2.4)

We shall also use the inequality (cf. [29, (3.5)]):

π(x)> x

logx for x>17. (2.5)

Lemma 2.1.For each of the following pairs of values of a and b, we have π(x)< bx/logxforx>a.

a 2 376 2090

b 1.25506 1.19768 1.15963 (2.6)

Proof: Fora= 2, the result is quoted in [29, (3.6))]. For the two other values ofa, we start from the inequality (cf. [7, Théorème 6.9]) valid forx>60184

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π(x)6 x

logx−1.1 6 x

(logx)(1−1.1/log 60184) 61.1114 x logx. Furthermore, if p andp⁰ are two consecutive primes, on the interval [p, p⁰), the function f(t) = π(t)(logt)/t is decreasing. For a = 376, we check that f(376)<1.19768 holds and that for all primepsatisfying376< p <60184, we also have f(p) < 1.19768. A similar computation shows the result for

a= 2090.

Under the Riemann hypothesis, we shall use the upper bounds (cf. [30, (6.2) and (6.3)])

|ψ(x)−x|6 1 8π

√xlog²x, for x>73.2 (2.7) and

|θ(x)−x|6 1 8π

√xlog²x, for x>599. (2.8) Let us introduce

δ(t) =

(0 if t610¹⁹

1 if t >10¹⁹. (2.9)

Then (2.1) and (2.8) imply θ(x)6x+δ(x)

8π

√xlog²x, for x >0. (2.10) Lemma 2.2.Let us denote bypi thei-th prime. Then, forpi>127, we have

pi+1−pi

p_i+1 6149−139

149 60.0672. (2.11)

We order the prime powers p^m, with m > 1, in a sequence (a_i)_i_>₁ = (2,3,4,5,7,8,9,11, . . .). Then, fora_i>127,

a_i+1−a_i ai+1

= 1− a_i

ai+1 60.0672. (2.12) Proof: In [7, Proposition 5.4], it is proved that, forx>89693, there exists a primepsatisfyingx < p6x(1 + 1/log³x). This implies that forpi>89693, we have pi+16pi+pi/log³pi and

pi+1−pi

p_i+1 6pi+1−pi

p_i 6 1

log³pi

6 1

log³89693 = 0.00067423. . . For26p_i<89693, the computation of(p_i+1−p_i)/p_i+1 completes the proof of (2.11).

Ifpj is the largest prime6ai thenai+16pj+1 holds, so that, by (2.11), forai>127,

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a_i+1−a_i ai+1

= 1− a_i

ai+1 61− p_j

pj+1 60.0672

holds, which proves (2.12).

2.2 The logarithmic integral

Forxreal >1, we defineli(x)as (cf. [1, p. 228]) li(x) =

Z x

0

− dt

logt = lim

ε→0⁺

Z 1−ε

0

+ Z x

1+ε

dt logt

= Z x

2

dt

logt+ li(2). (2.13) For0< x <1, (cf. [22, p. 1–3]),

li(x) =γ0+ log(−log(x)) +

∞

X

k=1

log^kx k·k!

whereγ₀ is the Euler constant. We have the following values:

x 0.5 1 1.45136. . . 1.96904. . . 2

li(x) −0.37867. . . −∞ 0 1 1.145163. . . (2.14) From the definition of li(x), it follows that

d

dxli(x) = 1

logx and d²

dx²li(x) =− 1

xlog²x. (2.15) The functionx7→li(x)is increasing forx >1. Forx >1the second derivative ofli(x)is negative and increasing. Therefore, from Taylor’s formula, ifa,cand hare three real numbers satisfyinga >1,a+h >1andc= min(a, a+h)>1, one has

li(a) + h

loga− h²

2clog²c 6li(a+h)6li(a) + h

loga. (2.16) We also have forx→ ∞

li(x) =

N

X

k=1

(k−1)!x (logx)^k +O

x (logx)^N⁺¹

. (2.17)

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2.3 Study of the function S(t) defined by (1.3)

Lemma 2.3.If a, b are fixed real numbers satisfying 16a < b <∞, and g any function with a continuous derivative on the interval [a, b], then

X

ρ

Z b

a

g(t)t^ρ ρ dt=

Z b

a

g(t)

t−ψ(t)−log(2π)−1 2log

1− 1

t²

dt, (2.18)

whereρruns over the non-trivial zeros of the Riemannζ function.

Proof: This is Théorème 5.8(b) of [10, p. 169] or Theorem 5.8(b) of [9, p.

162].

