The Landau function and the Riemann hypothesis

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The Landau function and the Riemann hypothesis

Marc Deléglise, Jean-Louis Nicolas

To cite this version:

The Landau function and the Riemann

hypothesis

Marc Deléglise, Jean-Louis Nicolas

July 8, 2019

Abstract

The Landau function 𝑔(𝑛) is the maximal order of an element of the sym-metric group 𝔖𝑛; it is also the largest product of powers of primes whose sum

is ≤ 𝑛. The main result of this article is that the property “ For all 𝑛 ≥ 1, log 𝑔(𝑛) < √li−1_{(𝑛)” (where li}−1_{denotes the inverse function of the}

loga-rithmic integral) is equivalent to the Riemann hypothesis.

2010 Mathematics Subject Classification: Primary 11A25; Secondary 11N37, 11N05, 11-04.

Keywords: distribution of primes, champion number, highly composite number, Landau function, Riemann hypothesis.

1

Introduction

Let 𝑛 be a positive integer. In [13], Landau introduced the function 𝑔(𝑛) as the maximal order of an element in the symmetric group 𝔖𝑛; he showed that

𝑔(𝑛) = max

𝓁(𝑀)≤𝑛𝑀 (1.1)

where 𝓁 is the additive function such that 𝓁(𝑝𝛼_{) = 𝑝}𝛼 for 𝑝 prime and 𝛼 ≥ 1.

In other words, if the standard factorization of 𝑀 is 𝑀 = 𝑞𝛼1

1 𝑞 𝛼2 2 ⋯ 𝑞 𝛼_𝑗 𝑗 we have 𝓁(𝑀) = 𝑞𝛼1 1 + 𝑞 𝛼₂ 2 +⋯ + 𝑞 𝛼_𝑗

𝑗 and 𝓁(1) = 0. He also proved that

log 𝑔(𝑛) ∼√𝑛log 𝑛, 𝑛→∞.

A function close to the Landau function is the function ℎ(𝑛) defined for 𝑛 ≥ 2 as the greatest product of a family of primes 𝑞1 < 𝑞2 < ⋯ < 𝑞𝑗 the sum of which

does not exceed 𝑛. If 𝜇 denotes the Möbius function, ℎ(𝑛) can also be defined by ℎ(𝑛) = max

𝓁(𝑀)≤𝑛

𝜇(𝑀)≠0

𝑀 . (1.2)

The above equality implies ℎ(1) = 1. Note that

𝓁(𝑔(𝑛))≤ 𝑛 and 𝓁(ℎ(𝑛)) ≤ 𝑛. (1.3)

From (1.2) and (1.1), it follows that

ℎ(𝑛)≤ 𝑔(𝑛), (𝑛≥ 1). (1.4)

Sequences (𝑔(𝑛))𝑛≥1and (ℎ(𝑛))𝑛≥1are sequences A000793 and A159685 in the

OEIS (On-line Encyclopedia of Integer Sequences). One can find results about 𝑔(𝑛)in [15, 17, 7, 11, 20], see also [18] and [4, §10.10]. In the introductions of [7, 11], other references are given. The three papers [8, 9, 10] are devoted to ℎ(𝑛). A fast algorithm to compute 𝑔(𝑛) (resp. ℎ(𝑛)) is described in [11] (resp. [8, §8]). In [9, (4.13)], it is shown that

log ℎ(𝑛)≤ log 𝑔(𝑛) ≤ log ℎ(𝑛) + 5.68 (𝑛 log 𝑛)1∕4_{, 𝑛≥ 1. (1.5)}

Let li denote the logarithmic integral and li−1its inverse function (cf. below §2.2).

In [15, Theorem 1 (i)], it is proved that log 𝑔(𝑛) =

√

li−1(𝑛) +(√𝑛exp(−𝑎√log 𝑛) )

(1.6) holds for some positive 𝑎. The asymptotic expansion of log 𝑔(𝑛) does coincide with the one of √li−1_(𝑛)(cf. [15, Corollaire, p. 225]) and also, from (1.5), with

the one of log ℎ(𝑛) : log ℎ(𝑛) log 𝑔(𝑛) √ li−1_(𝑛) ⎫ ⎪ ⎬ ⎪ ⎭ = √ 𝑛log 𝑛 ( 1 + log log 𝑛 − 1 2 log 𝑛 −

(log log 𝑛)2_{− 6 log log 𝑛 + 9 + 𝑜(1)}

8 log2𝑛

) (1.7) In [15, Théorème 1 (iv)], it is proved that under the Riemann hypothesis the inequality

log 𝑔(𝑛) < √

holds for 𝑛 large enough. In August 2009, the second author received an e-mail of Richard Brent asking whether it was possible to replace “𝑛 large enough” by “𝑛 ≥ 𝑛0” with a precise value of 𝑛0. The aim of this paper is to anwer this question

positively. For 𝑛 ≥ 2, let us introduce the sequences log 𝑔(𝑛) = √ li−1(𝑛) − 𝑎_𝑛(𝑛 log 𝑛)1∕4 i.e. 𝑎_𝑛 = √ li−1_{(𝑛) − log 𝑔(𝑛)} (𝑛 log 𝑛)1∕4 , (1.9) log ℎ(𝑛) = √ li−1(𝑛) − 𝑏𝑛(𝑛 log 𝑛) 1∕4 _{i.e. 𝑏} 𝑛 = √ li−1_{(𝑛) − log ℎ(𝑛)} (𝑛 log 𝑛)1∕4 , (1.10)

and the constant

𝑐 =∑

𝜌

1

|𝜌(𝜌 + 1)| = 0.046 117 644 421 509 … (1.11) where 𝜌 runs over the non trivial zeros of the Riemann 𝜁 function. The computation of the above numerical value is explained in [10, Section 2.4.2]. We prove

Theorem 1.1. Under the Riemann hypothesis,

(i) log 𝑔(𝑛) < √ li−1(𝑛) for 𝑛≥ 1, (ii) 𝑎_𝑛 ≥ 2 − √ 2 3 − 𝑐 − 0.43 log log 𝑛 log 𝑛 >0 for 𝑛≥ 2, (iii) 𝑎𝑛 ≤ 2 −√2 3 + 𝑐 + 1.02 log log 𝑛 log 𝑛 for 𝑛≥ 19425, (iv) 0.11104 < 𝑎_{𝑛≤ 𝑎2}= 0.9102 … for 𝑛≥ 2, (v) 0.149 … = 2 − √ 2

3 − 𝑐 ≤ lim inf 𝑎𝑛 ≤ lim sup 𝑎𝑛 ≤

2 −√2 3 + 𝑐 = 0.241 … (vi) When 𝑛→∞, (2 −√2 3 − 𝑐 )( 1 + log log 𝑛 +(1) 4 log 𝑛 ) ≤ 𝑎𝑛 ≤(2 − √ 2 3 + 𝑐 )( 1 + log log 𝑛 +(1) 4 log 𝑛 ) .

Corollary 1.3. Each of the six points of Theorem 1.1 is equivalent to the Riemann

hypothesis.

Proof. If the Riemann hypothesis fails, it is proved in [15, Theorem 1 (ii)] that there exists 𝑏 > 1∕4 such that

log 𝑔(𝑛) = √

li−1(𝑛) + Ω_±((𝑛 log 𝑛)𝑏) (1.12) which contradicts (i), (ii), …, (vi) of Theorem 1.1.

In the paper [10], the following theorem is proved:

Theorem 1.4. Under the Riemann hypothesis,

(i) log ℎ(𝑛) < √ li−1(𝑛) for 𝑛 ≥ 1, (ii) 𝑏₁₇= 0.49795 … _{≤ 𝑏𝑛 ≤ 𝑏1137}= 1.04414 … for 𝑛_{≥ 2,} (iii) 𝑏𝑛≥ 2 3 − 𝑐 − 0.23 log log 𝑛 log 𝑛 for 𝑛≥ 18, (iv) 𝑏𝑛≤ 2 3 + 𝑐 + 0.77 log log 𝑛 log 𝑛 for 𝑛≥ 4 422 212 326,

(v) 2∕3 − 𝑐 = 0.620 … _{≤ lim inf 𝑏𝑛 ≤ lim sup 𝑏𝑛 ≤ 2∕3 + 𝑐 = 0.712 …} (vi) and, when 𝑛→∞,

(₂ 3− 𝑐 ) ( 1 + log log 𝑛 +(1) 4 log 𝑛 ) ≤ 𝑏𝑛 ≤(2 3+ 𝑐 ) ( 1 + log log 𝑛 +(1) 4 log 𝑛 ) .

The main tools in the proof of Theorem 1.4 in [10] are the explicit formulas for ∑𝑝𝑚_≤𝑥𝑝and ∑_𝑝𝑚_≤𝑥log 𝑝.

We deduce Theorem 1.1 about 𝑔(𝑛) from Theorem 1.4 about ℎ(𝑛) by studying the difference log 𝑔(𝑛) − log ℎ(𝑛) in view of improving inequalities (1.5). More precisely, we prove

(i) For 𝑛 tending to infinity, log𝑔(𝑛) ℎ(𝑛) = √ 2 3 (𝑛 log 𝑛) 1∕4 (

1 + log log 𝑛 − 4 log 2 − 11∕3 4 log 𝑛 − 3 32(log log 𝑛) 2₋(3 log 2 4 + 15 16 )

log log 𝑛 + (log 2)2

2 + 29 log 2 12 + 635 288 (log 𝑛)2 ) + ( (log log 𝑛)3 (log 𝑛)3 ) (ii) log𝑔(𝑛) ℎ(𝑛) ≤ √ 2 3 (𝑛 log 𝑛) 1∕4 ( 1 + log log 𝑛 + 2.43 4 log 𝑛 ) for 𝑛≥ 3 997 022 083 663, (iii) log𝑔(𝑛) ℎ(𝑛) ≥ √ 2 3 (𝑛 log 𝑛) 1∕4 ( 1 + log log 𝑛 − 11.6 4 log 𝑛 ) for 𝑛≥ 4 230,

(iv) For 𝑛≥ 1 we have 𝑔(𝑛)

ℎ(𝑛) ≥ 1 with equality for 𝑛= 1, 2, 3, 5, 6, 8, 10, 11, 15, 17, 18, 28, 41, 58, 77. (v) For 𝑛 ≥ 1, we have log𝑔(𝑛)

ℎ(𝑛) ≤ 0.62066 … (𝑛 log 𝑛)

1∕4 _{with equality for}

𝑛= 2243.

