A class of Equivalent Problems Related to the Riemann Hypothesis
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PHD THESIS Sadegh Nazardonyavi A Class of Equivalent Problems Related to the Riemann Hypothesis Tese submetida `a Faculdade de Ciˆencias da Universidade do Porto para obten¸c˜ao do grau de Doutor em Matem´atica Departamento de Matem´atica Faculdade de Ciˆencias da Universidade do Porto 2013
To My Parents
Acknowledgments I would like to thank my supervisor Prof. Semyon Yakubovich for all the guidance and support he provided for me during my studies at the University of Porto. I would also like to thank Professors: Bagher Nashvadian-Bakhsh, Abdolhamid Riazi, Abdolrasoul Pourabbas, Jos´e Ferreira Alves for their teaching and supporting me in academic stuffs. Also I would like to thank Professors: Ana Paula Dias, Jos´e Miguel Urbano, Marc Baboulin, Jos´e Peter Gothen and Augusto Ferreira for the nice courses I had with them. I thank Professors J. C. Lagarias, C. Calderon, J. Stopple, and M. Wolf for useful discussions and sending us some relevant references. My sincere thanks to Professor Jean-Louis Nicolas for careful reading some parts of the manuscript, helpful comments and suggestions which rather improved the presentation of the last chapter. Also I sincerely would like to thank Professor Paulo Eduardo Oliveira for his kindly assistances and advices as a coordinator of this PhD program. My thanks goes to all my friends which made a friendly environment, in particular Mohammad Soufi Neyestani for what he did before I came to Portugal until now. My gratitude goes to Funda¸c˜ao Calouste Gulbenkian for the financial support during my PhD. I would like to thank all people who taught me something which are useful in my life but I have not mentioned their names. I thank to the members of the juries of my thesis especially Jos´e Assis Azevedo and Ant´onio Machiavelo for their careful reading the thesis and their useful comments. Last but no means least, I would like to thank my family, for their love and encouragement. ii
The scientific work was carried out in 2009-2012 at Department of Mathematics - University of Porto Supervisor Semyon Yakubovich
Resumo Durante um estudo da fun¸c˜ao zeta de Riemann, analizando alguns gr´aficos com ela rela¸cionadis e procurando rela¸c˜oes entre a hip´otese de Riemann e o tamanho (valor absoluto) da fun¸c˜ao-zeta, observamos uma rela¸c˜ao interessante entre esses valores, nomeadamente, que na faixa 0 < σ < 1/2 com |t| ≥ 6.5, se tem (onde, como ´e usual, s=σ+it): |ζ(1 −s)|≤|ζ(s)| Mais tarde, apercebemo-nos que este resultado tinha j´a sido demonstrado por Dixon-Schoenfeld e Spira na d´ecada de 1960. No entanto, a nossa demonstra¸c˜ao ´e diferente e tem a vantagem de envolver, em vez da f´ormula assimpt´otica de Stirling, algumas desigualdades relacionadas com um produto infinito de πe a fun¸c˜ao Γ de Euler. O resultado principal do primeiro cap´ıtulo ´e, pois, que |ζ(1 −s)|≤|ζ(s)|,for 0 < σ < 1 2, onde a igualdade ocorre s´o se ζ(s) = 0. No segundo cap´ıtulo d´a-se um refinamento de estimativas de algumas fun¸c˜oes relacionadas com a distribui¸c˜ao dos n´umeros primos, tais como as fun¸c˜oes ψeϑ de Chebyshev, usando uma nova regi˜ao livre de zeros e o c´alculo de novos zeros da fun¸c˜ao zeta, obtidas por Kadiri e Gourdon, respectivamente. No terceiro cap´ıtulo introduzimos e investigamos algumas sequˆencias relacionadas com um teorema de Robin, que afirma que, a hip´otese de Riemann ´e equivalente `a desigualdade σ(n)< eγnlog log npara todos n > 5040, onde σ(n) ´e a soma dos divisores de neγ´e a constante de Euler. Com base nesta desigualdade, introduzimos uma sequˆencia de n´umeros, que apelidamos de extremamente abundantes, e mostramos que a hip´otese de Riemann ´e verdadeira se e s´o se existe uma infinidade destes n´umeros. Investigamos ainda algumas das suas propriedades e a estrutura dos n´umeros extremamente abundantes, assim como algumas propriedades dos os n´umeros superabundantes e colossalmente abundantes. Finalmente apresentamos dois outros conjuntos de n´umeros, relacionados com os n´umeros extremamente abundantes, que mostram o quanto subtil ´e a hip´otese de Riemann. iv
Abstract While studying the Riemann zeta-function, observing some graphs related to it and looking for some relation between the Riemann hypothesis and the absolute value of the Riemann zeta function, we noted an interesting relationship between those values, namely, that in the strip 0 < σ < 1/2 with |t| ≥ 6.5, one has (as usual s=σ+it): |ζ(1 −s)|≤|ζ(s)| Later we found that this result had been proved by Dixon-Schoenfeld and Spira in 1960’s. Nevertheless, our proof was different and has the advantages of involving, instead of Stirling’s asymptotic formula, some inequalities related to an infinite product for πand Euler’s Γ-function. The main result of the first chapter is, thus, that |ζ(1 −s)|≤|ζ(s)|,for 0 < σ < 1 2,(0.1) where the equality takes place only if ζ(s) = 0. In the second chapter, we give an improvement for estimates of some functions related to the distribution of primes, such as Chebyshev’s ψand ϑfunctions, using some new zero-free region and computations of new zeros of the zeta-function, obtained by Kadiri and Gourdon respectively. In the third chapter, we introduce and investigate some sequences related to Robin’s theorem, which states that, the Riemann hypothesis is equivalent to the inequality σ(n)< eγnlog log nfor all n > 5040, where γis Euler’s constant. Inspired by this inequality, we introduce a sequence of numbers, that we call extremely abundant, and show that the Riemann hypothesis is true if and only if there are infinitely many of these numbers. Moreover, we investigate some of their properties and structure, as well as some properties of superabundant and colossally abundant numbers. Finally we introduce two other sets of numbers, related to extremely abundant numbers, that show how subtle the Riemann hypothesis is. v
Contents 0 Introduction 1 0.1 A Note About the Riemann Hypothesis . . . . . . . . . . . . . . . . 1 0.2 Equivalent Statements to RH . . . . . . . . . . . . . . . . . . . . . 4 0.3 Brief Description of the Chapters . . . . . . . . . . . . . . . . . . . 5 0.4 Ownpapers ............................... 7 1 Riemann Zeta-Function, Its size in Critical Strip and the RH 9 1.1 Preliminaries About the Riemann Zeta Function . . . . . . . . . . . 10 1.1.1 Complex-Valued Functions . . . . . . . . . . . . . . . . . . . 10 1.1.2 Representations of Entire Functions . . . . . . . . . . . . . . 10 1.1.3 Gamma Function . . . . . . . . . . . . . . . . . . . . . . . . 11 1.1.4 The Riemann Zeta Function . . . . . . . . . . . . . . . . . . 12 1.1.5 The Functional Equation . . . . . . . . . . . . . . . . . . . . 13 1.1.6 Values of ζ(s) in the Integers . . . . . . . . . . . . . . . . . 13 1.1.7 Zeros of the Riemann Zeta Function . . . . . . . . . . . . . 14 1.1.8 Bounds of ζ(s) ......................... 16 1.2 On an Inequality for the Riemann Zeta Function in the Critical Strip 16 1.2.1 Auxiliary Lemmas . . . . . . . . . . . . . . . . . . . . . . . 17 1.2.2 Proof of the Main Result . . . . . . . . . . . . . . . . . . . . 19 1.2.3 An Application to the RH . . . . . . . . . . . . . . . . . . . 23 2 Chebyshev’s Functions, Improved Bounds and the RH 26 2.1 Introduction and Preliminary Results . . . . . . . . . . . . . . . . . 26 2.1.1 Explicit Bounds for Distribution of Primes . . . . . . . . . . 29 2.1.2 Irregularities of the Distribution of Primes . . . . . . . . . . 33 2.2 A Sufficient Condition for the RH . . . . . . . . . . . . . . . . . . . 34 2.3 Improved Explicit Bounds for Chebyshev’s Functions . . . . . . . . 35 2.3.1 Improved Explicit Bounds for Large Values of x....... 35 vi
2.3.2 Improved Explicit Bounds for Moderate Values of x..... 38 2.4 Improved Explicit Bounds and Distribution of Primes . . . . . . . . 39 2.5 Explicit Estimates for Y p≤x (1 + 1/p) .................. 42 3 Extremely Abundant Numbers and the RH 47 3.1 Introduction and Background . . . . . . . . . . . . . . . . . . . . . 47 3.2 Extremely Abundant Numbers: Definition and Motivations . . . . . 50 3.3 AuxiliaryLemmas ........................... 53 3.4 Some Properties of SA, CA and XA Numbers . . . . . . . . . . . . 56 3.4.1 SANumbers........................... 57 3.4.2 CANumbers .......................... 68 3.4.3 XANumbers .......................... 71 3.5 DelicacyoftheRH ........................... 75 3.6 Numerical Experiments . . . . . . . . . . . . . . . . . . . . . . . . . 77 vii
List of Tables 1 Numerical verification of zeros of the zeta function ([6], p. 39) . . . 2 2.1 Prime counting function and logarithmic integral[23, p. 2] . . . . . 27 3.1 First 10 extremely abundant numbers (pk]=Qk j=1 pjis primorial of p) ................................... 79 3.2 |ψ(x)−x|< xε∗ 0,(x≥eb), for Theorem 2.20 . . . . . . . . . . . . 85 3.3 ηkfor the case |ψ(x)−x|< xε∗ 0.................... 86 3.4 ηkfor the case |ψ(x)−x|< xε∗ 0.................... 87 3.5 ηkfor the case |ψ(x)−x|< xε∗ 0.................... 88 viii