Definition 2.4.One definesσ2by σ2=X

ρ

1

ρ² = 1−π²

8 + 2γ1+γ²₀=−0.046 154 317 295 804...

where the coefficientsγ_mare defined by the Laurent expansion ofζ(s)around 1 (see [5, p. 206 and 272])

ζ(s) = 1 s−1+

∞

X

m=0

(−1)^mγ_m

m!(s−1)^m Lemma 2.5.Forx >1, one has

S(x) = Z x

1

t−ψ(t)

t dt−(log(2π)) logx+σ2+π² 24−

∞

X

j=1

1

4j²x^2j. (2.19) Proof: Applying Lemma 2.3 withg(t) = 1/t, a= 1, b=xyields

S(x) =X

ρ

x^ρ ρ² =

Z x

1

t−ψ(t)−log(2π)−1 2log

1− 1

t² dt

t +σ₂ (2.20) which, by expanding log(1−1/t²)/(2t)in power series, implies (2.19).

Lemma 2.6.Let a be a real number >1/2. The function Y :=t7→ −a/t− log(1−1/t²)/(2t)is increasing and negative fort>2.

Proof: We haveY =−a/t+P∞

j=11/(2jt^2j+1), Y⁰= a

t²−

∞

X

j=1

2j+ 1 2jt^2j+2 > a

t²−

∞

X

j=1

3

2t^2j+2 > a t²−

∞

X

j=1

3/2

2^2jt² = a−1/2 t² >0.

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Therefore, fort>2,Y in increasing and, aslim_t→∞Y = 0,Y is negative.

Lemma 2.7.Let S be defined by (1.3)andx andy be real numbers. Then, under the Riemann hypothesis,

S(x)−S(y)60.18(x−y) for 26y6x, (2.21) S(x)−S(y)60.0398|x−y|log²y

√y for 10⁸6y6x (2.22)

and

S(x)

log²x− S(y) log²y

60.04|x−y|

√y for 10⁸6y6x. (2.23) Proof: In a first step, we assume that x andy satisfyai 6y < x6ai+1

for somei>1, witha_i defined as in Lemma 2.2. Then, from (2.20), we get S(x)−S(y) =

Z x

y

1−ψ(ai)

t −log(2π)

t −log(1−1/t²) 2t

dt. (2.24) From Lemma 2.6, the above square bracket is increasing fort>2and (2.24) yields

S(x)−S(y6 Z x

y

1−ψ(ai)

a_i+1 −log(2π)

a_i+1 −log(1−1/a²_i+1) 2a_i+1

dt

= (x−y)

1−ψ(ai) ai+1

−log(2π) ai+1

−log(1−1/a²_i+1) 2ai+1

. (2.25) Fora1 = 26ai 6a43 = 127, one computes the parenthesis of the right- hand side of (2.25). The maximum0.1759. . . is attained fora1= 2.

Forai >a44= 128, we use the inequalityψ(ai)>ai−(√

ailog²ai)/(8π)>

a_i−(√

a_i+1log²a_i+1)/(8π)(cf. (2.7)), whence, from (2.25) and Lemma 2.6, S(x)−S(y)

x−y 61−ψ(a_i)

ai+1 61− a_i ai+1

+log²a_i+1 8π√

ai+1

and, from (2.12) and (1.18), S(x)−S(y)

x−y 60.0672 +log²128 8π√

128 = 0.149994. . .

In the general case, if2< y < xholds, we determine the two integersi>1 andj>1 such thatai6y6ai+16ai+j 6x6ai+j+1. We have

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S(x)−S(y)

=S(ai+1)−S(y) +

j−1

X

k=1

S(ai+k+1)−S(ai+k)

!

+S(x)−S(ai+j)

60.18 ai+1−y+

j−1

X

k=1

ai+k+1−ai+k

!

+x−ai+j

!

= 0.18(x−y), which completes the proof of (2.21).

From (2.7) and (1.18), for73.26y6x, one has

Z x

y

t−ψ(t) t dt

6

Z x

y

log²t 8π√

tdt6|x−y|

8π

log²y

√y

, (2.26) Z x

y

dt

t = log(x/y)6x/y−1 =|x−y|/y (2.27) and

Z x

y

−log(1−1/t²)

2t dt

= Z x

y

1 2t

∞

X

j=1

1

jt^2jdt6 |x−y|

2y

∞

X

j=1

1 jy^2j

6 |x−y|

2y

∞

X

j=1

1

y^2j = |x−y|

2y(y²−1), (2.28) whence, from (2.20), (2.26), (2.27) and (2.28), withy₀= 10⁸,

|S(x)−S(y)|6|x−y|

log²y 8π√

y +log(2π)

y + 1

2y(y²−1)

=|x−y|log²y

√y

1

8π+ 1

√ylog²y

log(2π) + 1 2(y²−1)

!

6|x−y|log²y

√y

1

8π+ 1

√y₀log²y0

log(2π) + 1 2(y₀²−1)

!