Remark 1.6. From the asymptotic expansion (i), it follows that, for 𝑛 very large,

the inequalitylog(𝑔(𝑛)∕ℎ(𝑛)) > (√2∕3)(𝑛 log 𝑛)1∕4 _{holds. But finding the largest}

𝑛for which

(log(𝑔(𝑛)∕ℎ(𝑛)))∕(𝑛 log 𝑛)1∕4 _{does not exceed}√_{2∕3 = 0.471 … seems difficult.} Theorem 1.7. For 𝑛≥ 373 623 863, √ 𝑛log 𝑛 ( 1 + log log 𝑛 − 1 2 log 𝑛 − (log log 𝑛)2 8 log2𝑛 ) ≤ log ℎ(𝑛) ≤ log 𝑔(𝑛) (1.13) and, for 𝑛≥ 4,

log ℎ(𝑛)≤ log 𝑔(𝑛) ≤√𝑛log 𝑛 (

1 + log log 𝑛 − 1 2 log 𝑛

)

The lower bound (1.13) of log ℎ(𝑛) improves on [9, Theorem 4] where it was shown that, for 𝑛 ≥ 77 615 268, we have log ℎ(𝑛) ≥ √𝑛log 𝑛(1 + log log 𝑛−1.16

2 log 𝑛

) . Inequality (1.14) improves on the result of [15, Corollaire, p. 225], where −1 was replaced by −0.975. From the common asymptotic expansion (1.7) of log ℎ(𝑛) and log 𝑔(𝑛), one can see that the constants 8 in (1.13) and −1 in (1.14) are optimal.

1.1

Notation

− 𝜋_𝑟(𝑥) = ∑

𝑝_≤𝑥

𝑝𝑟. For 𝑟 = 0, 𝜋(𝑥) = 𝜋₀(𝑥) = ∑

𝑝_≤𝑥

1 is the prime counting function. 𝜋− 𝑟(𝑥) = ∑ 𝑝<𝑥 𝑝𝑟. − 𝜃(𝑥) =∑ 𝑝≤𝑥

log 𝑝is the Chebichev function. 𝜃−_{(𝑥) =}∑

𝑝<𝑥log 𝑝.

− _{ = {2, 3, 5, …} denotes the set of primes. 𝑝1} = 2, 𝑝₂ = 3, … , 𝑝_𝑗is the 𝑗-th prime. For 𝑝 ∈  and 𝑛 ∈ ℕ, 𝑣𝑝(𝑛)denotes the largest exponent such that

𝑝𝑣𝑝(𝑛)divides 𝑛.

− P+_{(𝑛)denotes the largest prime factor of 𝑛.}

− li(𝑥) denotes the logarithmic integral of 𝑥 (cf. below Section 2.2). The inverse function is denoted by li−1_.

− If lim

𝑛→∞𝑢𝑛 = +∞, 𝑣𝑛 = Ω±(𝑢𝑛) is equivalent to lim sup_𝑛→∞ 𝑣𝑛∕𝑢𝑛 > 0 and

lim inf

𝑛→∞ 𝑣𝑛∕𝑢𝑛 <0.

− We use the following constants:

– 𝑥₁takes three values (cf. (2.3)),

– 𝑥₀= 1010+19is the smallest prime exceeding 1010, log(𝑥₀) = 23.025 850 …,

– 𝜈₀ = 2 220 832 950 051 364 840 = 2.22 … 1018 is defined below in

(3.17),

– log 𝜈₀ = 42.244414 … , log log 𝜈₀ = 3.743472 …

– The numbers (𝜆_𝑗)_𝑗≥2 described in Lemma 3.8 and (𝑥(0)_𝑗 )_{2≤𝑗≤29}defined in (3.18).

– For convenience, we sometimes write 𝐿 for log 𝑛, 𝜆 for log log 𝑛, 𝐿₀

We often implicitly use the following result : for 𝑢 and 𝑣 positive and 𝑤 real, the function

𝑡↦ (log 𝑡 − 𝑤)

𝑢

𝑡𝑣 is decreasing for 𝑡 > exp(𝑤 + 𝑢∕𝑣). (1.15)

Also, if 𝜖 and 𝜖0 are real numbers satisfying 0 ≤ 𝜖 ≤ 𝜖0 < 1we shall use the

following upper bound 1 1 − 𝜖 = 1 + 𝜖 1 − 𝜖 ≤ 1 + 𝜖 1 − 𝜖₀. (1.16)

Let us write 𝜎0= 0, 𝑁0= 1, and, for 𝑗 ≥ 1,

𝑁_𝑗 = 𝑝₁𝑝₂⋯ 𝑝𝑗 and 𝜎𝑗 = 𝑝1+ 𝑝2+⋯ + 𝑝𝑗 = 𝓁(𝑁𝑗). (1.17)

For 𝑛 ≥ 0, let 𝑘 = 𝑘(𝑛) denote the integer 𝑘 ≥ 0 such that

𝜎_𝑘 = 𝑝₁+ 𝑝₂+⋯ + 𝑝𝑘 ≤ 𝑛 < 𝑝1+ 𝑝2+⋯ + 𝑝𝑘+1 = 𝜎𝑘+1. (1.18)

In [8, Proposition 3.1], for 𝑗 ≥ 1, it is proved that

ℎ(𝜎_𝑗) = 𝑁_𝑗. (1.19)

In the general case, one writes 𝑛 = 𝜎𝑘+ 𝑚with 0 < 𝑚 < 𝑝𝑘+1and, from [8, Section

8], we have

ℎ(𝑛) = 𝑁_𝑘𝐺(𝑝_𝑘, 𝑚) (1.20)

where 𝐺(𝑝𝑘, 𝑚)can be calculated by the algorithm described in [11, Section 9].

1.2

Plan of the article

− In Section 2, we recall some effective bounds for the Chebichev function 𝜃(𝑥)and for 𝜋_𝑟(𝑥) =∑_𝑝≤𝑥𝑝𝑟. We give also some properties of the logarith-mic integral li(𝑥) and its inverse li−1.

− In Section 4 we present some methods used to compute efficiently: how to quickly enumerate the superchampion numbers, how to find the largest integer 𝑛 in a finite interval [𝑎, 𝑏] which does not satisfy a boolean property 𝑜𝑘(𝑛), by computing only a small number of values 𝑜𝑘(𝑛).

− In Section 5 we prove Theorem 1.7.

− In Section 6, in preparation to the proof of Theorem 1.5, we study the func-tion log 𝑔(𝑛) − log ℎ(𝑛) for which we give an effective estimate for 𝑛 ≥ 𝜈0

(defined in (3.17)) and also an asymptotic estimate.

− In Section 7, we prove Theorem 1.5, first for 𝑛 ≥ 𝜈₀, by using the results of Section 6, and further, for 𝑛 < 𝜈0, by explaining the required computation.

− In Section 8, we prove Theorem 1.1. For 𝑛 ≥ 𝜈₀, it follows from the reunion of the proofs of Theorem 1.4 (given in [10]) and of Theorem 1.5. For 𝑛 < 𝜈0,

some more computation is needed.

All computer calculations have been implemented in Maple and C++. Maple

pro-grams are slow but can be executed by anyone disposing of Maple. C++_programs

are much faster. They use real double precision, except for the demonstration of the Lemma 8.1 where we used the GNU-MPFR Library to compute with real num-bers with a mantissa of 80 bits. The most expensive computations are the proof of theorem 1.5.(ii) and the proof of Lemma 8.1 which took respectively 40 hours and 10 hours of CPU (with the C++ programs). The Maple programs can be loaded on

[27].

2

Useful results

2.1

Effective estimates

In [5], Büthe has proved

𝜃(𝑥) =∑

𝑝≤𝑥

log 𝑝 < 𝑥 for 𝑥 ≤ 1019 (2.1) while Platt and Trudgian in [21] have shown that

𝜃(𝑥) < (1 + 7.5⋅ 10−7) 𝑥for 𝑥 ≥ 2 (2.2) so improving on results of Schoenfeld [26]. Without any hypothesis, we know that

|𝜃(𝑥) − 𝑥] < 𝛼 𝑥

with 𝛼 = ⎧ ⎪ ⎨ ⎪ ⎩ 1 and 𝑥₁ = 89 967 803 (cf. [12, Theorem 4.2]) 0.5 and 𝑥₁ = 767 135 587 (cf. [12, Theorem 4.2]) 0.15 and 𝑥₁ = 19 035 709 163 (cf. [3, Theorem 1.1]) .

Lemma 2.1. Let us denote 𝜃−_{(𝑥) =}∑

𝑝<𝑥log 𝑝. Then 𝜃(𝑥)≥ 𝜃−_{(𝑥) ≥ 𝑥 − 0.0746} 𝑥 log 𝑥 (𝑥 > 48757) (2.4) 𝜃(𝑥) _{≤ 𝑥}(1 + 0.000079 log 𝑥 ) (𝑥 > 1) (2.5) 𝜋(𝑥) < 1.26 𝑥 log 𝑥 (𝑥 > 1). (2.6)

Proof. − From [26, Corollary 2*, p. 359], for 𝑥 ≥ 70877, we have 𝜃(𝑥) > 𝐹(𝑥)with 𝐹 (𝑡) = 𝑡(1 − 1∕(15 log 𝑡)). As 𝐹 (𝑡) is increasing for 𝑡 > 0, this implies that if 𝑥 > 70877 then 𝜃−_{(𝑥)≥ 𝐹 (𝑥) holds. Indeed, if 𝑥 is not prime,}

we have 𝜃−_{(𝑥) = 𝜃(𝑥)}while if 𝑥 > 70877 is prime then 𝑥−1 > 70877 holds

and we have 𝜃−_{(𝑥) = 𝜃(𝑦) > 𝐹 (𝑦)for 𝑦 satisfying 𝑥 − 1 < 𝑦 < 𝑥. When 𝑦}

tends to 𝑥, we get 𝜃−_{(𝑥) ≥ 𝐹 (𝑥) and, as 1∕15 < 0.0746 holds, this proves}

(2.4) for 𝑥 > 70877. Now, let us assume that 48757 < 𝑥 ≤ 70877 holds. For all primes 𝑝 satisfying 48757 ≤ 𝑝 < 70877 and 𝑝+ _{the prime following}

𝑝we consider the function 𝑓(𝑡) = 𝑡(1 − 0.0746∕ log 𝑡) for 𝑡 ∈ (𝑝, 𝑝+]. As 𝑓 is increasing, the maximum of 𝑓 is 𝑓(𝑝+₎and 𝜃−_(𝑡)is constant and equal to

𝜃(𝑝). So, to complete the proof of (2.4), we check that 𝜃(𝑝) ≥ 𝑓(𝑝+_)holds

for all these 𝑝’s.