0.3 Brief Description of the Chapters •The first chapter consists of two sections. In the first section, we give the definition of the Riemann zeta function and list some of its known behaviors such as its extension, functional equation, symmetric location of zeros, etc. In the second section, over a curious studying of the Riemann zeta function and looking for some problem related to the RH, we found an interesting problem about the size of it. It is stated as a main theorem of this section. The corresponding theorem below has been proved independently by Spira ([74]) and Dixon and Schoenfeld [19] in 1960’s. Our proof has an advantage, because it involves elementary calculus, new elementary inequalities and known formulas for some functions and constants. Moreover, we present a relation between the size of the Riemann zeta function and the RH. More precisely, Theorem. Let s=σ+it, where |t| ≥ 12. Then |ζ(1 −s)| ≤ |ζ(s)|,for 0< σ < 1 2,(0.6) where the equality takes place only if ζ(s)=0. As a corollary, one can prove that the strict inequality in (0.6) is equivalent to the RH. Besides, we prove the following result related to the partial derivative of |ζ(s)|2with respect to the real part of s Proposition. If ∂ ∂σ|ζ(s)|2<0,for (0 < σ < 1 2,|t|>6.5),(A) then the RH is true. We conclude the first chapter by the converse of the previous proposition as Conjecture. The condition (A) is also necessary for the validity of the RH. •The second chapter is devoted to an improvement of some results due to Rosser and Schoenfeld (R-S) ([68], [70]) related to the distribution of prime numbers in the sense of Chebyshev’s ψfunction. To do this we use new zero free region [48] and recently calculated zeros of ζ(s) function [31]. For instance, we prove the following results. 5
(i) Let ε0(x) = p8/πX1/2e−X. Then |ψ(x)−x|< xε0(x),(x≥3) and |ϑ(x)−x|< xε0(x),(x≥3). (ii) Let T0be defined as in (2.23) and satisfy T0≥D, where 2 ≤D≤A and Ais defined in 2.13. Let mbe a positive integer and let δ > 0. Then |ψ(x)−x|< ε∗ 0x, (x≥eb), where ε∗ 0= Ω∗ 1e−b/2+ Ω∗ 3+m 2δ+e−blog 2π, where Ω∗ 1and Ω∗ 3are defined in 2.24 and 2.25. For more details we refer to Theorem 2.20. We note that some of the results were mentioned by Dusart [21, p. 5]. However, our computed values in the tables at the end of the thesis are different from Dusart’s. our method is similar to Rosser and Schoenfeld (see details in [68], [70]). Then we infer some estimates for certain functions of distribution of primes. Also we establish a sufficient condition for the RH. We conclude this chapter by giving the estimates for the certain product over primes which is like a dual for Mertens’ third theorem. •The third chapter is based on Robin’s inequality (3.2) and his equivalence to the RH. Investigating the sum of divisors function and Robin’s inequality, we noticed that the first integer nwhich violates this inequality, if exists, should have the property that σ(n)/n > σ(m)/m for m<n. These numbers are called superabundant numbers. Akbary and Friggstad [2] proved that this is indeed the case. However, continuing to investigate Robin’s inequality and modifying the Robin’s inequality, we extracted and introduced a new subsequence of positive integers, which possibly can give a progress to the truth of the RH. Namely, in some way, it is a translation of Robin’s criterion to a different aspect using the Gronwall theorem. But in our opinion these numbers have their own interest. We call this sequence “extremely abundant numbers” (in short XA numbers) and present some of their properties. For instance, we prove 6
(i) If there is any counterexample to Robin’s inequality (3.2), then the least one is an XA number. (ii) The RH is true if and only if #XA =∞. (iii) If n∈XA, then p(n)<log n. Besides, we present a list of properties for the well known sequences of superabundant and colossally abundant and extremely abundant numbers. Finally, we state certain numerical results about superabundant and extremely abundant numbers, worth to be mentioned as well. Finally we demonstrate the delicacy of RH by defining a subset of superabundant which is is defined in a particular way (that is also a superset of extremely abundant number) and giving the proof for infinitude of the cardinality of this subset. 0.4 Own papers –Nazardonyavi, Sadegh; Yakubovich, Semyon, Another proof of Spira’s inequality and its application to the Riemann hypothesis, Journal of Mathematical Inequalities (7) No. 2 (2013), 167-174. –Nazardonyavi, Sadegh; Yakubovich, Semyon, Extremely abundant numbers and the Riemann hypothesis, (submitted) 7
Chapter 1 Riemann Zeta-Function, Its size in Critical Strip and the RH Contents 1.1 Preliminaries About the Riemann Zeta Function . . . 10 1.1.1 Complex-Valued Functions . . . . . . . . . . . . . . . . 10 1.1.2 Representations of Entire Functions . . . . . . . . . . . 10 1.1.3 Gamma Function . . . . . . . . . . . . . . . . . . . . . . 11 1.1.4 The Riemann Zeta Function . . . . . . . . . . . . . . . . 12 1.1.5 The Functional Equation . . . . . . . . . . . . . . . . . 13 1.1.6 Values of ζ(s) in the Integers . . . . . . . . . . . . . . . 13 1.1.7 Zeros of the Riemann Zeta Function . . . . . . . . . . . 14 1.1.8 Bounds of ζ(s) ....................... 16 1.2 On an Inequality for the Riemann Zeta Function in theCriticalStrip....................... 16 1.2.1 Auxiliary Lemmas . . . . . . . . . . . . . . . . . . . . . 17 1.2.2 Proof of the Main Result . . . . . . . . . . . . . . . . . 19 1.2.3 An Application to the RH . . . . . . . . . . . . . . . . . 23 9
1.1 Preliminaries About the Riemann Zeta Function 1.1.1 Complex-Valued Functions An entire function is a function which is analytic in the whole plane. For example polynomials, ez, sin z, and cos zare entire. If z0is an isolated singular point of f, then the Laurent series representation for f(z) in a punctured disk 0 <|z−z0|< R2 is: f(z) = ∞ X n=0 an(z−z0)n+∞ X n=1 bn (z−z0)n. If in the principal part of fat z0(i.e., the second series above) bm6= 0 and bm+1 =bm+2 =··· = 0 (m≥1), then the isolated singular point z0is called a pole of order m. If m= 1 then z0is a simple pole and b1as the residue of fat z0. A meromorphic function is a function which is analytic except for poles(cf. [8, pp. 73, 231, 241, 291]). For example Riemann ζfunction is a meromorphic in the hole plane except at s= 1 with residue 1. 1.1.2 Representations of Entire Functions Assume that f(z) is an entire function and has mzeros at the origin, and a1, a2, . . . , aN are non-zero zeros of f. We can write (see [1]) f(z) = zmeg(z) N Y n=1 1−z an. If there are infinitely many zeros, then f(z) = zmeg(z)∞ Y n=1 1−z an. This representation is valid if the infinite product converges uniformly on every compact set. We formulate here the famous Weierstrass theorem ([1, p. 194]). Theorem 1.1 ([1], p. 195).There exists an entire function with arbitrarily prescribed zeros anprovided that, in the case of infinitely many zeros, an→ ∞. Every entire function with these and no other zeros can be written in the form f(z) = zmeg(z)∞ Y n=1 1−z anez an+1 2(z an)2+···+1 mn(z an)mn, 10
where the product is taken over all an6= 0, the mnare certain integers, and g(z) is an entire function. The product ∞ Y n=1 1−z anez an+1 2(z an)2+···+1 h(z an)h converges and represent an entire function provided that the series P1/|an|h+1 converges. Assume that his the smallest integer for which this series converges (see [1, p. 196]). 1.1.3 Gamma Function The zeros of sin πz are the integers z=±n. Since P1/n diverges and P1/n2 converges, we must take h= 1. Then sin πz =πz ∞ Y n=1 1−z2 n2.(1.1) The function Γ(z) is called Euler’s gamma function ([1, p. 199]). It has the representation Γ(z) = e−γz z ∞ Y n=1 1 + z n−1ez/n, or equivalently Γ(z) = 1 z ∞ Y n=1 (1 + 1/n)z 1 + z/n .(1.2) It satisfies the following equations Γ(z+ 1) = zΓ(z), Γ(z)Γ(1 −z) = π sin πz (1.3) and √πΓ(2z) = 22z−1Γ(z)Γ(z+1 2),(Legendre’s duplication formula).(1.4) The function Γ(z) is a meromorphic function with poles at z= 0,−1,−2, . . . and has no zeros (cf. [1, pp. 197-200]). 11
1.1.4 The Riemann Zeta Function The Riemann zeta function ζ(s) is defined as ζ(s) = ∞ X n=1 1 ns,(<(s)>1). Riemann introduced the notation s=σ+it in his paper to denote a complex number. The following theorem is a useful tool in number theory to compute series and it is called Abel’s identity. Theorem 1.2 ([44], p. 18).Let λ1, λ2, . . . be a real sequence which increases (in the wide sense) and has the limit infinity, and let C(x) = X λn≤x cn, where the cnmay be real or complex, and the notation indicates a summation over the (finite) set of positive integers nfor which λn≤x. Then, if X≥λ1and φ(x) has a continuous derivative, we have X λn≤X cnφ(λn) = C(X)φ(X)−ZX λ1 C(x)φ0(x)dx. (1.5) If, further, C(X)φ(X)→0as X→ ∞, then ∞ X 1 cnφ(λn) = −Z∞ λ1 C(x)φ0(x)dx, provided that either side is convergent. Using this theorem, we can extend the domain of definition of the ζfunction to the left side of σ= 1. By Theorem 1.2, with λn=n,cn= 1, φ(x) = x−s, X n≤X 1 ns=sZX 1 bxc xs+1 dx +bXc Xs,(X≥1). where bxcis the largest integer not greater than x. Writing bxc=x−{x}, so that 0≤ {x}<1, we obtain X n≤X 1 ns=s s−1−s (s−1)Xs−1−sZX 1 {x} xs+1 dx +1 Xs−1−{X} Xs. Since|1/Xs−1|= 1/Xσ−1and |{X}/Xs|<1/Xσ, we deduce, making X→ ∞ ζ(s) = s s−1−sZ∞ 1 {x} xs+1 dx, (σ > 1). 12
The integral in the right-hand side of the latter equation is convergent for σ > 0. So that this equation gives an analytic continuation of ζ(s) over the half-plane σ > 0 (cf. [44, p. 26]). 1.1.5 The Functional Equation Riemann in his 1859 paper established the functional equation and used it to construct the analytic continuation of ζ(s) beyond the region σ > 1. Theorem 1.3 ([76], Th. 2.1).The function ζ(s)is regular (i.e., analytic) for all values of s except s= 1, where there is a simple pole with residue 1. It satisfies the functional equation ζ(s) = 2sπs−1sin 1 2πsΓ(1 −s)ζ(1 −s).(1.6) Functional equation (1.6) may be written in the form ζ(s) = χ(s)ζ(1 −s),(1.7) with χ(s)=2sπs−1sin 1 2πsΓ(1 −s). By (1.3) and (1.4) we get χ(s) =2sπs−1sin 1 2πsΓ(1 −s) =2sπs−1π Γ(s/2)Γ(1 −s/2)Γ(1 −s) =2sπs−1π Γ(s/2)Γ(1 −s/2)π−1/22−sΓ1 2−s 2Γ1−s 2 =πs−1/2Γ(1/2−s/2) Γ(s/2) ,(1.8) and by substitution χ(s)χ(1 −s) = 1. 1.1.6 Values of ζ(s)in the Integers The Bernoulli numbers Bncan be defined by the generating function (cf. [28, p. 41]) x ex−1=∞ X n=0 Bn n!xn. 13