60.0397. . .|x−y|log²y

√y

which proves (2.22).

To prove (2.23), one writes

S(x)

log²x− S(y) log²y

6|S(x)−S(y)|

log²x +

S(y) 1

log²x− 1 log²y

. (2.29) From (2.22), it follows

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|S(x)−S(y)|

log²x 6 |S(x)−S(y)|

log²y 60.0398|x−y|

√y . (2.30)

(1.4) and (1.5) imply|S(y)|6τ√

y60.0462√

y, whence, withy0= 10⁸,

S(y) 1

log²x− 1 log²y

=|S(y)|

Z x

y

2dt

t log³t 62|S(y)| |x−y|

y log³y 62τ |x−y|

√ylog³y 6 0.0924 log³y0

|x−y|

√y = 0.000014827. . . |x−y|

√y ,

which, together with (2.29) and (2.30), proves (2.23).

2.4 Four lemmas in calculus

Lemma 2.8.The function

f(t) = 1.52t^β³

logt −5.12√ t log³t

is positive for 7.38 < t < 1.1×10⁴⁰ and negative for 1 < t < 7.37 and t >1.11×10⁴⁰.

Proof : Let us write f(t) = (t^β³/logt)[1.52−5.12t^1/2−β³/log²t]. From (1.18), the above square bracket is maximal fort=t0= exp(2/(1/2−β3)) = 1.67. . .10¹⁰, is increasing for t < t0, decreasing for t > t0 and vanishes for

t= 7.3735. . . and1.10026. . .10⁴⁰.

Lemma 2.9.Let abe a positive real number and, for n>0, ϕn = log(n+ 1)−an.

(i) If, for some positive integerk,ais equal tolog(1 + 1/k), (i.e.k= 1/(e^a− 1)), then the sequence(ϕ_n)_n_>₀ attains its maximum on the two pointskand k−1. More precisely, for06n6k−2 orn>k+ 1,ϕ_n< ϕ_k−1=ϕ_k holds.

(ii) If log(1 + 1/(k+ 1)) < a < log(1 + 1/k) holds (log(1 + 1/0) = ∞ is assumed), then the maximum of ϕ_n is attained on only one point k = b1/(e^a−1)c. More precisely, for06n6k−1 orn>k+ 1,ϕ_n< ϕ_k holds.

Proof: Forn>1, we have∆n=ϕn−ϕ_n−1= log(1 + 1/n)−a.

Ifa= log(1 + 1/k)holds, then∆nis positive forn < k, vanishes forn=k and is negative forn > k, which implies (i).

Iflog(1+1/(k+1))< a <log(1+1/k)holds, then∆nis positive forn6k, and is negative forn > k, which implies (ii). Note thatk <1/(e^a−1)< k+ 1

holds and thus,k=b1/(e^a−1)c.

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Lemma 2.10.Let us set β_k = (log(1 + 1/k))/log 2 as in (1.2)above. The function

Φ(t) = 0.352 exp((β₃−β₂/2)t) + 0.9132 exp((β₄−β₂/2)t)−0.0143t² (2.31) is positive for t>0.

Proof: Leta=β3−β2/2 = 0.1225. . .andb=β4−β2/2 = 0.02944. . .. One hasΦ⁰= 0.352aexp(a t)+0.9132bexp(b t)−0.0286tandΦ⁰⁰= 0.352a²exp(a t)+

0.9132b²exp(b t)−0.0286, The third derivative is positive and thus, the second derivative is increasing. The variation of Φ⁰ and Φ is displayed in the array below (cf. [32]).

t 0 3.294 13.43 20.62 ∞

Φ⁰⁰ −0.02 − −0.019 − 0 + 0.039 + ∞

0.07 ∞

Φ⁰ & 0 & % 0 %

−0.12

1.37 ∞

Φ % & 0.6 & %

1.26 0.002

The minimum ofΦ(t)fort>0is>0.0019, which proves lemma 2.10.

Lemma 2.11.Let a, b, cbe three real numbers satisfying06a63,06b6 30andc>0.F(t)is defined by (1.7),S(t) by (1.3)andβk by (1.2).

(i) The function Gdefined by

G=G(a, b, c, t) =F(t)− a√ t log²t− b√

t

log³t+c t^β³

logt (2.32) is increasing fort>12.

(ii) The function

G₁(t) =G(2 +τ,0,0, t) =F(t)−(2 +τ)√ t

log²t (2.33)

is increasing and concave fort >1.

(iii) The function H defined by

H=H(a, b, c, t) =F(t)− a√ t

log²t− S(t) log²t − b√

t

log³t +c t^β³

logt (2.34) is continuous fort >1and increasing for t>12.