− (2.5) follows from (2.1) for 𝑥 ≤ 1019, while, for 𝑥 > 1019, from (2.3), we have 𝜃(𝑥)_{≤ 𝑥} ( 1 + 0.15 log3𝑥 ) ≤ 𝑥 ( 1 + 0.15 (log21019_{)(log 𝑥)} ) = 𝑥 ( 1 + 0.0000783 … log 𝑥 ) . − (2.6) is stated in [24, (3.6)].

Lemma 2.2. Let us set

𝑊(𝑥) =∑

𝑝≤𝑥

log 𝑝

Then, for 𝑥 >0, 𝑊(𝑥) 𝑥 ≤ 𝜔 = { 𝑊(7)∕7 = 1.045 176 … if 𝑥 ≤ 7.32 1.000014 if 𝑥 >7.32. (2.8)

Proof. First, we calculate 𝑊 (𝑝) for all primes 𝑝 < 106. For 11 ≤ 𝑝 < 106, 𝑊(𝑝) < 𝑝holds while, for 𝑝 ∈ {2, 3, 5, 7}, the maximum of 𝑊 (𝑝)∕𝑝 is attained for 𝑝 = 7. If 𝑝 and 𝑝+ _{are consecutive primes, 𝑊 (𝑥) is constant and 𝑊 (𝑥)∕𝑥 is}

decreasing on [𝑝, 𝑝+_{). As 𝑊 (7) = 7.316 … this proves (2.8) for 𝑥 ≤ 7.32 and}

𝑊(𝑥) < 𝑥for 7.32 < 𝑥 ≤ 106.

Let us assume now that 𝑥 > 𝑦 = 106_{holds. We have}

𝑊(𝑥) = 𝑊 (𝑦) + ∑ 𝑦<𝑝≤𝑥 log 𝑝 1 − 1∕𝑝 ≤ 𝑊 (𝑦) + 𝑦 𝑦− 1 ∑ 𝑦<𝑝≤𝑥 log 𝑝 = 𝑊 (𝑦) − 𝑦 𝑦− 1𝜃(𝑦) + 𝑦 𝑦− 1𝜃(𝑥) = 12.240 465 … + 106 106_{− 1}𝜃(𝑥) and, from (2.2), 𝑊(𝑥)≤ 12.241 𝑥 106 + 106 106_{− 1}(1 + 7.5 × 10 −7_{)𝑥 < 1.00001399 … 𝑥,}

which completes the proof of Lemma 2.2.

Lemma 2.3. Let 𝐾 ≥ 0 and 𝛼 > 0 be two real numbers. Let us assume that there exists 𝑋₀>1 such that, for 𝑥 _{≥ 𝑋0},

𝑥− 𝛼𝑥

log𝐾+1𝑥 ≤ 𝜃(𝑥) ≤ 𝑥 + 𝛼𝑥 log𝐾+1𝑥

. (2.9)

If 𝑎 is a positive real number satisfying 𝑎 < log𝐾+1𝑋₀, for 𝑥≥ 𝑋0, we have

𝜋 ( 𝑥+ 𝑎 𝑥 log𝐾𝑥 ) − 𝜋(𝑥)≥ 𝑏 𝑥 log𝐾+1𝑥 (2.10) with 𝑏 = ( 1 − 𝑎 log𝐾+1𝑋₀ ) ( 𝑎− 2𝛼 log 𝑋₀ − 𝛼 𝑎 log𝐾+1𝑋₀ ) .

and

0 < log 𝑥 < log 𝑦 < log 𝑥 + 𝑎

log𝐾𝑥 = (log 𝑥) ( 1 + 𝑎 log𝐾+1𝑥 ) ≤ (log 𝑥) ( 1 + 𝑎 log𝐾+1𝑋₀ ) < log 𝑥 1 − 𝑎∕ log𝐾+1𝑋₀. (2.11) Further, 𝜋(𝑦) − 𝜋(𝑥) = ∑ 𝑥<𝑝_≤𝑦 1≥ ∑ 𝑥<𝑝_≤𝑦 log 𝑝 log 𝑦 = 1 log 𝑦(𝜃(𝑦) − 𝜃(𝑥)) and, from (2.9), 𝜋(𝑦) − 𝜋(𝑥)≥ 1 log 𝑦 ( 𝑦− 𝑥 − 𝛼 𝑦 log𝐾+1𝑦 − 𝛼 𝑥 log𝐾+1𝑥 ) = 1 log 𝑦 ( 𝑎 𝑥 log𝐾𝑥 − 𝛼 𝑦 log𝐾+1𝑥 − 𝛼 𝑥 log𝐾+1𝑥 ) = 𝑥 (log 𝑦) log𝐾𝑥 ( 𝑎− 2𝛼 log 𝑥− 𝛼 𝑎 log𝐾+1𝑥 ) ≥ 𝑥 (log 𝑦) log𝐾𝑥 [ 𝑎− 2𝛼 log 𝑋₀ − 𝛼 𝑎 log𝐾+1𝑋₀ ] . If the above bracket is ≤ 0 then 𝑏 is also ≤ 0 and (2.10) trivially holds. If the bracket is positive then (2.10) follows from (2.11), which ends the proof of Lemma 2.3.

Corollary 2.4. For 𝑥_{≥ 𝑥0} = 1010_{+ 19,}

𝜋(𝑥(1 + 0.045∕ log2𝑥)) − 𝜋(𝑥)≥ 0.012√𝑥. (2.12) Proof. Since, for 𝑥 ≥ 𝑥₀, (2.3) implies (2.9) with 𝛼 = 1∕2, 𝐾 = 2 and 𝑋₀ = 𝑥₀, we may apply Lemma 2.3 that yields 𝜋(𝑥(1 + 0.045∕ log2_𝑥

)) − 𝜋(𝑥)≥ 𝑏 𝑥∕ log3𝑥 with 𝑏 = 0.001568 …

From (1.15), for 𝑥 ≥ 𝑥0,

√

𝑥∕ log3𝑥is increasing and√𝑥₀∕ log3𝑥₀ = 8.19 …, so that 𝜋(𝑥(1+0.045∕ log2

𝑥))−𝜋(𝑥)_{≥ 𝑏 𝑥∕ log}3𝑥_{≥ 8.19 𝑏}√𝑥_{≥ 0.012}√𝑥.

2.2

The logarithmic integral

From the definition of li(𝑥), it follows that 𝑑 𝑑𝑥li(𝑥) = 1 log 𝑥 and 𝑑2 𝑑𝑥2li(𝑥) = − 1 𝑥log2𝑥 . For 𝑥 → ∞, the logarithmic integral has the asymptotic expansion

li(𝑥) = 𝑁 ∑ 𝑘=1 (𝑘 − 1)!𝑥 (log 𝑥)𝑘 + ( 𝑥 (log 𝑥)𝑁+1 ) . (2.13)

The function 𝑡 ↦ li(𝑡) is an increasing bijection from (1, +∞) onto (−∞, +∞). We denote by li−1_(𝑦)its inverse function that is defined for all 𝑦 ∈ ℝ. Note that

li−1_{(𝑦) > 1}always holds.

To compute numerical values of li(𝑥), we used the formula, due to Ramanujan (cf. [6, p. 126-131]),

li(𝑥) = 𝛾₀+ log log 𝑥 +√𝑥

∞ ∑ 𝑛=1 𝑎_𝑛(log 𝑥)𝑛 with 𝑎_𝑛 = (−1) 𝑛−1 𝑛! 2𝑛−1 ⌊𝑛−1 2 ⌋ ∑ 𝑚=0 1 2𝑚 + 1. The computation of li−1_{𝑦is carried out by solving the equation li(𝑥) = 𝑦 by the}

Newton method.

2.3

Study of 𝜋

(𝑥) =

∑

𝑝

In the article [10], we have deduced from (2.3) the following proposition:

Proposition 2.5. Let 𝛼, 𝑥₁ = 𝑥₁(𝛼) be two real numbers such that 0 < 𝛼 ≤ 1, 𝑥_{1 ≥ 89 967 803 and}|𝜃(𝑥) − 𝑥| < 𝛼 𝑥∕ log3𝑥for 𝑥 ≥ 𝑥1. Then, for 𝑟 ≥ 0.6 and

The unique positive root 𝑟₀(𝛼) of the equation 3𝑟4 _{+ 8𝑟}3_{+ 6𝑟}2 _{− 24∕𝛼 − 1 =}

0 is decreasing on 𝛼 and satisfies 𝑟₀(1) = 1.1445 …, 𝑟₀(0.5) = 1.4377 … and 𝑟₀(0.15) = 2.1086 … For 0.06≤ 𝑟 ≤ 𝑟0(𝛼) and 𝑥≥ 𝑥1(𝛼), we have

𝜋_𝑟(𝑥)≥ ̂𝐶₀+ 𝑥 𝑟+1 (𝑟 + 1) log 𝑥 + 𝑥𝑟+1 (𝑟 + 1)2_log2_𝑥+ 2𝑥𝑟+1 (𝑟 + 1)3_log3_𝑥 − (2𝛼 𝑟 4_{+ 7𝛼 𝑟}3_{+ 9𝛼 𝑟}2_{+ 5𝛼 𝑟 + 𝛼 − 6)𝑥}𝑟+1 (𝑟 + 1)4_log4_𝑥 (2.16)

while, if 𝑟 > 𝑟₀(𝛼) and 𝑥_{≥ 𝑥1}(𝛼), we have 𝜋_𝑟(𝑥)_{≥ ̂}𝐶₀+ 𝑥 𝑟+1 (𝑟 + 1) log 𝑥 + 𝑥𝑟+1 (𝑟 + 1)2_log2_𝑥+ 2𝑥𝑟+1 (𝑟 + 1)3_log3_𝑥 − (51𝛼 𝑟 4_{+ 176𝛼 𝑟}3_{+ 222𝛼 𝑟}2_{+ 120𝛼 𝑟 + 23𝛼 − 168)𝑥}𝑟+1 24(𝑟 + 1)4_log4_𝑥 , (2.17) with ̂ 𝐶₀ = 𝜋_𝑟(𝑥₁) − 𝑥 𝑟 1𝜃(𝑥1) log 𝑥₁ + 3𝛼 𝑟4_{+ 8𝛼 𝑟}3_{+ 6𝛼 𝑟}2_{− 𝛼 − 24} 24 li(𝑥 𝑟+1 1 ) − (3𝛼 𝑟 3_{+ 5𝛼 𝑟}2_{+ 𝛼 𝑟 − 𝛼 − 24)𝑥}𝑟+1 1 24 log 𝑥₁ − 𝛼(3𝑟 2_{+ 2𝑟 − 1)𝑥}𝑟+1 1 24 log2𝑥₁ − 𝛼(3𝑟 − 1)𝑥 𝑟+1 1 12 log3𝑥₁ + 𝛼 𝑥 𝑟+1 1 4 log4𝑥₁ . (2.18)