From this one gets B0= 1, B1=−1 2, B2=1 6, B4=−1 30, B6=1 42, B2n+1 = 0,(n∈Z+), and in general are given by the double sum (see [40], [30], [78]) Bn= n X k=0 1 k+ 1 k X r=0 (−1)rk rrn,(n= 0,1,2, . . .). It is known that ζ(−n) = (−1)n n+ 1 Bn+1, n = 0,1,2,.... Since B2n+1 = 0 for n∈Z+, then ζ(−2n) = 0 for n∈Z+. ζ(0) = −1 2, ζ(−2n) = 0, ζ(1 −2n) = −1 2nB2n, n = 1,2,.... Using functional equation (1.6) for zeta and the values of Gamma function in integers, one deduces the formula ζ(2n) = (−1)n−1(2π)2n 2(2n)! B2n, n ∈Z+. 1.1.7 Zeros of the Riemann Zeta Function From Euler’s identity (0.3) we deduce that ζ(s) has no zeros for σ > 1. From the functional equation (1.6) we observe that ζ(s) has no zeros for σ < 0 except trivial zeros at s=−2n, (n∈Z+). Zeros, which lie inside the region 0 ≤ <(s)≤1 are called non-trivial. By the functional equation and the relation ζ(s) = ζ(s) (reflection principle) one sees that non-trivial zeros are symmetric with respect to the vertical line <(s) = 1 2and the real axis =(s) = 0. Hence, if ρis a zero of ζ, then ρ, 1−ρand 1−ρare. Also we mentioned in the Introduction, Riemann made a conjecture that <(ρ) = 1/2 for all non-trivial zeros ρ. In 1896 Hadamard and de la Vall´ee Poussin proved independently that ζ(s)6= 0 on the line σ= 1 ([6, p. 16]). Number of Zeros of ζ(s): Riemann-von Mangoldt Formula Let N(T), where T > 0, denote the number of zeros ρ=β+iγ of ζ(s), for which 0< β < 1 and 0 < γ ≤T. Let F(T) = T 2πlog T 2π−T 2π+7 8(1.9) 14
Hence |f(s)|= 21−2σ. Therefore, it is sufficient to show that for 0 < σ < 1 2and t≥12 |h(s)|<22σ−1.(1.26) Indeed, |h1(s)|is a decreasing function with respect to σand tfor 0 < σ < 1/2 and t > 0. Meanwhile |h2(s)|=∞ Y n=1 2n 2n+ 11−2σ 2n+ 1 −s 2n+s ,(1.27) is increasing with respect to σin the strip (σ, t)∈]0,1/2[×[1/2,∞[, and decreasing with respect to tin the strip (σ, t)∈]0,1/2[×R+. Denoting by h2,n(σ, t) = 2n 2n+ 11−2σ 2n+ 1 −(σ+it) 2n+ (σ+it) , the general term of the product and assuming for now h2,n(σ, t)<1,(0 < σ < 1 2, t ≥0),(1.28) we easily come out with the inequality N+1 Y n=1 h2,n(σ, t)< N Y n=1 h2,n(σ, t),(0 < σ < 1 2, t ≥0). To verify (1.28), we need to show that 1 + 1 2n1−2σ >s(2n+ 1 −σ)2+t2 (2n+σ)2+t2, t ≥0.(1.29) In fact, (2n+ 1 −σ)2+t2 (2n+σ)2+t2= 1 + (1 −2σ)(4n+ 1) (2n+σ)2+t2.(1.30) Hence inequality (1.29) yields 1 + 1 2n1−2σ >2n+ 1 −σ 2n+σ≥s(2n+ 1 −σ)2+t2 (2n+σ)2+t2.(1.31) However, 2n+ 1 −σ 2n+σ= 1 + 1−2σ 2n+σ. So the first inequality in (1.31) follows immediately from (1.19), letting x= 2n and a= 1 −2σ. Thus we get inequality (1.28). 21
Further, we show that {h2,n(σ, t)}∞ n=1 is an increasing sequence for any (σ, t)∈ ]0,1/2[×R. To do this, we consider the function H2(y) = h2,y(σ, t) and differentiate it with respect to y. Hence by straightforward calculations one derives H0 2(y) = 1−2σ y(2y+ 1) 2y 2y+ 11−2σ ((2y+σ)2+t2)2s(2y+ 1 −σ)2+t2 (2y+σ)2+t2 ×n(2y+ 1 −σ)(1 −σ)σ(2y+σ) +(1 + 6y(1 + 2y)−2(1 −σ)σ)t2+t4o. Since (2y+ 1 −σ)(1 −σ)σ(2y+σ) + (1 + 6y(1 + 2y)−2(1 −σ)σ)t2+t4 ≥(2y+ 1 −σ)(1 −σ)σ(2y+σ)>0, we find that the derivative is positive, and therefore H2(y) is increasing for y > 0. Now fixing t≥1/2 one justifies that h2,n(σ, t) is increasing by σ. Precisely, ∂ ∂σh2,n(σ, t) = 2n 2n+ 11−2σ / 2n+ 1 −(σ+it) 2n+ (σ+it) ×n−(1 + 4n)(4n2+ 2n+σ−σ2+t2) +2((2n+ 1 −σ)2+t2)((2n+σ)2+t2) log(1 + 1 2n)o and we achieve the goal, showing that the latter multiplier is positive. But this is true due to inequality (1.13), because it is greater than −(1 −2σ)2(2n+ 1 −σ)(2n+σ) + (8n(1 + 2n)+3−8(1 −σ)σ)t2+ 4t4 1+4n ≥1 + (1 −σ)σ(8n(1 + 2n)−3 + 4(1 −σ)σ) 1+4n>0,(0 < σ < 1/2, t ≥1/2). Returning to (1.27) we conclude that |h2(σ, t)|is increasing with respect to σfor 0< σ < 1 2and t≥1/2, and by (1.30) it is decreasing with respect to tfor 0< σ < 1 2and t > 0. Since |hN(s)|=|1−s s| N Y n=1 2n 2n+ 11−2σ 2n+ 1 −s 2n+s (1.32) 22
is decreasing by N, one has |h(s)| ≤ |hN(s)|. As |hN(s)|is decreasing by t, it is enough to show that |hN(s)|<22σ−1for (t= 12,and N= 3) and this has been established in (1.23). Moreover, since ζ(s) is reflexive with respect to the real axis, i.e., ζ(s) = ζ(s), inequality (1.11) holds also for t≤ −12. Therefore, Theorem 1.7 is proved. Remark 1.10.A computer simulation suggests that the main result is still valid for t∈]6.5,12[ (See Figure 1.1). However, a direct proof by this approach is more complicated, because to achieve the goal we should increase the number Nof terms in the product (1.32). Figure 1.1: The graph of |g(s)|for 6 <t<12. 1.2.3 An Application to the RH Motivating our study of the size of ζ(s) and similar to [74], we apply the results of this chapter to the Riemann hypothesis. We have Proposition 1.11. The RH is true if and only if |ζ(1 −s)|<|ζ(s)|,for (0 < σ < 1 2,|t|>6.5). 23
As it is known [20], zeros of the derivative ζ0(s) of Riemann’s zeta-function are connected with the behavior of zeros of ζ(s) itself. Indeed, Speiser’s theorem [73] states that the RH is equivalent to ζ0(s) having no zeros on the left side of the critical line. Thus, one can get further tools to study RH, employing these properties. We will formulate a sufficient condition for the RH to be true. Proposition 1.12. If ∂ ∂σ|ζ(s)|2<0,for (0 < σ < 1 2,|t|>6.5),(A) then the RH is true. Proof. In fact, if the RH were not true, then by Speiser’s theorem [73], there exists a number s∈]0,1/2[×R, such that ζ0(s) = 0. Hence ∂ ∂σ |ζ(s)|2= 0. Finally in this chapter we conjecture the necessity of (A). Conjecture. The condition (A) is also necessary for the validity of the RH. 24
Chapter 2 Chebyshev’s Functions, Improved Bounds and the RH Contents 2.1 Introduction and Preliminary Results . . . . . . . . . . 26 2.1.1 Explicit Bounds for Distribution of Primes . . . . . . . 29 2.1.2 Irregularities of the Distribution of Primes . . . . . . . . 33 2.2 A Sufficient Condition for the RH . . . . . . . . . . . . 34 2.3 Improved Explicit Bounds for Chebyshev’s Functions 35 2.3.1 Improved Explicit Bounds for Large Values of x. . . . 35 2.3.2 Improved Explicit Bounds for Moderate Values of x. . 38 2.4 Improved Explicit Bounds and Distribution of Primes 39 2.5 Explicit Estimates for Y p≤x (1 + 1/p)............. 42 2.1 Introduction and Preliminary Results An integer greater than 1 is called prime if it is not a multiple of any smaller integers greater than 1. These numbers are important since they are the building blocks for integers. The fundamental theorem of arithmetic states that every integer greater than 1 is a product of prime numbers and this factorization is unique up to rearrangement. There are certain questions arising in the studying of prime numbers. For example: How many primes are there? How many primes are there less than a given number x? What is the distribution of prime numbers? 26
xCount of primes < x Rdn log nDifference 500 000 41 556 41 606.4 50.4 1 000 000 78 501 78 627.5 126.5 1 500 000 114 112 114 263.1 151.1 2 000 000 148 883 149 054.8 171.8 2 500 000 183 016 183 245.0 229.0 3 000 000 216 745 216 970.6 225.6 Table 2.1: Prime counting function and logarithmic integral[23, p. 2] How many primes are there which the difference is 2 (twin prime conjecture)? Every even integer greater than 2 can be expressed as the sum of two primes (Goldbach’s conjecture)? Are there infinitely many Mersenne primes (primes of the form 2n−1)? etc (see [62, Introduction]). About 300 BC it was proved in Euclid’s Elements that there are infinitely many prime numbers. Euler was the first one who discovered fundamental formula (0.3) as an analytic version of the fundamental theorem of arithmetic, and as a corollary it gives X p 1 p=∞. Let π(x) denote as usual the number of primes not exceeding x. In 1808 Legendre conjectured that π(x)∼x log x−A(x), where limx→∞ A(x)≈1.08366. It was conjectured by Gauss that π(x) is asymptotically Li(x) (see [29]). Chebyshev ([12], [75]) proved the asymptotic estimate (A0+o(1)) x log x≤π(x)≤6 5A0+o(1)x log xas x→ ∞, with A0= log(21/231/351/5301/30)≈0.92129.(2.1) He also pointed out that lim inf x→∞ π(x) x/ log x≤1≤lim sup x→∞ π(x) x/ log x. Moreover, he proved the following result in a beautiful way 27
Theorem 2.1 ([12], p. 379).For all x > 1 ϑ(x)<6 5A0x−A0x1 2+5 4 log 6 log2x+5 2log x+ 2 ϑ(x)> A0x−12 5A0x1 2−5 8 log 6 log2x−15 4log x−3,(2.2) where A0is defined in (2.1) and ϑ(x)is the first Chebyshev’s function and it is defined by ϑ(x) = X p≤x log p. (2.3) As we mentioned in Chapter 1, in 1859 Riemann [63] started his paper with the fundamental formula of Euler (0.3). He defined the zeta function for complex numbers with real part σgreater than 1 and using analytic continuation to the whole complex plane except s= 1. In his paper he also gives an explicit formula which we will talk about later. Recall from §1.1.7 that J. Hadamard [33] and C. J. de la Vall´ee Poussin [18], independently and using methods of complex analysis, proved that there is no zeros on the line σ= 1. This fact implies the prime number theorem (PNT), i.e., lim x→∞ π(x) log x x= 1. Finally in 1949, an elementary proof (without using complex analysis) of PNT was given by Selberg ([72]) and Erd˝os ([25]). The PNT can be expressed in different ways. Namely Theorem 2.2 ([4], p. 79).The following relations are logically equivalent: lim x→∞ π(x) log x x= 1, lim x→∞ ϑ(x) x= 1, lim x→∞ ψ(x) x= 1, where ψ(x) = X pm≤x log p, The Riemann zeta function has a significant influence on the law of distribution of primes. Riemann introduced a tool which does this task and it is called explicit formula. This explicit formula gives a link between non-trivial zeros of the Riemann zeta function and Chebyshev’s ψfunction (see [45], [29]). Namely ψ(x) = x−X ρ xρ ρ−ζ0(0) ζ(0) −1 2log(1 −1 x2),(x > 1, x 6=pm),(2.4) 28
where X ρ xρ ρ= lim T→∞ X |γ|≤T xρ ρ,(ζ(ρ) = 0,0<<ρ < 1). and when x=pm, then in the left-hand side of (2.4) put ψ(x)−1 2Λ(x), where for any integer n≥1 Λ(n) = (log p, if n=pmfor some prime pand some m≥1; 0,otherwise. The explicit expression (2.4) was proved by H. von Mangoldt in 1895. As we see in the explicit formula (2.4), the size of the error term in PNT has a link to Θ = sup{<ρ:ζ(ρ) = 0}.(2.5) By the functional equation for zeta function we know that the non-trivial zeros are in the critical strip, symmetric about the line σ= 1/2 and therefore 1/2≤Θ≤1. Also Θ = 1/2 if and only if RH is true ([44, p. 82]). Until now no upper bound Θ≤1−δwith δ > 0 is known. As a first application of explicit formula (2.4), one has Theorem 2.3 (cf. [44], Th. 30). ψ(x) = x+O(xΘlog2x), π(x) = li(x) + O(xΘlog x). 2.1.1 Explicit Bounds for Distribution of Primes Also, mathematicians have worked on the numerical verification of the RH and finding better zero-free region for the Riemann zeta function. The proof of Hadamard and de la Vall´ee Poussin (see [75], [6]) gives that all non-trivial zeros of zeta function lie in the region σ≤1−c log9(3 + |t|) for some c > 0. Later de la Vall´ee Poussin improved this result to σ≤1−c log(3 + |t|). Vinogradov and Korobov extended this zero free region and showed that σ≤1−c0 (log(3 + |t|))2/3(log log(3 + |t|))1/3. 29
Rosser and Schoenfeld are among the mathematicians who have made much efforts on determining zeros and an explicit zero free region of the Riemann zeta function. They give a explicit error term in prime number theorem employing the computation of the zeros on critical line and zero free region in 1975 and 1976 (see [68], [70]). More precisely, they determined that the first 3 502 500 zeros lie on the critical line and proved Theorem 2.4. There is no zeros on the region σ≥1−1 Rlog |t/17|, R = 9.645 908 801.(2.6) Then they employed some estimates and deduced the explicit error term in the prime PNT given by Theorem 2.5 ([68], Th. 2).If log x≥105, then |ψ(x)−x|< xε(x), where one may take either ε(x) = 0.257634 1 + 0.96642 XX3/4e−X, X =plog x/R, where Ris defined in (2.6), or simply by replacing plog x/R ε(x)=0.110123 1 + 3.0015 √log x(log x)3/8e−√(log x)/R. The above bounds were improved by Schoenfeld. Precisely Theorem 2.6 (cf. [70], Th. 11).Let R= 9.645 908 801. Then |ψ(x)−x|< xε0(x),(x≥17), |ϑ(x)−x|< xε0(x),(x≥101), where ε0(x) = r8 17πX1/2e−X, X =plog x/R. and Ris defined in (2.6). In 2010 Dusart [21] proved the explicit estimates for the functions of distribution of primes. Some of them will be employed in the sequel. Namely it has 30
Also for positive ν, positive integer m, and non-negative reals T1and T2, define Rm(ν) ={(1 + ν)m+1 + 1}m,(2.16) S1(m, ν) =2 X β≤1/2 0<γ≤T1 2 + mν 2|ρ|,(2.17) S2(m, ν) =2 X β≤1/2 γ>T1 Rm(ν) νm|ρ(ρ+ 1) ···(ρ+m)|,(2.18) S3(m, ν) =2 X β>1/2 0<γ≤T2 (2 + mν) exp(−X2/log γ) 2|ρ|,(2.19) S4(m, ν) =2 X β>1/2 γ>T2 Rm(ν) exp(−X2/log γ) νm|ρ(ρ+ 1) ···(ρ+m)|,(2.20) and φm(y) = e−X2/log y ym+1 , q(y) = 0.137 log y+ 0.443 ylog ylog(y/2π).(2.21) The following lemma is the basis for finding estimates for Chebyshev’s function in the proof of Rosser and Scoenfeld [68], [70]. Lemma 2.18 ([68], Lemma 8).Let T1and T2be non-negative real numbers. Let mbe a positive integer. Let x > 1and 0< δ < (x−1)/(xm). Then 1 xψ(x)−{x−log 2π−1 2log 1−1 x2}(2.22) ≤1 √x{S1(m, δ) + S2(m, δ)}+S3(m, δ) + S4(m, δ) + mδ 2. Therefore, to compute the estimates for error term we need only to minimize the terms in the right-hand side of (2.22). Note that if we replace 17 with 1 in the proof of Theorem 2.6 and adjust the terms in its proof when it is necessary, we could prove in a similar manner the following theorem Theorem 2.19. Let ε0(x) = p8/πX1/2e−X. 37
Then |ψ(x)−x|< xε0(x),(x≥3) and |ϑ(x)−x|< xε0(x),(x≥3). As you may observe the error term in Theorem 2.6 has the coefficient p8/(17π) which is smaller than p8/π in the next theorem and therefore gives a better bound. However, as we use a better zero-free region, we will get a better bound when x≥e255. 2.3.2 Improved Explicit Bounds for Moderate Values of x As in Theorem 2.19, the role of Athe verified height of RH (defined in (2.13)) was not vigorous, but it has more efficient role in estimating of the Chebyshev’s function for moderate values of xusing the next theorem. Let T0=1 δ2Rm(δ) 2 + mδ 1/m ,(2.23) G(D) = X 0<γ≤D 1 (γ2+ 1/4)1/2−1 4π(log D 2π−12 + 1) +1 D0.137 log D+ 0.443 log log D+1 log D+ 2.6−N(D), and C(D) = 4π0.137 + 0.443 log D. Theorem 2.20 ([70], Lemma 9∗).Let T0be defined as above and satisfy T0≥D, where 2≤D≤A. Let mbe a positive integer and let δ > 0. Then S1(m, δ) + S2(m, δ)<Ω∗ 1, where Ω∗ 1=2 + mδ 4π(log T0 2π+1 m2 + 4πG(D) + 1 m2−mC(D) (m+ 1)T0)(2.24) and G(D)and C(D)are defined as above. Moreover, if Ω∗ 3=1 2πh3(T2) + e3(T2), T2≥A, (2.25) 38
where h3(T) = 2 + mδ 2ZT A φ0(y) log y 2πdy +Rm(δ) δmZ∞ T φm(y) log y 2πdy and e3(T) =q(T)−2 + mδ 2ZT A φ0(y) log y 2πdy +Rm(δ) δmZ∞ T φm(y) log y 2πdy +R(T)φ0(T){2 + mδ + 2Rm(δ) (δT)m}, then |ψ(x)−x|< ε∗ 0x, (x≥eb), where ε∗ 0= Ω∗ 1e−b/2+ Ω∗ 3+m 2δ+e−blog 2π. (2.26) Table 3.2 is made from the above Theorem. 2.4 Improved Explicit Bounds and Distribution of Primes Using Theorem 2.20 we can get a little better estimate for the first Chebyshev’s function (2.3). Proposition 2.21. Let xk≥8·1011. Then |ϑ(x)−x|< η0 k x logkx,(x≥xk), where k0 1 2 3 4 η0 k0.00002945957104 0.00082486799 0.0230963037 0.6466965035 1230 Proof. Let eb≤x<eb+1. Appealing to Proposition 2.7, we treat ϑ(x)−xin the following way ϑ(x)−x=ϑ(x)−ψ(x) + ψ(x)−x <−0.9999√x+xε∗ 0 =−0.9999logkx √x+ε∗ 0logkxx logkx. 39
In the same manner we find ϑ(x)−x=ϑ(x)−ψ(x) + ψ(x)−x >−1.00007√x−1.78 3 √x−xε∗ 0 =−1.00007logkx √x−1.78logkx 3 √x2−ε∗ 0logkxx logkx. To estimate η0 k, it is enough to choose x=eb+1 in each parenthesis. For instance, to estimate η0 1in the interval [8 ·1011, e28), we have ε∗ 0= 0.0000284888 (see computations just before Table 3.2 at the end of thesis), and ϑ(x)−x < −0.9999 28 √e28 + 0.0000284888(28)x log x<0.000774406 x log x, ϑ(x)−x > −1.00007 28 √e28 −1.78 28 3 √e2·28 −0.0000284888(28)x log x >−0.00082486799 x log x. Continuing this process for all intervals [eb, eb+1) where b= 28,29, . . . up to x= e5200, we get the desired results. Remark 2.22.The number 5200 in Table 3.2 is chosen as the last number, since for x≥e5204 we obtain ε0< ε∗ 0, therefore we can apply then Theorem 2.19. Applying the previous proposition, we obtain the estimates for the function π(x). Proposition 2.23. Let x≥8·1011. Then π(x)<x log x1 + 1.0796 log x, π(x)<x log x1 + 1 log x+2.2703 log2x. Proof. By Abel’s identity (Theorem 1.2) π(x) = ϑ(x) log x+Zx 2 ϑ(y) ylog2ydy <x log x1 + η0 k logkx+Zx 2 1 log2y1 + η0 k logkydy. (2.27) We are looking for inequality of this type: π(x)< A2(x),(x≥8·1011), 40
where A2(x) = x log x1 + c log x, and cis a constant which will be determined in the following. Let A1(x) be the right-hand side of (2.27). Therefore we must have A1(x)< A2(x) for x≥8·1011. To have this inequality it is enough to have A1(x0)≤A2(x0) with x0= 8 ·1011 and A0 1(x)< A0 2(x) for x≥x0. Indeed, A0 1(x) = 1 log x+2η0 kx logk+1 x−η0 k(−1 + x+kx) logk+2 x and A0 2(x) = 1 log x+−1 + c log2x−2c log3x. We apply the case η0 1in Proposition 2.21, and get for x≥8·1011 π(x)<x log x1 + 1.0796 log x or if we let A2(x) = x log x1 + 1 log x+c0 log x, by a similar method we arrive at π(x)<x log x1 + 1 log x+2.2703 log2x. Note that if the values of the function li(x) can be calculated in some way, we could use the following formula li(x)−li(2) = Zx 2 1 log ydy =y log y+y log2y+2!y log3y+3!y log4y+···+j!y logj+1 yx 2 + (j+ 1)! Zx 2 1 logj+2 ydy, (j= 0,1, . . .) to compute the integral in (2.27) instead of the method of differential calculus which we applied above. In the next proposition we will determine the length of intervals which contain at least one prime. Proposition 2.24. For all x≥492 227, there exist at least one prime psuch that x<p≤x1 + 0.0297139 log2x. 41
Proof. Since the first Chebyshev’s function has a jump of the size log pon a prime p, for having a prime in the interval [x, y) it is enough to find ysuch that ϑ(y)−ϑ(x)> 0. Assume y=x1 + αk logkxwhere k= 1,2,3,4 and αkis a constant. Hence, ϑ(y)−ϑ(x)> y 1−η0 k logky−x1 + η0 k logkx >x logkxαk1−η0 k logkx−2η0 k. If αk>2η0 k 1−η0 k/logkx, then we get the desired condition. Now according to [70], p. 355 pn+1 −pn≤652,for all pn≤2.686 ·1012. On the other hand, ε∗ 0= 0.0000170896 (defined 2.26) for x≥x0= 2.686 ·1012 . From here we have |ϑ(x)−x|<0.0148566 x log2x,(x≥x0). Therefore, α2>2(0.0148567) 1−(0.0148567)/log2x0≈0.0297139. For 5 254 433 ≤x < 2.686 ·1012 we note that 0.0297139 x log2x>652. For 492 227 ≤x < 5 254 433 we check it by computer. 2.5 Explicit Estimates for Y p≤x (1 + 1/p) In this subsection we give bounds for Qp≤x(1 + 1/p). First we determine some values for which we encounter later. Let S(x) = X p>x log 1 + 1 p−1 p=∞ X n=2 (−1)n−1 nX p>x 1 pn. Hence, X p>x 1 2p2−1 3p3<−S(x)<X p>x 1 2p2. 42
Using Abel’s identity (Theorem 1.2) and estimates for ϑ(x) in Proposition 2.8 one obtains X p>x 1 2p2<1 xlog x and X p>x 1 2p2−1 3p3>1 2xlog x−5 xlog2x. From Y p1 + 1 p=Y p1−1 p2/Y p1−1 p and definition of Bin (2.7) we have X plog 1 + 1 p−1 p= log 6 π2+γ−B. (2.28) Therefore, X p≤x 1 p−B=X p≤x log 1 + 1 p+X p>x log 1 + 1 p−1 p−log 6 π2−γ. Now by (2.9) X p≤x log 1 + 1 p+S(x)−log 6 π2−γ−log log x < Ck(x),(2.29) where Ck(x) is the right-hand side of (2.9). Expanding terms inside absolute value (2.29), we get X p≤x log 1 + 1 p<log 6 π2+γ+ log log x+Ck(x)−S(x), X p≤x log 1 + 1 p>log 6 π2+γ+ log log x−Ck(x)−S(x). We take exponential of both sides in each inequality, and using the estimate 1 + t<et<1 1−t,(t < 1) (2.30) and noting that −S(x) is very small compared to the difference between the two sides of latter estimate when treplace with Ck(x)<1, so that it is negligible. Thus we arrive at 43
Proposition 2.25. We have Y p≤x1 + 1 p<6eγ π2 1 1−Ck(x)log x, (2.31) Y p≤x1 + 1 p>6eγ π2{1−Ck(x)}log x. (2.32) for all x≥xkwhere xkdepends on ηk. We can treat the proof of the Proposition 2.25 in a different way. In this method we do not use the estimates for Qp(1 −1/p2), Qp(1 −1/p) or Qp≤x(1 −1/p). Recall that for t > 0 we have the inequality (cf. 1.15) 1 t+ 1/2<log 1 + 1 t<1 21 t+1 t+ 1. Let X p≤xlog 1 + 1 p−1 21 p+1 p+ 1=−ax, X p≤xlog 1 + 1 p−1 p+ 1/2=bx. It is clear that ax+bx=1 2X p≤x1 p−2 p+ 1/2+1 p+ 1. Therefore, log Y p≤x1 + 1 p=X p≤x log 1 + 1 p=1 2X p≤x1 p+1 p+ 1−ax =X p≤x 1 p−1 2X p≤x 1 p(p+ 1) −ax <log log x+B+Ck(x)−1 2X p≤x 1 p(p+ 1) −ax, log Y p≤x1 + 1 p=X p≤x log 1 + 1 p=X p≤x 1 p+ 1/2+bx =X p≤x 1 p−1 2X p≤x 1 p(p+ 1/2) +bx >log log x+B−Ck(x)−1 2X p≤x 1 p(p+ 1/2) +bx. 44