Corollary 2.6. For 𝑥_{≥ 110 117 910, we have}

𝜋₁(𝑥)≤ 𝑥 2 2 log 𝑥+ 𝑥2 4 log2𝑥 + 𝑥2 4 log3𝑥 + 107 𝑥2 160 log4𝑥 (2.19) and, for 𝑥≥ 905 238 547, 𝜋₁(𝑥)_≥ 𝑥 2 2 log 𝑥 + 𝑥2 4 log2𝑥 + 𝑥 2 4 log3𝑥 + 3 𝑥 2 20 log4𝑥 . (2.20)

Proof. It is Corollary 2.7 of [10], cf. also [2, Theorem 6.7 and Proposition 6.9].

and for 𝑥 ≥ 60 297, 𝜋₂(𝑥)_≤ 𝑥 3 3 log 𝑥 ( 1 + 0.385 log 𝑥 ) (2.22) while, for 𝑥≥ 1 091 239, we have

𝜋₂(𝑥)≥ 𝑥 3 3 log 𝑥 + 𝑥3 9 log2𝑥 + 2𝑥3 27 log3𝑥− 1069𝑥3 648 log4𝑥 . (2.23)

and for 𝑥 >32 321, with 𝜋− 2(𝑥) = ∑ 𝑝<𝑥𝑝 2_, 𝜋₂(𝑥)≥ 𝜋− 2(𝑥)≥ 𝑥3 3 log 𝑥 ( 1 + 0.248 log 𝑥 ) . (2.24)

Proof. From (2.3), the hypothesis |𝜃(𝑥) − 𝑥| ≤ 𝛼𝑥∕ log3𝑥is satisfied with 𝛼 = 1 and 𝑥1 = 89 967 803. By computation, we find 𝜋2(𝑥1) = 13 501 147 086 873 627

946 348, 𝜃(𝑥₁) = 89 953 175.416 013 726 … and 𝐶₀, defined by (2.15) with 𝑟 = 2 and 𝛼 = 1 is equal to −1.040 … × 1018 _{< 0so that (2.21) follows from}

(2.14) for 𝑥 ≥ 𝑥1. From (1.15), the right-hand side of (2.21) is increasing on 𝑥

for 𝑥 ≥ 𝑒4∕3 _{= 3.79 …We check that (2.21) holds when 𝑥 runs over the primes}

𝑝 satisfying 60209 ≤ 𝑝 ≤ 𝑥₁ but not for 𝑝 = 60169. For 𝑥 = 60172.903 …, the right-hand side of (2.21) is equal to 𝜋2(60169), which completes the proof of

(2.21).

Let us set 𝑥2 = 315 011. From (2.21), for 𝑥 ≥ 𝑥2, we have

𝜋₂(𝑥)≤ 𝑥 3 3 log 𝑥 ( 1 + 1 log 𝑥 ( 1 3+ 2 9 log 𝑥₂ + 1181 216 log2𝑥₂ )) ≤ 𝑥 3 3 log 𝑥 ( 1 + 0.385 log 𝑥 ) which proves (2.22) for 𝑥 ≥ 𝑥2. Further, we check that (2.22) holds when 𝑥 runs

over the primes 𝑝 such that 60317 ≤ 𝑝 ≤ 𝑥2but does not hold for 𝑝 = 60293.

Solv-ing the equation 𝜋2(60293) = 𝑡3∕(3 log 𝑡)(1+0.385∕ log 𝑡)yields 𝑡 = 60296.565 …

which completes the proof of (2.22).

Similarly, ̂𝐶₀ defined by (2.18) is equal to 8.022 … × 1018 _{> 0}which implies

(2.23) from (2.17) for 𝑥 ≥ 𝑥1. Let us define 𝐹 (𝑡) =

𝑡3 3 log 𝑡+ 𝑡3 9 log2𝑡+ 2𝑡3 27 log3𝑡− 1069𝑡3 648 log4𝑡. We have 𝐹′_{(𝑡) =} 𝑡2 648 log5𝑡(648 log 4_𝑡

− 3351 log 𝑡 + 4276)which is positive for 𝑡 > 1 and thus, 𝐹 (𝑡) is increasing for 𝑡 > 1. For all primes 𝑝 satisfying 1 091 239 ≤ 𝑝 ≤ 𝑥₁, we denote by 𝑝+the prime following 𝑝 and we check that 𝜋

2(𝑝)≥ 𝐹 (𝑝+), which

Let us set 𝑓(𝑡) = 𝑡3 3 log 𝑡(1 + 0.248 log 𝑡), so that 𝐹 (𝑡) − 𝑓(𝑡) = 𝑡3 81000 log4𝑡(2304 log 2_𝑡

+ 6000 log 𝑡 − 133625). The largest root of the trinomial on log 𝑡 is 6.424 … so that 𝐹 (𝑡) > 𝑓(𝑡) holds for 𝑡 ≥ 618 > exp(6.425), which, as (2.23) holds for 𝑥_{≥ 1 091 239, proves (2.24) for 𝑥 ≥ 1 091 239.}

After that, we check that 𝜋2(𝑝) ≥ 𝑓 (𝑝+) for all pairs (𝑝, 𝑝+) of consecutive

primes satisfying 32321 ≤ 𝑝 < 𝑝+ _{≤ 1 091 239, which, for 𝑥 ≥ 32321, proves}

that 𝜋2(𝑥) ≥ 𝑓 (𝑥) and (2.24) if 𝑥 is not prime. For 𝑥 prime and 𝑥 > 32321, we

have 𝑥 − 1 > 32321 and we consider 𝑦 satisfying 𝑥 − 1 < 𝑦 < 𝑥. We have 𝜋₂−(𝑥) = 𝜋₂(𝑦)≥ 𝑓 (𝑦), which proves (2.24) when 𝑦 tends to 𝑥.

Corollary 2.8. We have 𝜋₃(𝑥)_{≤ 0.271} 𝑥 4 log 𝑥 for 𝑥≥ 664, (2.25) 𝜋₄(𝑥)_{≤ 0.237} 𝑥 5 log 𝑥 for 𝑥≥ 200, (2.26) 𝜋₅(𝑥)≤ 0.226 𝑥 6 log 𝑥 for 𝑥≥ 44 (2.27) 𝜋_𝑟(𝑥)≤ log 3 3 ( 1 + (₂ 3 )𝑟) 𝑥𝑟+1

log 𝑥 for 𝑥 > 1 and 𝑟≥ 5. (2.28) Proof. First, from (2.15), with 𝑟 ∈ {3, 4, 5}, 𝛼 = 1 and 𝑥₁ = 89 967 803, we calculate

𝐶₀(3) = −1.165 … 1026, 𝐶₀(4) = −1.171 … 1034, 𝐶₀(5) = −1.123 … 1042.

As these three numbers are negative, from (2.14), for 𝑟 ∈ {3, 4, 5} and 𝑥 ≥ 𝑥1, we

have 𝜋_𝑟(𝑥)_≤ 𝑥 𝑟+1 (𝑟 + 1) log 𝑥 ( 1 + 1 (𝑟 + 1) log 𝑥₁ + 2 (𝑟 + 1)2_log2_𝑥 1 + 51𝑟 4_{+ 176𝑟}3_{+ 222𝑟}2_{+ 120𝑟 + 191} 24(𝑟 + 1)3_log3_𝑥 1 ) (2.29) and 𝜋₃(𝑥)≤ 0.254 𝑥 4 log 𝑥, 𝜋4(𝑥)≤ 0.203 𝑥5 log 𝑥 and 𝜋5(𝑥)≤ 0.169 𝑥6 log 𝑥. If 𝑝 and 𝑝+ _{are two consecutive primes, for 𝑟 ≥ 3, it follows from (1.15) that the}

function 𝑡 ↦ 𝜋_𝑟(𝑡) log 𝑡

𝑡𝑟+1 is decreasing on 𝑡 for 𝑝 ≤ 𝑡 < 𝑝

proof of (2.25), one checks that 𝜋₃(𝑝) log 𝑝

𝑝4 ≤ 0.271 holds for 673 ≤ 𝑝 ≤ 𝑥1 and we

solve the equation 𝜋3(𝑡) log 𝑡 = 0.271 𝑡4 on the interval [661, 673) whose root is

663.35 …The proof of (2.26) and (2.27) are similar.

Since the mapping 𝑡 ↦ (log 𝑡)∕𝑡6 is decreasing for 𝑡 ≥ 2, the maximum of

𝜋₅(𝑥)(log 𝑥)∕𝑥6is attained on a prime 𝑝. In view of (2.27), calculating 𝜋₅(𝑝)(log 𝑝)∕𝑝6 for 𝑝 = 5, 7, … , 43 shows that the maximum for 𝑥 ≥ 5 is attained for 𝑥 = 5 and is equal to 0.350 … < (log 3)∕3. Let 𝑟 be a number ≥ 5. By applying the trivial inequality

𝜋_𝑟(𝑥)_{≤ 𝑥}𝑟−5𝜋₅(𝑥), 𝑟_{≥ 5,} we deduce that 𝜋𝑟(𝑥) < (log 3) 𝑥

𝑟+1_{∕(3 log 𝑥)} for 𝑥 ≥ 5, and, by calculating

𝜋_𝑟(2)(log 2)∕2𝑟+1 _{= (log 2)∕2 and 𝜋}

𝑟(3)(log 3)∕3 𝑟+1 ₌ (_{1 +}(2 3 )𝑟) log 3 3 , we ob-tain (2.28).