Taking exponential of both side in each inequality and using (2.30) we get the bounds in the proposition.1 Corollary 2.26. We have Y x<p≤y1 + 1 p<log y log x1 1−Ck(x)−Ck(y),(x≥xk) and Y x<p≤y1 + 1 p>log y log x{1−Ck(x)−Ck(y)},(x≥xk), where xkdepends on ηk. Note that in this corollary, for simplicity, we used the estimates of first method in the proof of Proposition 2.25. 1Continuing the second method we arrive at log Y x<p≤y1 + 1 p<log log y−log log x+Ck(x) + Ck(y)−1 2X x<p≤y 1 p(p+ 1) −(ay−ax) and log Y x<p≤y1 + 1 p>log log y−log log x−Ck(x)−Ck(y)−1 2X x<p≤y 1 p(p+ 1/2) + (by−bx), which are slightly better than the bounds in Corollary 2.26. 45
Necessity. On the other hand, if RH is true, then inequality (3.2) is true. If #XA is finite, then there exists an msuch that for every n > m,f(n)≤f(m). Then lim sup n→∞ f(n)≤f(m)< eγ, which is a contradiction to Theorem 3.2. There are some primes which cannot be the largest prime factors of any XA number. For example, referring to Table 3.1, suggests that there is no XA number with the largest prime factor p(n) = 149 (one can prove this using Proposition 3.14). Do there exist infinitely many such primes? 3.3 Auxiliary Lemmas Before we state several properties of SA, CA and XA numbers, we give the following lemmas which will be needed in the sequel. We note that inequality (1.12) or by changing variable x= 1/t t 1 + t<log(1 + t)< t, (t > 0),(3.11) will be employed frequently. Lemma 3.8. Let a, b be positive constants and x, y ∈R+for which log x > a, and x1−a log x< y < x 1 + b log x. Then y1−c log y< x < y 1 + d log y, where c≥b 1−b−b log x log x+b!, d ≥a 1 + a+b log x log x−a!. Proof. Dividing by x, inverting both sides and multiplying by y, we get y 1 + b/ log x< x < y 1−a/ log x. We are looking for constants cand dsuch that 1−c log y<1 1 + b/ log x, 53
or equivalently c > (log y)b log x+b, and 1 1−a/ log x<1 + d log y, or equivalently d > (log y)a log x−a. First we determine c. Since y < x 1 + b log x, then log y < log x+ log 1 + b log x<log x+b log x. So that if c > log x+b log xb log x+b =blog x+b−b+b log x1 log x+b =b 1−b−b log x log x+b!, then c > log yb log x+b, and hence x>y1−c log y. Similarly, if d > log x+b log xa log x−a =log x−a+a+b log xa log x−a =a 1 + a+b log x log x−a!, then d > log ya log x−a, and therefore x<y1 + d log y. 54
Similarly one can show Lemma 3.9. Let a, b be positive constants and x, y ∈R+for which log2x > a, and x1−a log2x< y < x 1 + b log2x, Then y1−c log2y< x < y 1 + d log2y, where c≥b 1−b−2b log x−b2 log4x log2x+b!, d ≥a 1 + a+2b log x+b2 log4x log2x−a!. By elementary differential calculus one proves also Lemma 3.10. Let h(x) = log log x. Then g(y) = yh(y)−xh(x) (y−x)h(x),(y > x > e). is increasing. In particular, if c > 1and e < x < y < c x, we have g(y)< g(c x). We will need in the sequel the following inequality 1 c−1clog log cx log log x−1<1 + c c−1 log c log xlog log x,(x > e, c > 1).(3.12) Indeed 1 c−1clog log cx log log x−1=1 c−1clog log cx −log log x log log x+c−1 =1 c−1c log log xlog 1 + log c log x+c−1 <1 c−1c log log xlog c log x+c−1 =1 + c c−1 log c log xlog log x. Lemma 3.11. Let x≥11. Then, for y > x the following inequality holds log log y log log x<√y √x. 55
Recall that the prime number theorem is equivalent to ψ(x)∼x, (3.13) where ψ(x) is Chebyshev’s function (Theorem 2.2; see also [36], Th. 434; [44], Th. 3, 12). The following result is a corollary of 2.1 which we will use in the sequel. Of course we could use the explicit bounds which we got in the previous chapter, but for historical point of view (due to Chebyshev) we use the following corollary in one of our results. Corollary 3.12. We have ϑ(x)>log 2 2x, (x≥3). Proof. First we prove that the right-hand side of (2.2) is greater than log 2 2xin [x0,∞) for some x0. Let g(x) = A0x−12 5A0x1 2−5 8 log 6 log2x−15 4log x−3−log 2 2x, where A0is defined in (2.1). Then g0(x) = (A0−log 2 2)−6 5A0x−1 2−5 4 log 6 log x x−15 4x. As log x < 3 4x1 2and x1 2< x for x > 1, g0(x)>(A0−log 2 2)−6 5A0+15 16 log 6 +15 4x−1 2. So, for x>x0=6 5A0+15 16 log 6 +15 42 /(A0−log 2 2)2≈87.591 we have g0(x)>0; i.e., g(x) is increasing. Also g(x0)>0 and therefore g(x)>0, for x≥x0. Now for 3 ≤x < x0, verify ϑ(x)>log 2 2xusing direct computation. 3.4 Some Properties of SA, CA and XA Numbers This section is divided into three subsections, for which we will exhibit several properties of SA, CA and XA numbers, respectively. In the following, when there is no ambiguity, we simply denote by pthe largest prime factor of n. 56
3.4.1 SA Numbers As the starting point, we show that for any real positive x≥1, there is at least one SA number in the interval [x, 2x). In other words Proposition 3.13. Let n<n0be two consecutive SA numbers. Then n0 n≤2. Proof. Let n=Qp q=2 qkq. We compare nwith 2n. In fact σ(2n)/(2n) σ(n)/n =2k2+2 −1 2k2+2 −2>1. Hence, n0≤2n. Alaoglu and Erd˝os [3] proved that If n= 2k2···pkpis a superabundant number then k2≥. . . ≥kpand the exponent of greatest prime factor of nis 1 except n= 4,36. Proposition 3.14 ([3], Th. 2).Let qand rbe prime factors of n∈SA such that q < r and β:= kqlog q log r, where kqis the exponent of q. Then kr(the exponent of r) has one of the three values : β−1,β+ 1,β. As we observe, the above proposition determines the exponent of each prime factor of a SA number with error of at most 1 in terms of smaller prime factor of that number. In the next theorem we give a lower bound for the exponent kq related to the largest prime factor of n. Theorem 3.15. Let n∈SA and 2≤q≤p(where pis the greatest prime factor of n) be a prime factor of n. Then log p log q≤kq. Proof. If q=p, it is trivial. Let q < p and kq=kand suppose that k≤ [log p/ log q]−1. Hence qk+1 < p. (3.14) 57
Now we compare values of σ(s)/s, taking s=nand s=m=nqk+1/p. Since σ(s)/s is multiplicative, we restrict our attention to the factors qand p. But nis SA and m < n, then 1<σ(n)/n σ(m)/m =q2k+2 −qk+1 q2k+2 −11 + 1 p=1 1+1/qk+1 1 + 1 p. Consequently, p<qk+1, which contradicts (3.14). The following proposition gives the asymptotic relation between prime factors of SA numbers. Proposition 3.16 ([3], p. 453).Let δdenote δ=(log log p)2 log plog q,(q1−θ<log p), δ=log p q1−θlog q,(q1−θ>log p), where θ≥5/8is the number which was discussed just before Lemma 2.9. Then log qk+1 −1 qk+1 −q>log q log plog 1 + 1 p{1 + O(δ)},(3.15) log qk+2 −1 qk+2 −q<log q log plog 1 + 1 p{1 + O(δ)}.(3.16) Corollary 3.17. Let n∈SA and 2≤q≤p(where pis the greatest prime factor of n) be a fixed prime factor of n. Then there exist two positive constants cand c0 (depending on q) such that c plog p log q< qkq< c0plog p log q. Proof. By inequality (3.11) log qk+1 −1 qk+1 −q= log 1 + q−1 qk+1 −q<q−1 qk+1 −q≤1 qk and (3.15), there exists a c0>0 such that qk< c 0plog p log q. On the other hand, again from inequality (3.11) log qk+2 −1 qk+2 −q= log 1 + q−1 qk+2 −q>q−1 qk+2 −1>1 2qk+1 and (3.16), there exists a c > 0 such that qk> c plog p log q. 58
Corollary 3.18. For large enough SA number n= 2k···p p < 2k−1.(3.17) Corollary 3.19. Let n= 2k···pbe a SA number. Then for large enough n klog 2 log p= 1. Proof. By Corollary 3.17 for q= 2 we have log(c plog p log 2)< k log 2 <log(c0plog p log 2). Hence, for large enough p 1<1 + log(cp log p/ log 2) log p<klog 2 log p<1 + log(c0plog p/ log 2) log p<2. Therefore, klog 2 log p= 1. Remark 3.20.In [3] it was proved that qkq<2k2+2 and, in p. 455 it was remarked that for large SA n,qkq<2k2for q > 11. Proposition 3.21 ([3], Th. 7).If n= 2k···p∈SA , then p∼log n. From Corollary 3.17 and Proposition 3.21 it follows that Proposition 3.22. For large enough n∈SA log n < 2k2. Proof. We use Remark 3.20, Theorem 2.10 and Corollary 3.18 to get log n 2k2=Plog qkq 2k2 <5 log 2k2+2 + (π(p(n)) −5) log 2k2) 2k2 =π(p(n))log 2k2 2k2+10 log 2 2k2 <p(n) log p(n)1 + 1.2762 log p(n)log 2k2 2k2+10 log 2 2k2 =p(n) 2k2 log 2k2 log p(n)1 + 1.2762 log p(n)+10 log 2 2k2 <1, where p(n) = pis the greatest prime factor of n. 59
Proposition 3.23. Let n= 2k2···qkq···p∈SA. Then ψ(p)≤log n. (3.18) Moreover, lim n→∞ ψ(p) log n= 1.(3.19) Proof. In fact, by Theorem 3.15 ψ(p) = X q≤plog p log qlog q≤X q≤p kqlog q= log n. In order to prove (3.19) we appeal to (3.13) and Proposition 3.21. Proposition 3.24 ([3], Lemma 4).If qis the greatest prime of exponent k, and if q1−θ>log p(where θ≥5/8), then all primes between qand q+qθhave exponent k−1. Remark 3.25.From the above proposition we observe that there is some n0such that for any superabundant number n > n0there exists a prime factor of nwith exponent 2 and there exists a prime factor of nwith exponent 3. Proposition 3.26 ([3], Th. 4).If qis either the greatest prime of exponent kor the least prime of exponent k−1, and if q1−θ>log p, then qk=plog p log q1 + Olog p q1−θlog q. From Remark 3.25 and Proposition 3.26 we get Corollary 3.27. Let xk(with k= 2,3) denote the greatest prime factor of exponent kor the least prime of exponent k−1in decomposition of n∈SA and x1−θ k>log p. Then for large enough n∈SA r3 2p<x2<3 2√p, and 3 r5 2p<x3<3 2 3 √p, where pis the greatest prime factor of n. Lemma 3.28. For large enough n= 2k2···qkq···p∈SA log n ϑ(p)<1 + 3 2√p1 + 4η1 log p, where η1is defined in (2.8) or (2.21). 60
Proof. Let x2be the largest prime factor with exponent 2. From Corollary 3.27 for large enough n∈SA log n ϑ(p)−1 = 1 ϑ(p)(X 2≤q≤x3 (kq−1) log q+ϑ(x2)−ϑ(x3)) <1 ϑ(p){ϑ(x2)+(k2−2)ϑ(x3)} <1 ϑ(p)ϑ(3 2√p) + log 2plog p log 2 −2ϑ(3 2 3 √p) <1 p(1 −η1/log p) 3√p 2(1 + η1 log 3 2√p+log plog p 2 log 2 · 3 √p √p1 + η1 log 3 23 √p) <3 2√p1 + 4η1 log p, p > p0for some p0 where η1is that in Theorem 2.8. In Proposition 3.21 it was proved that the log nis asymptotic to p(n). In the next proposition we give better bounds for this approximation. Proposition 3.29. For n= 2k···p∈SA we have log n>p1−η1 log p and for large enough n∈SA log n<p1 + 2η1 log p,(3.20) where η1is defined in (2.8) or (2.21). Proof. The first inequality holds by (3.18) and Theorem 2.8 or Proposition 2.21 (for p > 8·1011). Concerning the second inequality, we find log n p=log n ϑ(p) ϑ(p) p <1 + 3 2√p1 + 4η1 log p1 + η1 log p <1 + 2η1 log p,(p>p0for some p0). From Lemma 3.9 and Proposition 3.29, we conclude 61
Corollary 3.30. For large enough n= 2k···p∈SA, we have log n1−2η1 log log n<p<log n1 + 2η1 log log n, where η1is defined in (2.8) or (2.21). As we mentioned in the Introduction, two functions σ(n)/n and n/φ(n) are close functions (see (3.1) and (3.6)). Here we will show how close they are for SA numbers. In §18.3 and §18.4 of [36], it is proved that 6 π2<σ(n)ϕ(n) n2<1, and lim n→∞ σ(n)ϕ(n) n2=6 π2,lim n→∞ σ(n)ϕ(n) n2= 1. Proposition 3.31. For n= 2k2···qkq···p∈SA, we have σ(n) n>{1−ε(p)}n ϕ(n), where ε(p) = 6 √plog p1 + 1 log p. Proof. We show that σ(n) n·ϕ(n) n=Y q≤p1−1 qkq+1 >1−6 √plog p1 + 1 log p.(3.21) Hence, using logarithmic inequality (1.12) and Theorem 3.15 and Theorem 2.10, we obtain log Y q≤p1−1 qkq+1 =X q≤p log 1−1 qkq+1 >−X q≤p 1 qkq+1 −1 =−X q≤x2 1 qkq+1 −1−X x2<q≤p 1 q2−1 >−X q≤x2 1 qlog p/ log q−1−2√2 √plog p =−X q≤x2 1 p−1−2√2 √plog p =−π(x2) p−1−2√2 √plog p >−1 p−1 3√p 2 log(3/2)√p1 + 1.2762 log(3/2)√p−2√2 √plog p >−6 √plog p1 + 1 log p, 62
If ε /∈E, then the function σ(n)/n1+εattains its maximum at a single point Nε whose prime decomposition is Nε=Ypαp(ε), αp(ε) = $log p1+ε−1 pε−1 log p%−1 (3.30) or if prefer αp(ε) = (k, xk+1 < p < xk, k ≥1; 0, p > x =x1. If ε∈E, then by theorem of six exponentials at most two xk’s are prime (see [11], [3], [26], [64]). Hence, there are either two or four CA numbers of parameter ε, is defined by Nε= K Y k=1 Y p<xk or p≤xk p. (3.31) In fact formula (3.31) gives all possible values of a CA number for a parameter ε in or not in E. If Nis the largest CA number of parameter ε, then F(p, 1) = ε⇒p(N) = p, (3.32) where p(N) is the largest prime factor of N. It was proved by Robin ([64], Proposition 1) that the maximum order of the function fdefined in (3.3) is attained by CA numbers. Using this fact, one has Proposition 3.42. Let 3≤N < n < N0, where Nand N0are two successive CA numbers. Then f(n)<max{f(N), f(N0)}.(3.33) Proof. Robin [64, Prop. 1] proved the inequality f(n)≤max{f(N), f(N0)}. But, in fact, due to the strict convexity of the function t7→ εt −log log t, Robin’s proof naturally extends to the strict inequality (3.33). This fact shows, that if there is any counterexample to (3.2), then there exists at list one CA number which violates it. Corollary 3.43. Let N < N0be two consecutive CA numbers. If there exists an XA number n > 10080 satisfying N < n < N0, then N0is also an XA. 69
Proof. Let us set X={m∈XA :N < m < N0}. By the assumption n∈XA, then we have X6=∅. Let n0= max X. Since n0∈XA and n0> N then f(n0)> f(N). From inequality (3.33) we must have f(n0)< f(N0). Hence N0∈XA. Remark 3.44.In the case N < n = 10080 < N0, we have N= 5040, N0= 55440 and f(N)≈1.790 973 367, f(n)≈1.755 814 339, f(N0)≈1.751 246 515. Hence inequality (3.33) is satisfied with f(n)< f(N) = max{f(N), f(N0)}. Theorem 3.45. If RH holds, then there exist infinitely many CA numbers that are also XA. Proof. If RH holds, then by Theorem 3.7, #XA =∞. Let nbe in XA. Since #CA =∞(see [3], [26]), there exist two successive CA numbers N, N0such that N < n ≤N0. If N0=nthen it is readily in XA, otherwise N0belongs to XA via Corollary 3.43. It will be seen that there exist infinitely many CA numbers Nfor which the largest prime factor pis greater than log N. For this purpose, we will use the following Lemma 3.46 ([11], Lemma 3).Let Nbe a CA number of parameter ε<F(2,1) = log(3/2)/log 2 and define x=x(ε)by (3.29). Then (i) for some constant c > 0 log N≤ϑ(x) + c√x. (ii) Moreover, if Nis the largest CA number of parameter ε, then ϑ(x)≤log N≤ϑ(x) + c√x. The following lemma is a corollary of Littlewood oscillation for Chebyshev’s ϑ function (Corollary 2.14). 70
Lemma 3.47 ([11]).There exists a constant c > 0such that for infinitely many primes pwe have ϑ(p)< p −c√plog log log p, (3.34) and for infinitely many other primes pwe have ϑ(p)> p +c√plog log log p. These results give Theorem 3.48. There are infinitely many CA numbers Nε, such that log Nε< p(Nε). Proof. We choose plarge enough as in (3.34) and Nεthe largest CA number of parameter ε=F(p, 1). Then, from (3.32), one has p(Nε) = p. By Lemma 3.46(ii) log Nε−ϑ(p)< c√p, (for some c > 0). On the other hand, by Lemma 3.47 there exists a constant c0>0 such that ϑ(p)−p < −c0√plog log log p, (c0>0). Hence log Nε−p < {c−c0log log log p}√p < 0, and this is the desired result. 3.4.3 XA Numbers Returning to XA numbers, here we present some of their properties and describe the structure of these numbers. Theorem 3.49. Let n= 2k2···pbe an XA number. Then p < log n. Proof. For n= 10080 we have p(10080) = 7 <9.218 <log(10080). Let n > 10080 be an XA number and m=n/p. Then m > 10080, since for all primes pwe have ϑ(p)>log 2 2p > p 3(this follows from Corollary 3.12). Therefore, 71
for a number n∈SA we have log n≥ϑ(p(n)) > p(n)/3 and m=n/p(n)> n/(3 log n)>10080 if n≥400 000. For n < 400 000 we can check by computation. Hence by Definition 3.4 1 + 1 p=σ(n)/n σ(m)/m >log log n log log m. So 1 + 1 p>log log n log log m⇒1 p>log(1 + log p/ log m) log log m. Using inequality (1.12) we have 1 p>log p log nlog log m>log p log nlog log n⇒p < log n. We mention a similar result proved by Choie et al. ([13], Lemma 6.1) Proposition 3.50. Let t≥2be fixed. Suppose that there exists a t-free integer exceeding 5040 that does not satisfy Robin’s inequality. Let n= 2k2···pbe the smallest such integer. Then p < log n. In the previous section we showed that, if RH holds, then there exist infinitely many CA numbers that are also XA. Next theorem is a conclusion of Theorems 3.48 and 3.49 which is independent of RH. Theorem 3.51. There exist infinitely many CA numbers that are not XA. We know that by Definition 3.4, for n∈XA the function σ(n)/n is strictly increasing and φ(n)/n is decreasing. Next theorem compares the increase and decrease power by adding these two functions. Theorem 3.52. Let g(n) = σ(n) + ϕ(n) n.(3.35) For two consecutive XA numbers n= 2k···pand n0= 2k0···p0, if p0≥pand log(n0/n)>1/(3 log p0), then g(n)< g(n0)for large enough n, n0∈XA. Proof. If the largest primes of nand n0are equal, it is clear. Let p0=pk+1 > pk=p. If n > 10080 is XA, then (see (3.3)) f(n)> f(10080) >1.75. 72
Using inequality (1.12), Proposition 3.29, Lemma 2.12 and Lemma 2.9, we deduce for large enough n σ(n0) n0+ϕ(n0) n0−σ(n) n−ϕ(n) n >σ(n) n log log n0−log log n log log n−1 pk+1 k Y j=1 1−1 pj >1.75 log log n0 log n−1 pk+1 k Y j=1 1−1 pj >1.75log(n0/n) log n0−1 pk+1 k Y j=1 1−1 pj >1.75log(n0/n) log n0−1 pk+1 e−γ log pk1 + 0.2 log2pk >1.75 log(n0/n) pk+1(1 + 2η1 log pk+1 )−1 pk+1 e−γ log pk1 + 0.2 log2pk >1 pk+1 (1.75 3 log pk+1(1 + 2η1 log pk+1 )−e−γ log pk1 + 0.2 log2pk) >0. Remark 3.53.We checked that (3.35) (without further assumptions in the theorem) is increasing up to 8150-th element of XA. Structure of XA We can describe the structure of XA numbers (for large enough ones). Next theorem will determine the exponents of the prime factors of an XA numbers (for large enough XA) with an error at most 1. Theorem 3.54. Let n= 2k2···qkq···p∈XA, and αq(p) = logq1+(q−1)plog p qlog q.(3.36) Then for large enough n∈XA we have |kq−αq(p)| ≤ 1. Proof. Let kq=kand k−αq(p)≥2. Then qk≥qαq(p)+2 > q 1+(q−1)plog p qlog q.(3.37) 73
Now compare f(n) with f(m) where m=n/q. Since n∈XA we must have σ(n)/n σ(m)/m =qk+1 −1 qk+1 −q>log log n log log m, or using inequality (1.12) qk<1+(q−1)log nlog log m qlog q.(3.38) From (3.37) and (3.38) we get log nlog log m−qp log p>qlog q, and this is a contradiction with (3.20). Now assume k−αq(p)≤ −2. Then qk+2 −1 q−1≤plog p qlog q. Put m=nq/p. We show that under the assumption k−αq(p)≤ −2 we have that f(n)< f(m) or simply σ(n)/n σ(m)/m =1−q−1 qk+2 −11 + 1 p<1 + log p/q log nlog log m. It is enough to show that 1 p−qlog q plog p<log p/q log nlog log m.(3.39) If the left-hand side of (3.39) is negative, then clearly the inequality holds. Suppose that the left-hand side is positive. Then by (3.20) we have log nlog log m plog p<1 + (q−1) log q log p−qlog q,(n>n1). Hence for n > n1 σ(n)/n σ(m)/m <1−q−1 qk+2 −1+1 p<1 + log p/q log nlog log m<log log n log log m, which is a contradiction with the definition of n∈XA. We conclude this subsection by the following interesting conjecture. Conjecture. Let n= 2k2···qkq···p∈XA and αq(p)is defined by (3.36). Then for all n∈XA we have |kq−αq(p)| ≤ 1. 74
3.5 Delicacy of the RH We already proved that under the RH the number of XA numbers are infinite. Here we present an interesting theorem which demonstrates the delicacy of the RH by showing the infinitude of some superset of XA numbers which is defined by an inequality which is quite close to that (i.e., (3.9)) in the definition of XA numbers, independent of RH. Lemma 3.55. If m≥3, then there exists n>msuch that σ(n)/n σ(m)/m >1 + log n/m log nlog log m.(3.40) Proof. Given m≥3. Then by (3.5) σ(m) m≤eγ+0.648214 (log log m)2log log m, (3.41) Since log log m log log m01 + log m0/m log m0log log m<1 and decreasing for m0> m and tends to 0 as m0goes to infinity, then for some m0> m we have log log m log log m01 + log m0/m log m0log log meγ+0.648214 (log log m)2=eγ−ε, (3.42) where ε > 0. Hence by Gronwall’s theorem there is n≥m0such that σ(n) n>(eγ−ε) log log n =log log m log log m01 + log m0/m log m0log log meγ+0.648214 (log log m)2log log n ≥1 + log n/m log nlog log mσ(m) m, where the last inequality holds by (3.41) and (3.42). Definition 3.56. Given n1= 10080. Let nk+1 to be the first integer greater than nksuch that σ(nk+1)/nk+1 σ(nk)/nk >1 + log nk+1/nk log nk+1 log log nk ,(k= 1,2, . . .). We define X0to be the set of all n1, n2, n3, . . .. 75