3

𝓁-superchampion numbers

3.1

Definition of

𝓁-superchampion numbers

Definition 3.1. An integer 𝑁 is said 𝓁-superchampion (or more simply super-champion) if there exists 𝜌 > 0 such that, for all integer 𝑀 _{≥ 1}

𝓁(𝑀) − 𝜌 log 𝑀 ≥ 𝓁(𝑁) − 𝜌 log 𝑁. (3.1)

When this is the case, we say that 𝑁 is a𝓁-superchampionassociated to 𝜌. Geometrically, if we represent log 𝑀 in abscissa and 𝓁(𝑀) in ordinate for all 𝑀 _{≥ 1, the vertices of the convex envelop of all these points represent the} 𝓁-superchampion numbers (cf. [11, Fig. 1, p. 633]). If 𝑁 is an 𝓁-𝓁-superchampion, the following property holds (cf. [11, Lemma 3]):

𝑁 = 𝑔(𝓁(𝑁)). (3.2)

Similar numbers, the so-called superior highly composite numbers were first introduced by S. Ramanujan (cf. [22]). The 𝓁-superchampion numbers were also used in [19, ?, 16, 15, 17, 20, 7]. Let us recall the properties we will need. For more details, cf. [11, Section 4].

Lemma 3.2. Let 𝜌 satisfy 𝜌 ≥ 5∕ log 5 = 3.11 …. Then, depending on 𝜌, there exists an unique decreasing sequence (𝜉𝑗)𝑗_≥1 such that 𝜉1 > exp(1) and, for all

𝑗 _{≥ 2,} 𝜉_𝑗 > 1 and 𝜉_𝑗𝑗− 𝜉_𝑗𝑗−1 log 𝜉_𝑗 = 𝜉₁ log 𝜉₁ = 𝜌. (3.3)

Definition 3.3. For each prime 𝑝∈ , let us define the sets 𝑝= { 𝑝 log 𝑝, 𝑝2_{− 𝑝} log 𝑝 ,… , 𝑝𝑖+1_{− 𝑝}𝑖 log 𝑝 ,… } , _{ =}⋃ 𝑝∈ 𝑝. (3.4)

Remark 3.4. Note that all the elements of _𝑝 are distinct at the exception, for 𝑝= 2, of 2

log 2 =

22_{− 2}

log 2 and that, for 𝑝 ≠ 𝑞, 𝑝∩𝑞 = ∅ holds.

Furthermore if 𝜌 ∈ _𝑝, there is an unique ̂𝑗 = ̂𝑗(𝜌) _{≥ 1 such that 𝜉}𝑗̂is an

integer; this integer is 𝑝 and ̂𝑗 is given by ̂

𝑗 = {

1 if 𝜌 = 𝑝∕ log 𝑝

𝑗 if 𝜌 = (𝑝𝑗_{− 𝑝}𝑗−1_{)∕ log 𝑝 (𝑗 ≥ 2).} (3.5) Proposition 3.5. Le 𝜌 >5∕ log 5, 𝜉_𝑗 = 𝜉_𝑗(𝜌) defined by(3.3) and 𝑁_𝜌, 𝑁+

𝜌 defined by 𝑁_𝜌 =∏ 𝑗≥1 ( ∏ 𝑝<𝜉_𝑗 𝑝)= ∏ 𝑗≥1 ( ∏ 𝜉_𝑗+1≤𝑝<𝜉𝑗 𝑝𝑗) (3.6) and 𝑁_𝜌+ =∏ 𝑗_≥1 ( ∏ 𝑝≤𝜉𝑗 𝑝)= ∏ 𝑗_≥1 ( ∏ 𝜉_𝑗+1<𝑝≤𝜉𝑗 𝑝𝑗) (3.7) Then, 1. If 𝜌∉_{, 𝑁𝜌}= 𝑁+

𝜌 is the unique superchampion associated to 𝜌.

2. If 𝜌 ∈ 𝑝, 𝑁𝜌 and 𝑁𝜌+ are two consecutive superchampions, they are the

only superchampions associated to 𝜌, and

𝑁_𝜌+ = 𝑝𝑁𝜌= 𝜉𝑗̂(𝜌)𝑁𝜌. (3.8)

From (3.3) we deduce that the upper bound for 𝑗 in (3.6) and (3.7) is ⌊𝐽⌋ with 𝐽 defined by

2𝐽 _{− 2}𝐽−1

log 2 = 𝜌 = 𝜉

log 𝜉 i.e. 𝐽 =

log 𝜉 + log(2 log 2) − log log 𝜉

log 2 <

log 𝜉 log 2,

(3.9) as 𝜉 ≥ 5 is assumed.

Definition 3.6. Let us suppose 𝑛 _{≥ 7. Depending on 𝑛, we define 𝜌, 𝑁}′_{, 𝑁}′′_{, 𝑛}′_,

𝑛′′_{, 𝜉, and(𝜉}

𝑗)𝑗≥1.

1. 𝜌 is the unique element of such that

𝓁(𝑁𝜌)≤ 𝑛 < 𝓁(𝑁

+

2. 𝑁′_{, 𝑁}′′_{, 𝑛}′_{, 𝑛}′′_{are defined by}

𝑁′= 𝑁_𝜌, 𝑁′′ = 𝑁_𝜌+ 𝑛′ =𝓁(𝑁′), and 𝑛′′ =𝓁(𝑁′′). (3.11) 3. For 𝑗 ≥ 1, 𝜉𝑗 is defined by(3.3) and 𝜉 is defined by 𝜉 = 𝜉1i.e. log 𝜉∕𝜉 = 𝜌. Proposition 3.7. Let us suppose 𝑛≥ 7, 𝜌, 𝑁′_{, 𝑁}′′_{, 𝑛}′_{, 𝑛}′′_{and 𝜉 defined by Definiton}

3.6. Then

𝑛′ _{≤ 𝑛 < 𝑛}′′ and 𝑁′= 𝑔(𝑛′)_{≤ 𝑔(𝑛) < 𝑁}′′= 𝑔(𝑛′′) = 𝑝𝑁′. (3.12) 𝓁(𝑁′) − 𝜌 log 𝑁′ = 𝓁(𝑁′′) − 𝜌 log 𝑁′′. (3.13)

𝑁′′ _{≤ 𝜉𝑁}′. (3.14)

𝓁(𝑁′′) −𝓁(𝑁′)≤ 𝜉. (3.15)

Proof. − (3.12) results of (3.10), (3.11) and (3.8).

− By applying (3.1) first to 𝑀 = 𝑁′ , 𝑁 = 𝑁′′ and further to 𝑀 = 𝑁′′,

𝑁 = 𝑁′ we get (3.13). − Equality (3.8) gives 𝑁′′ _{= 𝜉}

̂

𝑗(𝜌)𝑁′; with the decreasingness of (𝜉𝑗)and the

definiton 𝜉 = 𝜉1we get (3.14).

− By using (3.13) and (3.14) we have 𝓁(𝑁′′_{) − 𝓁(𝑁}′_{) = 𝜌 log(𝑁ε∕𝑁}′_{)) ≤}

𝜌log 𝜉 = 𝜉.

In the array of Fig. (1), for a small 𝑛, one can read the value of 𝑁′_{, 𝑁}′′_{, 𝜌}and

𝜉 as given in Definition 3.6. For instance, for 𝑛 = 45, we have 𝑁′ = 60 060, 𝑁′′ = 180 180, 𝜌 = 6∕ log 3 and 𝜉 = 14.667 … We also can see the values of the parameter associated to a superchampion number 𝑁. For instance, 𝑁 = 360 360 is associated to all values of 𝜌 satisfying 4∕ log 2 ≤ 𝜌 ≤ 17∕ log 17.

As another example, let us consider 𝑥0 = 1010 + 19, the smallest prime

ex-ceeding 1010, and the two 𝓁-superchampion numbers 𝑁′

𝑛 𝑁 𝓁(𝑁) 𝜌 𝜉 12 = 22_{⋅ 3 7} 7..11 5∕ log 5 = 3.11 5 60 = 22⋅ 3 ⋅ 5 12 12..18 7∕ log 7 = 3.60 7 420 = 22_{⋅ 3 ⋅ 5 ⋅ 7 19} 19..29 11∕ log 11 = 4.59 11 4 620 = 22_{⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 30} 30..42 13∕ log 13 = 5.07 13 60 060 = 22⋅ 3 … 13 43 43..48 (9 − 3)∕ log 3 = 5.46 14.66 180 180 = 22_{⋅ 3}2_{⋅ 5 … 13 49} 49..52 (8 − 4)∕ log 2 = 5.77 16 360 360 = 23_{⋅ 3}2_{⋅ 5 … 13 53} 53..69 17∕ log 17 = 6.00 17 6 126 120 = 23⋅ 32_{⋅ 5 … 17 70} 70..88 19∕ log 19 = 6.45 19 116 396 280 = 23_{⋅ 3}2_{⋅ 5 … 19 89}

Figure 1: The first superchampion numbers and, for 2 ≤ 𝑗 ≤ 29, 𝜉𝑗 = 𝑥 (0) 𝑗 with 𝑥(0)₂ = 69588.8 … , 𝑥(0)₃ = 1468.8 … , 𝑥(0)₄ = 220.2 … , 𝑥(0)₅ = 71.5 … and 𝑥(0) 29 = 2.0 … (3.18)

A complete table of values of 𝑥(0)

𝑗 is given in [27].

Let 𝑛 be an integer, and 𝜉 = 𝜉(𝑛) (Definition 3.6). Let us suppose 𝑛 ≥ 𝜈0 =

𝓁(𝑁′

0); then, by (3.10), 𝜌 ≥ 𝜌0, ie 𝜉∕ log 𝜉 ≥ 𝑥0∕ log 𝑥0. So that

𝜉_{≥ 𝑥0}. (3.19)

3.2

Estimates of 𝜉

defined by

(3.3)

Lemma 3.8. (i) For 𝜉 ≥ 5 and 𝑗 ≥ 2, we have

𝜉_{𝑗 ≤ 𝜉}1∕𝑗. (3.20)

with 𝜆₂ = 80, 𝜆₃ = 586, 𝜆₄ = 6381, 𝜆₅ = 89017, 𝜆₆= 1 499 750, 𝜆₇ = 29 511 244, 𝜆₈ = 663 184 075.

(iii) For 𝜉 ≥ 5 and 𝑗 such that 𝜉𝑗 ≥ 𝑥

(0) 𝑗 ≥ 2 (where 𝑥 (0) 𝑗 is defined in(3.18)), we have 𝜉_{𝑗 ≤} ( 𝜉 𝑗(1 − 1∕𝑥(0)_𝑗 ) )1∕𝑗 ≤ ( 2𝜉 𝑗 )1∕𝑗 . (3.22)

Proof. (i) As the function 𝑡 ↦ 𝑡𝑗−𝑡𝑗−1

log 𝑡 is increasing, it suffices to show that

(𝜉1∕𝑗₎𝑗_{− (𝜉}1∕𝑗₎𝑗−1 log(𝜉1∕𝑗₎ = 𝜉− 𝜉(𝑗−1)∕𝑗 (1∕𝑗) log 𝜉 ≥ 𝜉 log 𝜉 which is equivalent to 𝜉 _≥ ( 1 + 1 𝑗− 1 )𝑗 . (3.23)

But the sequence (1+1∕(𝑗 −1))𝑗decreases from 4 to exp(1) when 𝑗 increases from

2to ∞, and 𝜉 ≥ 5 is assumed, which implies (3.23) and (3.20). (ii) Here, we have to prove

𝜉∕𝑗 − (𝜉∕𝑗)(𝑗−1)∕𝑗 (1∕𝑗) log(𝜉∕𝑗) ≥ 𝜉 log 𝜉 which is equivalent to 𝜉1∕𝑗 log 𝜉 ≥ 𝑗1∕𝑗 log 𝑗. (3.24) For 2 ≤ 𝑗 ≤ 8, we have 𝜆𝑗 > 𝑒

𝑗 and, from (1.15), the function 𝜉 ↦ 𝜉1∕𝑗_{∕ log 𝜉 is}

increasing for 𝜉 ≥ 𝜆𝑗 and its value for 𝜉 = 𝜆𝑗exceeds 𝑗1∕𝑗∕ log 𝑗so that inequality

(3.24) is satisfied.

(iii) Let us suppose that 𝜉 ≥ 5 and 𝜉𝑗 ≥ 𝑥

(0)

𝑗 ≥ 2 hold. From (3.3) and (3.20),

we have 𝜉_𝑗𝑗 = 𝜉log 𝜉𝑗 log 𝜉(1 − 1∕𝜉_𝑗) ≤ 𝜉 𝑗(1 − 1∕𝜉_𝑗) ≤ 𝜉 𝑗(1 − 1∕𝑥(0)_𝑗 ) ≤ 2𝜉 𝑗 which proves (3.22).

Corollary 3.9. For 𝑛 _{≥ 7, 𝜌 = 𝜌(𝑛), 𝜉=𝜉(𝑛) defined in Definition 3.6, the powers}

𝑝𝑗 _{of primes dividing 𝑁}′ _{= 𝑁}

𝜌or 𝑁′′= 𝑁𝜌+in(3.6) or (3.7) do not exceed 𝜉.

3.3

Convexity

In this paragraph we prove Lemma 3.13 which is used to speed-up the computa-tions done in Section 5.5 to prove Inequality (1.14) of Theorem 1.7, and and in Section 8.4 to prove Theorem 1.1 (iv).

Lemma 3.10. The function 𝑡 ↦ √li−1_{(𝑡) is concave for 𝑡 > li(𝑒}2_{) = 4.954 ….}

Let 𝑎 ≤ 1 be a real number. Then, for 𝑡 ≥ 31, the function 𝑡 ↦ √li−1_{(𝑡) −}

𝑎(𝑡(log 𝑡))1∕4 is concave.

Proof. The proof is a good exercise of calculus: cf. [10, Lemma 2.5].

Lemma 3.11. Let 𝑢 be a real number,0_{≤ 𝑢 ≤ 𝑒. The function Φ𝑢}defined by Φ_𝑢(𝑡) =√𝑡log 𝑡 ( 1 + log log 𝑡 − 1 2 log 𝑡 − 𝑢 (log log 𝑡)2 log2𝑡 ) (3.25) is increasing and concave for 𝑡≥ 𝑒𝑒 _{= 15.15 …}

Proof. Let us write 𝐿 for log 𝑡 and 𝜆 for log log 𝑡. One calculates (cf. [27]) 𝑑Φ_𝑢 𝑑𝑡 = 1 4𝐿2√_𝑡𝐿 ( 2𝐿(𝐿2− 𝑢𝜆2) + 𝐿(𝐿 − 𝜆 + 3) + 𝜆(𝐿2− 2𝑢) + 6𝑢𝜆(𝜆 − 1)) 𝑑2Φ𝑢 𝑑𝑡2 = −1 8𝐿2_(𝑡𝐿)3∕2(𝐿 2_(𝐿2_−2𝑢𝜆2_)+𝐿3_{(𝜆−1)+𝐿(2𝐿−3𝜆)+(𝐿}4_{+11𝐿−22𝑢𝜆)} + 2𝑢(15𝜆2− 21𝜆 + 8)). For 𝑡 ≥ 𝑒𝑒_{, we have 𝐿 ≥ 𝑒, 𝜆 ≥ 1, 𝐿 = 𝑒}𝜆 _{> 𝑒𝜆} ≥ 𝑢𝜆, so that Φ′ 𝑢is positive. In Φ′′ 𝑢, 𝐿 4_{+11𝐿≥ (𝑒}3_{+11)𝐿≥ (𝑒}3_{+11)𝑢𝜆 > 22𝑢𝜆}, the trinomial 15𝜆2_−21𝜆+8

is always positive and the three first terms of the parenthesis are also positive, so that Φ′′

𝑢 is negative.

Lemma 3.12. Let 𝑛′ _{= 𝓁(𝑁}′_{), 𝑛}′′ _{= 𝓁(𝑁}′′_{) where 𝑁}′ _{and 𝑁}′′ _{are two}

consec-utive𝓁-superchampion numbers associated to the same parameter 𝜌. Let Φ be a concave function on[𝑛′_{, 𝑛}′′_{] such that log 𝑁}′ _{≤ Φ(𝑛}′_{) and log 𝑁}′′ _{≤ Φ(𝑛}′′_{). Then,}

For 𝑛∈ [𝑛′, 𝑛′′], log 𝑔(𝑛)_{≤ Φ(𝑛).}

Proof. From (3.2), it follows that 𝑁′ _{= 𝑔(𝑛}′₎and 𝑁′′_{= 𝑔(𝑛}′′₎. Let us set 𝑁 = 𝑔(𝑛)

so that, from (1.3), we have 𝑛 ≥ 𝓁(𝑁). From the definition (3.1) of superchampion numbers and (3.13), we have

Now, we may write 𝑛= 𝛼𝑛′+ 𝛽𝑛′′ with 0 ≤ 𝛼 ≤ 1 and 𝛽 = 1 − 𝛼 (3.27) and (3.26) implies log 𝑁 ≤ 1 𝜌(𝑛 − (𝑛 ′_{− 𝜌 log 𝑁}′₎₎ = 1 𝜌(𝛼𝑛 ′_{+ 𝛽𝑛}′′_{− 𝛼(𝑛}′_{− 𝜌 log 𝑁}′_{) − 𝛽(𝑛}′′_{− 𝜌 log 𝑛}′′₎₎ = 𝛼 log 𝑁′+ 𝛽 log 𝑁′′. (3.28) From the concavity of Φ, log 𝑁′ _{≤ Φ(𝑛}′_{)and log 𝑁}′′ _{≤ Φ(𝑛}′′_{), (3.28) and (3.27)}

imply

log 𝑔(𝑛) = log 𝑁 _{≤ 𝛼 Φ(𝑛}′) + 𝛽 Φ(𝑛′′)_{≤ Φ(𝛼𝑛}′+ 𝛽𝑛′′) = Φ(𝑛), which completes the proof of Lemma 3.12.

For 𝑛 ≥ 2, let us define 𝑧𝑛by

log 𝑔(𝑛) =√𝑛log 𝑛 ( 1 + log log 𝑛 − 1 2 log 𝑛 − 𝑧𝑛 (log log 𝑛)2 log2𝑛 ) . (3.29)

Lemma 3.13. Let 𝑁′_{and 𝑁}′′ _{be two consecutive𝓁-superchampion numbers and}

𝓁(𝑁′_{)≤ 𝑛 ≤ 𝓁(𝑁}′′_).

(i) If 𝑛′ _{≥ 43 and if 𝑎}

𝑛′ and 𝑎_𝑛′′ (defined by(1.9)) both belong to [0, 1] then 𝑎_{𝑛 ≥ min(𝑎𝑛}′, 𝑎_𝑛′′).

(ii) If 𝑛′ _{≥ 19 and if 𝑧}

𝑛′ and 𝑧_𝑛′′ (defined by(3.29)) both belong to [0, 𝑒] then 𝑧_{𝑛 ≥ min(𝑧𝑛}′, 𝑧_𝑛′′).

Proof. From (3.2), it follows that 𝑁′ _{= 𝑔(𝑛}′₎and 𝑁′′_{= 𝑔(𝑛}′′₎. Let us set 𝑁 = 𝑔(𝑛)

and

Φ(𝑡) = √

li−1(𝑡) − min(𝑎_𝑛′, 𝑎_𝑛′′)(𝑡 log 𝑡)1∕4.

From Lemma 3.10, Φ is concave on [𝑛′_{, 𝑛}′′_]. Moreover, from the definition (1.9)

of 𝑎𝑛′ and 𝑎_𝑛′′, we have log 𝑁′ ≤ Φ(𝑛′)and log 𝑁′′ ≤ Φ(𝑛′′)which, from Lemma

3.12, implies log 𝑔(𝑛) ≤ Φ(𝑛) and, from (1.9), 𝑎𝑛 ≥ min(𝑎𝑛′, 𝑎_𝑛′′) holds, which

proves (i).