XA ⊂X0⊂S. (3.43) Now we are going to state the main theorem of this paper which is the second step towards the delicacy of the RH, i.e., Theorem 3.57. The set X0has infinite number of elements. Proof. If the RH is true, then the set X0has infinite elements by (3.43). If RH is not true, then there exists m0≥10080 such that σ(m0)/m0 σ(m)/m >log log m0 log log m,for all m≥10080. By Lemma 3.44 there exists m0> m0such that m0satisfies the inequality σ(m0)/m0 σ(m0)/m0 >1 + log m0/m0 log m0log log m0 . Let nbe the first number greater than m0which satisfies σ(n)/n σ(m0)/m0 >1 + log n/m0 log nlog log m0 . Then n∈X0. Lemma 3.58. If m≥3, then there exists n>msuch that σ(n)/n σ(m)/m >1 + 2 log n/m (log m+ log n) log log m.(3.44) Proof. The proof is similar to that of Lemma 3.55. Definition 3.59. Given n1= 10080. Let nk+1 to be the first integer greater than nksuch that σ(nk+1)/nk+1 σ(nk)/nk >1 + 2 log nk+1/nk (log nk+ log nk+1) log log nk ,(k= 1,2, . . .). We define X00 to be the set of all n1, n2, n3, . . .. It is easily seen that XA ⊂X00 ⊂X0. In a similar method of the proof of Theorem 3.57 one can prove that Theorem 3.60. The set X00 has infinite number of elements. 76
Note that #XA = 9240,#X00 = 9279,#X0= 9535. up to the 300 000th element of S(we used the list of SA numbers tabulated in [58]) and #(X00 −XA) = 39,#(X0−XA) = 295. We list here the elements of X00\XA up to s300 000: X00\XA ={s55, s62, s91, s106, s116, s127, s128, s137, s138, s149, s181, s196, s212, s219, s224, s231, s232, s246, s247, s259, s260, s263, s272, s273, s276, s288, s294, s299, s305, s311, s317, s330, s340, s341, s343, s354, s65343, s271143, s271151} We conclude this section by formulating another criterion for the RH (using Robin’s theorem) with Chebyshev’s ψfunction. Proposition 3.61. The RH is true if and only if σ(lcm(n)) lcm(n)< eγlog ψ(n),(n≥11) (3.45) where lcm(n) = lcm(1,2, . . . , n)is the least common multiples of the first npositive integers. Proof. If the RH is true, then by Robin’s theorem inequality (3.45) holds. On the other hand, if the RH is not true, then according to Proposition 2.15 and noting that σ(lcm(n)) lcm(n)>Q1 2√n<p≤n(1 −1/p2) Qp≤n(1 −1/p),(n≥121). inequality (3.45) does not hold. 3.6 Numerical Experiments In this section we give some numerical results mainly for the set of XA numbers up to its 13770-th element, which is less than C1=s500 000 (i.e., 500 000-th SA number) basing on the list provided by T. D. Noe [58]. We examined Property 3.62 to 3.65 and Remark 3.66 below for the corresponding XA numbers extracted from the list. Property 3.62. Let n= 2k2···qkq···rkr···pbe an XA number, where 2≤q < r≤p. Then for 10080 < n ≤C1 77
(i) log n<qkq+1, (ii) rkr< qkq+1 < rkr+2, (iii) qkq< kqp, (iv) qkqlog q < log nlog log n<qkq+2. Property 3.63. Let n= 2k2···xk k.···pbe an XA number. Then √p<x2<p2p, for 10080 < n ≤C1. Property 3.64. Let n= 2α2···qαq···pand n0= 2β2···qβq···p0be two consecutive XA numbers greater than 10080. Then for 10080 < n ≤C1 αq−βq∈ {−1,0,1},for all 2≤q≤p0. Property 3.65. If m, n are XA and m < n, then for 10080 < n ≤C1 (i) p(m)≤p(n), (ii) d(m)≤d(n). Remark 3.66.We note that Property 3.65 is not true for SA numbers. For example s47 = (19])(3])22, s48 = (17])(5])(3])23, p(s48) = 17 <19 = p(s47). and s173 = (59])(7])(5])(3])223, s174 = (61])(7])(3])222,d(s173) d(s174)=36 35 >1, where skdenotes k-th SA number. Using Table of SA and CA numbers in [58] we have #{n∈XA :n<C}= 24 875, #{n∈CA :n < C}= 21 187, #{n∈CA ∩XA :n < C}= 20 468, #{n∈CA \XA :n < C}= 719, #{n∈XA \CA :n < C}= 4407, where C=s1000,000. The following properties have been checked up to C2(250,000-th element of SA numbers) and for 8150-th element of XA numbers in this domain. 78
Table 3.2: |ψ(x)−x|< xε∗ 0,(x≥eb), for Theorem 2.20 b m δ ε b m δ ε 18.42 1 4.78(-4) 1.14790(-3) 900 22 2.08(-12) 2.39881(-11) 18.43 1 4.76(-4) 1.14336(-3) 950 21 2.15(-12) 2.36469(-11) 18.44 1 4.74(-4) 1.13884(-3) 1000 21 2.12(-12) 2.33114(-11) 18.45 1 4.71(-4) 1.13434(-3) 1050 21 2.09(-12) 2.29819(-11) 18.5 1 4.61(-4) 1.11208(-3) 1100 20 2.16(-12) 2.26511(-11) 18.7 1 4.22(-4) 1.02723(-3) 1150 20 2.13(-12) 2.23185(-11) 19.0 1 3.70(-4) 9.11615(-4) 1200 20 2.09(-12) 2.19902(-11) 19.5 1 2.96(-4) 7.46453(-4) 1250 19 2.17(-12) 2.16664(-11) 20 1 2.37(-4) 6.10561(-4) 1300 19 2.13(-12) 2.13331(-11) 21 1 1.52(-4) 4.07253(-4) 1350 19 2.10(-12) 2.10050(-11) 22 1 9.68(-5) 2.70618(-4) 1400 19 2.07(-12) 2.06828(-11) 23 1 6.17(-5) 1.79207(-4) 1450 18 2.14(-12) 2.03552(-11) 24 1 3.93(-5) 1.18314(-4) 1500 18 2.11(-12) 2.00268(-11) 25 1 2.51(-5) 7.79224(-5) 1550 18 2.07(-12) 1.97045(-11) 26 1 1.61(-5) 5.12515(-5) 1600 17 2.15(-12) 1.93836(-11) 27 1 1.06(-5) 3.37385(-5) 1650 17 2.12(-12) 1.90541(-11) 28 1 7.22(-6) 2.23274(-5) 1700 17 2.08(-12) 1.87301(-11) 29 1 5.26(-6) 1.49727(-5) 1750 17 2.05(-12) 1.84126(-11) 30 2 1.26(-6) 9.41428(-6) 1800 16 2.13(-12) 1.80866(-11) 35 2 1.22(-7) 1.05471(-6) 1850 16 2.09(-12) 1.77616(-11) 40 3 7.81(-9) 1.16290(-7) 1900 16 2.05(-12) 1.74427(-11) 45 4 5.60(-10) 1.23408(-8) 1950 15 2.14(-12) 1.71251(-11) 50 7 3.45(-11) 1.30541(-9) 2000 15 2.10(-12) 1.67987(-11) 75 26 2.20(-12) 3.32667(-11) 2100 15 2.02(-12) 1.61646(-11) 100 26 2.18(-12) 3.25398(-11) 2200 14 2.07(-12) 1.55206(-11) 150 26 2.16(-12) 3.13387(-11) 2300 13 2.12(-12) 1.48944(-11) 200 26 2.13(-12) 3.03713(-11) 2400 13 2.04(-12) 1.42535(-11) 250 25 2.18(-12) 2.95752(-11) 2500 12 2.10(-12) 1.36270(-11) 300 25 2.15(-12) 2.88982(-11) 2600 12 2.00(-12) 1.29976(-11) 350 25 2.13(-12) 2.83142(-11) 2700 11 2.06(-12) 1.23732(-11) 400 25 2.10(-12) 2.78000(-11) 3000 10 1.92(-12) 1.05303(-11) 450 24 2.16(-12) 2.73267(-11) 3200 9 1.86(-12) 9.32308(-12) 500 24 2.13(-12) 2.68923(-11) 3500 7 1.89(-12) 7.53761(-12) 550 24 2.10(-12) 2.64897(-11) 3700 6 1.83(-12) 6.39612(-12) 600 23 2.16(-12) 2.61010(-11) 4000 5 1.60(-12) 4.78674(-12) 650 23 2.14(-12) 2.57273(-11) 4500 3 1.23(-12) 2.46504(-12) 700 23 2.11(-12) 2.53666(-11) 4700 2 1.20(-12) 1.77229(-12) 750 22 2.17(-12) 2.50149(-11) 5000 2 6.51(-13) 9.76476(-13) 800 22 2.14(-12) 2.46639(-11) 5100 2 5.34(-13) 8.00754(-13) 850 22 2.11(-12) 2.43220(-11) 5200 2 4.38(-13) 6.56727(-13) 85
Table 3.3: ηkfor the case |ψ(x)−x|< xε∗ 0 b η1η2η3η4 18.42 0.0211558 0.389901 7.18587 132.436 18.43 0.0210836 0.388781 7.16912 132.199 18.44 0.0210116 0.387664 7.15241 131.962 18.45 0.0209853 0.388227 7.18220 132.871 18.5 0.0207960 0.388884 7.27214 135.989 18.7 0.0195173 0.370829 7.04574 133.869 19.0 0.0177765 0.346641 6.75951 131.81 19.5 0.0149291 0.298581 5.97162 119.432 20 0.0128218 0.269258 5.65441 118.743 21 0.00895956 0.19711 4.33643 95.4014 22 0.00622421 0.143157 3.29261 75.73 23 0.00430097 0.103223 2.47736 59.4567 24 0.00295785 0.0739463 1.84866 46.2165 25 0.00202598 0.0526755 1.36956 35.6087 26 0.00138379 0.0373624 1.00878 27.2372 27 0.000944678 0.026451 0.740628 20.7376 28 0.000647495 0.0187774 0.544543 15.7918 29 0.000449182 0.0134755 0.404264 12.1279 30 0.0003295 0.0115325 0.403637 14.1273 35 0.0000421884 0.00168754 0.0675015 2.70006 40 5.23306(-6) 0.000235488 0.0105969 0.476863 45 6.17042(-7) 0.0000308521 0.00154261 0.0771303 50 9.79061(-8) 7.34296(-6) 0.000550722 0.0413041 75 3.32667(-9) 3.32667(-7) 0.0000332667 0.00332667 100 4.88096(-9) 7.32145(-7) 0.000109822 0.0164733 150 6.26774(-9) 1.25355(-6) 0.00025071 0.0501419 200 7.59281(-9) 1.8982(-6) 0.000474551 0.118638 250 8.87255(-9) 2.66177(-6) 0.00079853 0.239559 300 1.01144(-8) 3.54003(-6) 0.00123901 0.433654 350 1.13257(-8) 4.53027(-6) 0.00181211 0.724843 400 1.251(-8) 5.62949(-6) 0.00253327 1.13997 450 1.36634(-8) 6.83168(-6) 0.00341584 1.70792 500 1.47907(-8) 8.13491(-6) 0.0044742 2.46081 550 1.58938(-8) 9.5363(-6) 0.00572178 3.43307 600 1.69656(-8) 0.0000110277 0.00716799 4.65919 650 1.80091(-8) 0.0000126064 0.00882445 6.17712 700 1.90249(-8) 0.0000142687 0.0107015 8.02615 750 2.0012(-8) 0.0000160096 0.0128076 10.2461 800 2.09643(-8) 0.0000178197 0.0151467 12.8747 850 2.18898(-8) 0.0000197008 0.0177307 15.9577 86
Table 3.4: ηkfor the case |ψ(x)−x|< xε∗ 0 b η1η2η3η4 900 2.27887(-8) 0.0000216493 0.0205668 19.5385 950 2.36469(-8) 0.0000236469 0.0236469 23.6469 1000 2.4477(-8) 0.0000257009 0.0269859 28.3352 1050 2.52801(-8) 0.0000278081 0.0305889 33.6478 1100 2.60488(-8) 0.0000299561 0.0344495 39.617 1150 2.67822(-8) 0.0000321386 0.0385663 46.2796 1200 2.74877(-8) 0.0000343597 0.0429496 53.687 1250 2.81663(-8) 0.0000366162 0.0476011 61.8814 1300 2.87997(-8) 0.0000388796 0.0524875 70.8582 1350 2.94071(-8) 0.0000411699 0.0576378 80.693 1400 2.999(-8) 0.0000434855 0.063054 91.4283 1450 3.05328(-8) 0.0000457993 0.0686989 103.048 1500 3.10416(-8) 0.0000481145 0.0745774 115.595 1550 3.15272(-8) 0.0000504435 0.0807096 129.135 1600 3.1983(-8) 0.000052772 0.0870738 143.672 1650 3.2392(-8) 0.0000550663 0.0936128 159.142 1700 3.27777(-8) 0.000057361 0.100382 175.668 1750 3.31427(-8) 0.0000596569 0.107382 193.288 1800 3.34602(-8) 0.0000619014 0.114518 211.857 1850 3.3747(-8) 0.0000641193 0.121827 231.471 1900 3.40133(-8) 0.0000663259 0.129336 252.204 1950 3.42503(-8) 0.0000685005 0.137001 274.002 2000 3.52773(-8) 0.0000740823 0.155573 326.703 2100 3.55621(-8) 0.0000782366 0.172121 378.665 2200 3.56974(-8) 0.0000821041 0.188839 434.331 2300 3.57465(-8) 0.0000857917 0.2059 494.16 2400 3.56337(-8) 0.0000890842 0.22271 556.776 2500 3.54303(-8) 0.0000921187 0.239509 622.722 2600 3.50936(-8) 0.0000947527 0.255832 690.747 2700 3.71195(-8) 0.000111358 0.334075 1002.23 3000 3.3697(-8) 0.00010783 0.345057 1104.18 3200 3.26308(-8) 0.000114208 0.399727 1399.04 3500 2.8266(-8) 0.000105998 0.397491 1490.59 3750 2.4491(-8) 0.0000979639 0.391855 1567.42 4000 2.01043(-8) 0.0000844381 0.35464 1489.49 4200 1.69978(-8) 0.0000764902 0.344206 1548.93 4500 1.15857(-8) 0.0000544527 0.255928 1202.86 4700 8.86144(-9) 0.0000443072 0.221536 1107.68 5000 4.98003(-9) 0.0000253981 0.12953 660.606 5100 4.16392(-9) 0.0000216524 0.112592 585.48 5200 3.42352(-9) 0.0000178468 0.0930354 484.993 87