The proof of (ii) is similar. We set 𝑢 = min(𝑧𝑛′, 𝑧_𝑛′′). From Lemma 3.11, Φ_𝑢

is concave on [𝑛′_{, 𝑛}′′_], log 𝑔(𝑛′_{) ≤ Φ}

𝑢(𝑛′) and log 𝑔(𝑛′′) ≤ Φ𝑢(𝑛′′) so that, from

Lemma 3.12, we have log 𝑔(𝑛) ≤ Φ𝑢(𝑛) and, from (3.29), 𝑧𝑛 ≥ 𝑢 holds, which

3.4

Estimates of 𝜉

defined by

(3.3)

By iterating the formula 𝜉2 =

√

𝜉log 𝜉2

log 𝜉 + 𝜉2(cf. (3.3)), for any positive integer 𝐾,

we get 𝜉₂ = √ 𝜉 2 ( 1 + 𝐾−1 ∑ 𝑘=1 𝛼_𝑘 log𝑘𝜉 +_ ( 1 log𝐾𝜉 )) , 𝜉→∞, (3.30) with 𝛼₁ = −log 2 2 , 𝛼2= − (log 2)(log 2 + 4) 8 , 𝛼3 = −

(log 2)(log22 + 8 log 2 + 8) 16

Proposition 3.14. We have the following bounds for 𝜉₂:

𝜉₂ < √ 𝜉 2 ( 1 − log 2 2 log 𝜉 ) < √ 𝜉 2 ( 1 − 0.346 log 𝜉 ) for 𝜉 ≥ 31643 (3.31) 𝜉₂ > √ 𝜉 2 ( 1 − 0.366 log 𝜉 ) for 𝜉 ≥ 4.28 × 109_{. (3.32)}

Proof. Let us suppose 𝜉 ≥ 4 and 𝑎 ≤ 0.4. We set

Φ = Φ_𝑎(𝜉) = √ 𝜉 2 ( 1 − 𝑎 log 𝜉 ) ≥ √ 4 2 ( 1 − 0.4 log 4 ) = 1.006 … > 1 and 𝑊 = 𝑊_𝑎(𝜉) = (Φ 2_{− Φ) log 𝜉} 𝜉 − log Φ = (log Φ)(log 𝜉) 𝜉 ( Φ2_{− Φ} log Φ − 𝜉 log 𝜉 ) = (log Φ)(log 𝜉) 𝜉 ( Φ2_{− Φ} log Φ − 𝜉₂2− 𝜉₂ log 𝜉₂ ) .

As Φ > 1 and 𝑡 ↦ (𝑡2_{− 𝑡)∕ log 𝑡is increasing, we have}

𝑊_𝑎(𝜉) > 0 ⟺ 𝜉₂< Φ(𝜉) = √ 𝜉 2 ( 1 − 𝑎 log 𝜉 ) . (3.33)

By the change of variable

𝜉= exp(2𝑡), 𝑡= 1

with the help of Maple (cf. [27]), we get 𝑊 = −𝑎 +𝑎

2

4𝑡 + 3

2log 2 + log 𝑡 − log(2𝑡 − 𝑎) − 𝑒

−𝑡 √ 2 2 (2𝑡 − 𝑎), 𝑊′= 𝑑𝑊 𝑑𝑡 = 𝑈 4𝑡2_{(2𝑡 − 𝑎)} with 𝑈 = −2𝑎(𝑎 + 2)𝑡 + 𝑎3+ 2√2𝑒−𝑡𝑉(4𝑡4− 4(𝑎 + 1)𝑡3+ 𝑎(𝑎 + 2)𝑡2), 𝑈′ = 𝑑𝑈 𝑑𝑡 = −2𝑎(𝑎 + 2) + 2√2𝑒−𝑡(−4𝑡4+ 4(𝑎 + 5)𝑡3− (𝑎2+ 14𝑎 + 12)𝑡2+ 2𝑎(𝑎 + 2)𝑡) 𝑈′′ = 𝑑 2_𝑈 𝑑𝑡2 = 2 √ 2𝑒−𝑡𝑉 with 𝑉 = 4𝑡4_{− 4(𝑎 + 9)𝑡}3_{+ (𝑎}2_{+ 26𝑎 + 72)𝑡}2_{− 4(𝑎}2_{+ 8𝑎 + 6)𝑡 + 2𝑎(𝑎 + 2).}

The sign of 𝑈′′ _{is the same than the sign of the polynomial 𝑉 that is easy to find.}

For 𝑎 fixed and 𝑡 ≥ log 2 (i.e. 𝜉 ≥ 4), one can successively determine, the variation and the sign of 𝑈′, 𝑈 and 𝑊 .

For 𝑎= (log 2)∕2 = 0.346 … 𝑡 log 2 0.96 2.39 4.23 6.38 ∞ 𝑈′′ + + 0 − − 0 + 16.09 −1.62 𝑈′ ↗ 0 ↗ ↘ 0 ↘ ↗ −2.79 −10.03 𝑡 log 2 0.96 1.49 4.23 8.0 ∞ 𝑈′ _{− 0 + + 0 − −} −1.76 29.6 𝑈 ↘ ↗ 0 ↗ ↘ 0 ↘ −2.18 −∞ 𝑡 log 2 1.49 5.18 8.0 ∞ 𝑈 or 𝑊′ − 0 + + 0 − −0.036 0.021 𝑊 ↘ ↗ 0 ↗ ↘ −0.27 0 (3.34) The root of 𝑊 is 𝑤0 = 5.1811243 …and exp(2𝑤0) = 31642.25 …. Therefore,

For 𝑎= 0.366 𝑡 log 2 0.97 2.39 4.23 6.39 ∞ 𝑈′′ + + 0 − − 0 + 15.91 −1.731 𝑈′ ↗ 0 ↗ ↘ 0 ↘ ↗ −2.957 −10.09 𝑡 log 2 0.9797 1.5208 4.231 7.892 ∞ 𝑈′ − 0 + + 0 − − −1.830 29.04 𝑈 ↘ ↗ 0 ↗ ↘ 0 ↘ −2.308 −∞ 𝑡 log 2 1.52 6.37 7.89 11.08 ∞ 𝑊′ − 0 + + 0 − − −0.025 0.004 𝑊 ↘ ↗ 0 ↗ ↘ 0 ↘ −0.28 −0.019 (3.35) The root 𝑤0 of 𝑊 is equal to 11.08803 … and exp(2𝑤0) = 4.27505 … × 109 so

that (3.32) follows from (3.33) and from array (3.35).

Remark 3.15. By solving the system 𝑊 = 0, 𝑈 = 0 on the two variables 𝑡 and 𝑎, one finds

𝑎= 𝑎₀ = 0.370612465 … , 𝑡= 𝑡₀ = 7.86682407 … and for 𝑎 = 𝑎₀, 𝑡= 𝑡₀is a double root of 𝑊_𝑎

0. By studying the variation of 𝑊𝑎0, we find an array close to(3.35), but 𝑊𝑎₀(𝑡0) = 𝑊𝑎′₀(𝑡0) = 0, so that 𝑊𝑎₀ is nonpositive

for 𝑡≥ log 2, which proves 𝜉_{2 ≥} √ 𝜉 2 ( 1 − 𝑎0 log 𝜉 ) > √ 𝜉 2 ( 1 − 0.371 log 𝜉 ) for 𝜉_{≥ 4.}

Corollary 3.16. If 𝜉 _{≥ 𝑥0} = 1010_{+ 19 holds and 𝜉}

and

1 log 𝜉₂ ≥

2

(log 𝜉)(1 − (log 2)∕ log 𝜉) ≥ 2 log 𝜉 ( 1 + log 2 log 𝜉 ) ≥ 2 log 𝜉. On the other hand, (3.32) implies

log 𝜉₂ ≥ 1 2log 𝜉 − 1 2log 2 + log ( 1 − 0.366 log 𝜉 ) ≥ 1 2log 𝜉 − 1 2log 2 + log ( 1 − 0.366 log 𝑥₀ ) ≥ 1 2log 𝜉 − 0.363 = log 𝜉 2 ( 1 − 0.726 log 𝜉 ) ≥ log 𝜉 2 ( 1 − 0.726 log 𝑥₀ ) ≥ log 𝜉 2.07 and by (1.16), 1 log 𝜉₂ ≤ 2 (log 𝜉)(1 − 0.726∕ log 𝜉) ≤ 2 log 𝜉 ( 1 + 0.726 (log 𝜉)(1 − 0.726∕ log 𝑥₀) ) ≤ 2 log 𝜉 ( 1 + 0.75 log 𝜉 ) ≤ 2 log 𝜉 ( 1 + 0.75 log 𝑥₀ ) ≤ 2.07 log 𝜉, which completes the proof of (3.36).

Corollary 3.17. If 𝜉 ≥ 𝑥0 = 1010+ 19 holds and 𝜉2is defined by(3.3), then

√ 𝜉 2 ( 1 − 0.521 log 𝜉 ) ≤ 𝜃−_(𝜉 2) = ∑ 𝑝<𝜉2 log 𝑝≤ 𝜃(𝜉2)≤ √ 𝜉 2 ( 1 − 0.346 log 𝜉 ) . (3.37) Proof. We have 𝜉₂ > 𝑥(0)

2 >69588and (2.4), (3.36) and (3.32) imply

𝜃−(𝜉₂) _{≥ 𝜉2} ( 1 − 0.0746 log 𝜉₂ ) ≥ 𝜉2 ( 1 − 0.0746 × 2.07 log 𝜉 ) ≥ 𝜉2 ( 1 − 0.155 log 𝜉 ) ≥ √ 𝜉 2 ( 1 − 0.366 log 𝜉 ) ( 1 − 0.155 log 𝜉 ) ≥ √ 𝜉 2 ( 1 − 0.521 log 𝜉 ) . Similarly, for the upper bound, we use (2.5), (3.36) and (3.31) to get

Corollary 3.18. Let 𝜉 ≥ 𝑥0 = 1010+ 19 be a real number and 𝜉2 be defined by (3.3). Then 𝜉3∕2 3√2 log 𝜉 ( 1 + 0.122 log 𝜉 ) ≤ 𝜋− 2(𝜉2) = ∑ 𝑝<𝜉₂ 𝑝2 _{≤ 𝜋2}(𝜉₂) ≤ 𝜉 3∕2 3√2 log 𝜉 ( 1 + 0.458 log 𝜉 ) . (3.38)

Proof. First, from (3.3), we observe that 𝜉3 2 log 𝜉₂ = 𝜉2 ( 𝜉2 2 − 𝜉2 log 𝜉₂ ) + 𝜉 2 2 log 𝜉₂ = 𝜉2 𝜉 log 𝜉 + 𝜉2 2 log 𝜉₂ ≥ 𝜉2 𝜉 log 𝜉, (3.39) whence, from (2.24) and (3.36), since 𝜉 ≥ 𝑥0and 𝜉2 ≥ 𝑥

(2) 0 are assumed, 𝜋₂−(𝜉₂)≥ 𝜉 3 2 3 log 𝜉₂ ( 1 + 0.248 log 𝜉₂ ) ≥ 𝜉2 𝜉 3 log 𝜉 ( 1 + 0.496 log 𝜉 ) and, from (3.32), 𝜋₂−(𝜉₂)≥ 𝜉 3∕2 3√2 log 𝜉 ( 1 + 0.496 log 𝜉 ) ( 1 − 0.366 log 𝜉 ) = 𝜉 3∕2 3√2 log 𝜉 ( 1 + 0.13 log 𝜉 − 0.496 × 0.366 log2𝜉 ) ≥ 𝜉 3∕2 3√2 log 𝜉 ( 1 + 0.13 log 𝜉 − 0.496 × 0.366 (log 𝑥₀) log 𝜉 ) ≥ 𝜉 3∕2 3√2 log 𝜉 ( 1 + 0.122 log 𝜉 ) ,

which proves the lower bound of (3.38).