Table 3.5: ηkfor the case |ψ(x)−x|< xε∗ 0 b m δ ε η1η2η3η4 3800 6 1.67(-12) 5.86122(-12) 2.23312(-8) 0.000085082 0.324163 1235.06 3810 5 1.94(-12) 5.80739(-12) 2.21842(-8) 0.0000847437 0.323721 1236.61 3820 5 1.92(-12) 5.74859(-12) 2.20171(-8) 0.0000843255 0.322967 1236.96 3830 5 1.90(-12) 5.69039(-12) 2.18511(-8) 0.0000839082 0.322207 1237.28 3840 5 1.88(-12) 5.63277(-12) 2.16862(-8) 0.0000834918 0.321443 1237.56 3850 5 1.86(-12) 5.57575(-12) 2.15224(-8) 0.0000830764 0.320675 1237.8 3860 5 1.84(-12) 5.51930(-12) 2.13597(-8) 0.000082662 0.319902 1238.02 3870 5 1.82(-12) 5.46344(-12) 2.11982(-8) 0.0000822488 0.319126 1238.21 3880 5 1.80(-12) 5.40817(-12) 2.10378(-8) 0.0000818369 0.318346 1238.36 3890 5 1.78(-12) 5.35348(-12) 2.08786(-8) 0.0000814265 0.317563 1238.5 3900 5 1.77(-12) 5.29927(-12) 2.07201(-8) 0.0000810158 0.316772 1238.58 3910 5 1.75(-12) 5.24559(-12) 2.05627(-8) 0.0000806058 0.315975 1238.62 3920 5 1.73(-12) 5.19249(-12) 2.04065(-8) 0.0000801975 0.315176 1238.64 3930 5 1.71(-12) 5.13998(-12) 2.02515(-8) 0.000079791 0.314376 1238.64 3940 5 1.70(-12) 5.08798(-12) 2.00975(-8) 0.0000793852 0.313572 1238.61 3950 5 1.68(-12) 5.03642(-12) 1.99442(-8) 0.0000789791 0.312757 1238.52 3960 5 1.66(-12) 4.98545(-12) 1.97922(-8) 0.0000785752 0.311944 1238.42 3970 5 1.64(-12) 4.93509(-12) 1.96417(-8) 0.0000781739 0.311132 1238.31 3980 5 1.63(-12) 4.88504(-12) 1.94913(-8) 0.0000777704 0.310304 1238.11 3990 5 1.61(-12) 4.83561(-12) 1.93424(-8) 0.0000773697 0.309479 1237.92 4000 5 1.60(-12) 4.78674(-12) 1.91948(-8) 0.0000769713 0.308655 1237.71 88
Bibliography [1] Lars V. Ahlfors. Complex analysis. McGraw-Hill Book Co., New York, third edition, 1978. An introduction to the theory of analytic functions of one complex variable, International Series in Pure and Applied Mathematics. [2] Amir Akbary and Zachary Friggstad. Superabundant numbers and the Riemann hypothesis. Amer. Math. Monthly, 116(3):273–275, 2009. [3] L. Alaoglu and P. Erd¨os. On highly composite and similar numbers. Trans. Amer. Math. Soc., 56:448–469, 1944. [4] Tom M. Apostol. Introduction to analytic number theory. Springer-Verlag, New York, 1976. Undergraduate Texts in Mathematics. [5] E. Bombieri. Prime territory: Exploring the infinite landscape at the base of the number system, 1992. [6] Peter Borwein, Stephen Choi, Brendan Rooney, and Andrea Weirathmueller, editors. The Riemann hypothesis. CMS Books in Mathematics/Ouvrages de Math´ematiques de la SMC. Springer, New York, 2008. A resource for the afficionado and virtuoso alike. [7] Keith Briggs. Abundant numbers and the Riemann hypothesis. Experiment. Math., 15(2):251–256, 2006. [8] Churchill R. V. Brown, J. W. Complex variables and applications. McGrawHill International Edition, 8th edition, 2009. [9] H. M. Bui, Brian Conrey, and Matthew P. Young. More than 41% of the zeros of the zeta function are on the critical line. Acta Arith., 150(1):35–64, 2011. [10] Geoffrey Caveney, Jean-Louis Nicolas, and Jonathan Sondow. Robin’s theorem, primes, and a new elementary reformulation and the Riemann hypothesis. Integers, 11:A33, 10, 2011. 90
[11] Geoffrey Caveney, Jean-Louis Nicolas, and Jonathan Sondow. On SA, CA, and GA numbers. Ramanujan J., 29(1-3):359–384, 2012. [12] P. L. Chebyshev. M´emoir sur les nombres premiers. J. math. pures appl., 17(1):366–390, 1852. [13] YoungJu Choie, Nicolas Lichiardopol, Pieter Moree, and Patrick Sol´e. On Robin’s criterion for the Riemann hypothesis. J. Th´eor. Nombres Bordeaux, 19(2):357–372, 2007. [14] J. B. Conrey. At least two-fifths of the zeros of the Riemann zeta function are on the critical line. Bull. Amer. Math. Soc. (N.S.), 20(1):79–81, 1989. [15] J. B. Conrey. More than two fifths of the zeros of the Riemann zeta function are on the critical line. J. Reine Angew. Math., 399:1–26, 1989. [16] J. B. Conrey and D. W. Farmer. Equivalences to the Riemann hypothesis. http://aimath.org/pl/rhequivalences. [17] N. G. de Bruijn. Asymptotic methods in analysis. Second edition. Bibliotheca Mathematica, Vol. IV. North-Holland Publishing Co., Amsterdam, 1961. [18] Ch.-J. de la Vall´ee Poussin. Sur la fonction ζ(s) de Riemann et le nombre des nombres premiers inf´erieurs `a une limite donn´ee. In Colloque sur la Th´eorie des Nombres, Bruxelles, 1955, pages 9–66. Georges Thone, Li`ege, 1956. [19] R. D. Dixon and Lowell Schoenfeld. The size of the Riemann zeta-function at places symmetric with respect to the point 1 2.Duke Math. J., 33:291–292, 1966. [20] Eduardo Due˜nez, David W. Farmer, Sara Froehlich, C. P. Hughes, Francesco Mezzadri, and Toan Phan. Roots of the derivative of the Riemann-zeta function and of characteristic polynomials. Nonlinearity, 23(10):2599–2621, 2010. [21] P. Dusart. Estimates of some functions over primes without R.H. arXiv:1002.0442. [22] P. Dusart. Autour de la fonction qui compte le nombre de nombres premiers. PhD thesis, Universit´e de Limoges, 1998. 91
[23] H. M. Edwards. Riemann’s zeta function. Academic Press [A subsidiary of Harcourt Brace Jovanovich, Publishers], New York-London, 1974. Pure and Applied Mathematics, Vol. 58. [24] William John Ellison. Les nombres premiers. Hermann, Paris, 1975. En collaboration avec Michel Mend`es France, Publications de l’Institut de Math´ematique de l’Universit´e de Nancago, No. IX, Actualit´es Scientifiques et Industrielles, No. 1366. [25] P. Erd¨os. On a new method in elementary number theory which leads to an elementary proof of the prime number theorem. Proc. Nat. Acad. Sci. U. S. A., 35:374–384, 1949. [26] Paul Erd˝os and Jean-Louis Nicolas. R´epartition des nombres superabondants. Bull. Soc. Math. France, 103(1):65–90, 1975. [27] L. Euler. Variae observationes circa, series infinitas. Commentarii academiae scientiarum Petropolitanae, 9:160–188, 1744. [28] Steven R. Finch. Mathematical constants, volume 94 of Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 2003. [29] L. J. Goldstein. A history of the prime number theorem. Amer. Math. Monthly, 80:599–615, 1973. [30] H. W. Gould. Explicit formulas for Bernoulli numbers. Amer. Math. Monthly, 79:44–51, 1972. [31] X. Gourdon. The 1013 first zeros of the Riemann zeta function, and zeros computation at very large height. http://numbers.computation.free.fr/ Constants/Miscellaneous/zetazeros1e13-1e24.pdf. [32] T. H. Gronwall. Some asymptotic expressions in the theory of numbers. Trans. Amer. Math. Soc., 14(1):113–122, 1913. [33] J. Hadamard. Sur la distribution des z´eros de la fonction ζ(s) et ses cons´equences arithm´etiques. Bull. Soc. Math. France, 24:199–220, 1896. [34] G. H. Hardy. Sur les z´eros de la fonction ζ(s) de Riemann. C. R. Acad. Sci. Paris, 158:1012–1014, 1914. 92
[35] G. H. Hardy and J. E. Littlewood. The zeros of Riemann’s zeta-function on the critical line. Math. Z., 10(3-4):283–317, 1921. [36] G. H. Hardy and E. M. Wright. An introduction to the theory of numbers. Oxford University Press, Oxford, sixth edition, 2008. Revised by D. R. HeathBrown and J. H. Silverman, With a foreword by Andrew Wiles. [37] Julian Havil. Gamma. Princeton University Press, Princeton, NJ, 2003. Exploring Euler’s constant, With a foreword by Freeman Dyson. [38] D. R. Heath-Brown. Simple zeros of the Riemann zeta function on the critical line. Bull. London Math. Soc., 11(1):17–18, 1979. [39] Hans Heilbronn. ¨ Uber den Primzahlsatz von Herrn Hoheisel. Math. Z., 36(1):394–423, 1933. [40] James Higgins. Double series for the Bernoulli and Euler numbers. J. London Math. Soc. (2), 2:722–726, 1970. [41] G. Hoheisel. Primzahl probleme in der analysis. S.-B. Preuss. Akad. Wiss., pages 580–588, 1930. [42] M. N. Huxley. On the difference between consecutive primes. Invent. Math., 15:164–170, 1972. [43] A. E. Ingham. On the difference between consecutive primes. Quarterly J. Math. (Oxford), 8:255–266, 1937. [44] A. E. Ingham. The distribution of prime numbers. Cambridge Mathematical Library. Cambridge University Press, Cambridge, 1990. Reprint of the 1932 original, With a foreword by R. C. Vaughan. [45] Aleksandar Ivi´c. The Riemann zeta-function. A Wiley-Interscience Publication. John Wiley & Sons Inc., New York, 1985. The theory of the Riemann zeta-function with applications. [46] Henryk Iwaniec and Matti Jutila. Primes in short intervals. Ark. Mat., 17(1):167–176, 1979. [47] Henryk Iwaniec and J´anos Pintz. Primes in short intervals. Monatsh. Math., 98(2):115–143, 1984. 93
[48] Habiba Kadiri. Une r´egion explicite sans z´eros pour la fonction ζde Riemann. Acta Arith., 117(4):303–339, 2005. [49] D. Kilminster. Table of n, a(n) for n=0..2000. http://oeis.org/A004394/ b004394.txt. [50] K. Knopp. Theory and application of infinite series. Blackie & Son Limited, 1951. [51] Jeffrey C. Lagarias. An elementary problem equivalent to the Riemann hypothesis. Amer. Math. Monthly, 109(6):534–543, 2002. [52] Edmund Landau. Handbuch der Lehre von der Verteilung der Primzahlen. 2 B¨ande. Chelsea Publishing Co., New York, 1953. 2d ed, With an appendix by Paul T. Bateman. [53] Norman Levinson. More than one third of zeros of Riemann’s zeta-function are on σ= 1/2. Advances in Math., 13:383–436, 1974. [54] D. S. Mitrinovi´c. Elementary inequalities. In cooperation with E. S. Barnes, D. C. B. Marsh, J. R. M. Radok. Tutorial Text, No. 1. P. Noordhoff Ltd., Groningen, 1964. [55] H. L. Montgomery. Zeros of L-functions. Invent. Math., 8:346–354, 1969. [56] Jean-Louis Nicolas. Petites valeurs de la fonction d’Euler. J. Number Theory, 17(3):375–388, 1983. [57] Jean-Louis Nicolas. Small values of the Euler function and the Riemann hypothesis. Acta Arith., 155(3):311–321, 2012. [58] T. D. Noe. First 1,000,000 superabundant numbers. http://oeis.org/ A004394. [59] S. Ramanujan. Highly composite numbers [Proc. London Math. Soc. (2) 14 (1915), 347–409]. In Collected papers of Srinivasa Ramanujan, pages 78–128. AMS Chelsea Publ., Providence, RI, 2000. [60] Srinivasa Ramanujan. Highly composite numbers. Ramanujan J., 1(2):119– 153, 1997. Annotated and with a foreword by Jean-Louis Nicolas and Guy Robin. 94