To prove the upper bound, as (3.20) implies 𝜉2 ≤

√

𝜉and 𝜉₂∕ log 𝜉₂ ≤ 2√𝜉∕ log 𝜉, from (3.39), we observe that

and, from (3.31) and (3.36), 𝜉₂3 log 𝜉₂ ≤ 𝜉 log 𝜉 ( 1 + _√2 𝜉 + √ 𝜉 2 ( 1 − log 2 2 log 𝜉 )) = 𝜉 3∕2 √ 2 log 𝜉 ( 1 − 1 log 𝜉 ( log 2 2 − √ 2(1 + 2∕√𝜉) log2𝜉 √ 𝜉 )) ≤ 𝜉 3∕2 √ 2 log 𝜉 ( 1 − 1 log 𝜉 ( log 2 2 − √ 2(1 + 2∕√𝑥₀) log2𝑥₀ √ 𝑥 0 )) ≤ 𝜉 3∕2 √ 2 log 𝜉 ( 1 − 0.339 log 𝜉 ) .

Further, from (3.36), we have 0.385∕ log 𝜉2 ≤ 2.07 × 0.385∕ log 𝜉 ≤ 0.797∕ log 𝜉,

whence from (2.22), 𝜋₂(𝜉₂)≤ 𝜉 3 2 3 log 𝜉₂ ( 1 + 0.385 log 𝜉₂ ) ≤ 𝜉 3∕2 3√2 log 𝜉 ( 1 − 0.339 log 𝜉 ) ( 1 + 0.797 log 𝜉 ) ≤ 𝜉 3∕2 3√2 log 𝜉 ( 1 + 0.458 log 𝜉 ) , which completes the proof of Corollary 3.18.

3.5

The additive excess and the multiplicative excess

3.5.1 The additive excess

Let 𝑁 be a positive integer and 𝑄(𝑁) = ∏𝑝∣𝑁𝑝be the squarefree part of 𝑁. The

additive excess 𝐸(𝑁) of 𝑁 is defined by

𝐸(𝑁) =𝓁(𝑁) − 𝓁(𝑄(𝑁)) = 𝓁(𝑁) −∑

𝑝∣𝑁

𝑝=∑

𝑝∣𝑁

(𝑝𝑣𝑝(𝑁)_{− 𝑝).} (3.40)

If 𝑁′_{and 𝑁}′′_{are two consecutive superchampion numbers of common parameter}

𝜌= 𝜉∕ log 𝜉 (cf. Proposition 3.5), then, from (3.40) and (3.6), 𝓁(𝑁′) = ∑

𝑝∣𝑁′

𝑝+ 𝐸(𝑁′) = ∑

𝑝<𝜉

𝑝+ 𝐸(𝑁′). (3.41)

Proposition 3.19. Let 𝑛 be an integer satisfying 𝑛 ≥ 𝜈0(defined by(3.17)) and 𝑁′

and 𝜉 defined by Defintion 3.6 so that 𝜉 ≥ 𝑥0 = 1010+ 19 holds, then the additive

Proof. With 𝐽 defined by (3.9), from (3.6), we have 𝐸(𝑁′) = ∑ 𝑝∣𝑁′ ( 𝑝𝑣𝑝(𝑁′)_{− 𝑝})₌ 𝐽 ∑ 𝑗=2 ∑ 𝜉_{𝑗+1≤𝑝<𝜉𝑗} (𝑝𝑗− 𝑝). (3.43) For an asymptotic estimates of 𝐸(𝑁′_{)see below (6.13).}

The lower bound. From (3.43), we deduce

𝐸(𝑁′)≥ ∑ 𝑝<𝜉₂ 𝑝2−∑ 𝑝_≤𝜉2 𝑝= 𝜋₂−(𝜉₂) − 𝜋₁(𝜉₂). (3.44) As 𝜉2 ≥ 𝑥 (0)

2 > 69588 holds, from (3.21) we have 𝜉 2

2 ≤ 𝜉∕2 and from (2.6),

𝜋₁(𝜉₂)≤ 𝜉2𝜋(𝜉2)≤ 1.26𝜉22∕ log 𝜉2≤ 0.63𝜉∕ log 69588 ≤ 0.057𝜉. From (3.44) and

(3.38), it follows that 𝐸(𝑁′)_≥ 𝜉 3∕2 3√2 log 𝜉 ( 1 + 0.122 log 𝜉 ) − 0.057𝜉 = 𝜉 3∕2 3√2 log 𝜉 ( 1 + 1 log 𝜉 ( 0.122 −0.057 × 3 √ 2 log2𝜉 √ 𝜉 )) ≥ 𝜉 3∕2 3√2 log 𝜉 ( 1 + 1 log 𝜉 ( 0.122 −0.057 × 3 √ 2 log2𝑥₀ √ 𝑥₀ )) ≥ 𝜉 3∕2 3√2 log 𝜉 ( 1 + 0.12 log 𝜉 )

which proves the lower bound of (3.42).

The upper bound. Let us consider an integer 𝑗₀, 3 ≤ 𝑗₀ ≤ 29, that will be fixed later; (3.43) implies 𝐸(𝑁′) = ∑ 𝑝<𝜉_𝑗0 ( 𝑝𝑣𝑝(𝑁′)_{− 𝑝})₊ 𝑗₀−1 ∑ 𝑗=2 ∑ 𝜉_𝑗+1≤𝑝<𝜉𝑗 (𝑝𝑗 − 𝑝)_{≤ 𝑆1}+ 𝑆₂. (3.45) with 𝑆₁ = ∑ 𝑝<𝜉_𝑗0 𝑝𝑣𝑝(𝑁′) and 𝑆 2 = 𝑗_∑0−1 𝑗=2 𝜋_𝑗(𝜉_𝑗).

Let 𝜌 = 𝜉∕ log 𝜉 be the common parameter of 𝑁′_{and 𝑁}′′_{. From (3.6), for 𝑝 < 𝜉}

𝑗0,

we have 𝑝𝑣_𝑝(𝑁′₎

𝜔𝜌𝜉_𝑗

0 with 𝜔 = 1.000014 if 𝑗0 ≤ 10 and 𝜔 = 1.346 if 𝑗0 ≥ 11, since 𝑥

(0) 11 =

6.55 < 7.32 < 𝑥(0)₁₀ = 7.96. In view of applying (3.21) and (3.22), we set 𝛽_𝑗 = 𝑗 for 2 ≤ 𝑗 ≤ 8 and 𝛽𝑗 = 𝑗(1 − 1∕𝑥

(0)

𝑗 )(with 𝑥

(0)

𝑗 defined by (3.18)) for 9 ≤ 𝑗 ≤ 29.

Therefore, from Lemma 2.2, (3.21) and (3.22), we get 𝑆_{1 ≤ 𝜔} 𝜉𝜉_𝑗 0 log 𝜉 ≤ 𝜔𝜉1+1∕𝑗0 𝛽1∕𝑗0 𝑗0 log 𝜉 = 𝜉 3∕2 3√2 log2𝜉 ⎛ ⎜ ⎜ ⎝ 3√2 𝜔 log 𝜉 𝛽1∕𝑗0 𝑗0 𝜉 1∕2−1∕𝑗0 ⎞ ⎟ ⎟ ⎠ ≤ 𝜉 3∕2 3√2 log2𝜉 ⎛ ⎜ ⎜ ⎝ 3√2 𝜔 log 𝑥₀ 𝛽1∕𝑗0 𝑗₀ 𝑥 1∕2−1∕𝑗0 0 ⎞ ⎟ ⎟ ⎠ . (3.46)

In view of applying (2.25)–(2.28), we set 𝛼3 = 0.271, 𝛼4 = 0.237, 𝛼5 = 0.226and,

for 𝑗 ≥ 6, 𝛼𝑗 = (1 + (2∕3)

𝑗_{)(log 3)∕3. Therefore, for 3 ≤ 𝑗 ≤ 29, it follows from}

(2.25)–(2.28), (3.21) and (3.22) that 𝜋𝑗(𝜉𝑗)≤ 𝛼𝑗𝜉 𝑗+1 𝑗 ∕ log 𝜉𝑗, 𝜉𝑗 ≤ (𝜉∕𝛽𝑗)1∕𝑗 and 𝜋_𝑗(𝜉_𝑗)_{≤ 𝛼𝑗} 𝜉_𝑗𝑗+1 log 𝜉_𝑗 ≤ 𝛼_𝑗(𝜉∕𝛽_𝑗)1+1∕𝑗 (1∕𝑗) log(𝜉∕𝛽_𝑗) = 𝑗𝛼_𝑗(𝜉∕𝛽_𝑗)1+1∕𝑗

(log 𝜉)(1 − (log 𝛽_𝑗)∕ log 𝜉) ≤ 𝛾𝑗 𝜉1+1∕𝑗 log 𝜉 (3.47) with 𝛾_𝑗 = 𝑗𝛼𝑗 𝛽_𝑗1+1∕𝑗(1 − (log 𝛽_𝑗)∕ log 𝑥₀) . Further, for 3 ≤ 𝑗 ≤ 𝑗0− 1, we have

𝛾_𝑗𝜉 1+1∕𝑗 log 𝜉 3√2 log2𝜉 𝜉3∕2 = 3√2𝛾_𝑗log 𝜉 𝜉1∕2−1∕𝑗 ≤ 𝛿𝑗 with 𝛿𝑗 = 3√2𝛾_𝑗log 𝑥₀ 𝑥1∕2−1∕𝑗₀ which implies from (3.47)

𝜋_𝑗(𝜉_𝑗)≤ 𝛿𝑗𝜉

3∕2

3√2 log2𝜉

. (3.48)

From the definition of 𝑆2, from (3.38) and from (3.48), one gets