On numbers satisfying Robin's inequality, properties of the next counterexample and improved specific bounds

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On numbers satisfying Robin’s inequality, properties of the next counterexample and improved specific bounds

Robert Vojak

To cite this version:

Robert Vojak. On numbers satisfying Robin’s inequality, properties of the next counterexample and

improved specific bounds. 2020. �hal-02085868v2�

On numbers satisfying Robin’s inequality, properties of the next counterexample and improved specific bounds

Robert VOJAK

Croatia & France

Abstract

Defines(n) := n⁻¹σ(n) (σ(n) :=P

d|nd) andω(n) is the number of prime divisors ofn. One of the properties ofs plays a central role:s(p^a)> s(q^b) ifp < q are prime numbers, with no special condition on a, bother than a, b > 1. This result, combined with the Multiplicity Permutation theorem, will help us establish properties of the next counterexample (say c) to Robin’s inequalitys(n)< e^γlog logn. The number cis superabundant, and ω(c) must be greater than a number close to one billion. In addition, the ratiopω(c)/logchas a lower and upper bound. At mostω(c)/14 multiplicity parameters are greater than 1. Last but not least, we apply simple methods to sharpen Robin’s inequality for various categories of numbers.

1. Introduction

The following notations will be used throughout the article:

• γ: Euler’s constant,

• ⌊x⌋denotes the largest integer not exceedingx

• pi: i-th prime number,

• σ(n) =P

d|nd,

• ϕ(n) counts the positive integers up to a given integernthat are relatively prime ton

• ω(n) is the number of distinct primes dividingn,

• N_i=Q

j6ip_j (primorial),

• rad(n) = Q

p|n p prime

p, radical ofn.

Email address:vojakrob@gmail.com (Robert VOJAK).

Defines(n) = σ(n)

n ,f(n) = n ϕ(n).

A numbernsatisfiesRobin’s inequality[3] if and only ifn>5041, and s(n)< e^γlog logn

Definition1.1 A number nis superabundant (SA) [1] if for all m < n, we haves(m)< s(n).

Definition1.2 A numbernis a Hardy-Ramanujan number (HR) [6] ifn= Q

i6ω(n)

p^a_iⁱ witha_i>ai+1>1 for alli6ω(n)−1.

All the results presented hereafter concern Robin’s inequality, which is known to be equivalent to Rie- mann’s Hypothesis[3]. We start by establishing an important result about the multiplicities in the prime factorization of a numbern, and how switching these multiplicities affectsnands(n).

Next, we give some categories of numbers satisfying Robin’s inequality, and for those who do not, we exhibit some of their properties (theorem 1.6).

The last part is devoted to sharpening Robin’s inequality in specific cases.

In the sequel, the notation q1, q2, ... will be used to define a finite strictly increasing sequence of prime numbers. The following theorems represent our main results (the proofs will be given later on):

Theorem 1.3 Letn=

ω(n)

Q

i=1

q^a_iⁱ be a positive integer. If – ω(n)>18and#{i6ω(n) ; a_i6= 1}> ^ω(n)

2 , or – ω(n)>39and#{i6ω(n) ; ai6= 1}> ^ω(n)

3 , or – ω(n)>969672728and#{i6ω(n) ; a_i6= 1}> ^ω(n)₁₄ thenn satisfies Robin’s inequality.

Theorem 1.4 Letn= Q

i6k

q_i^ai. For all positive integersk, q, define

M(k) =e^e−γf(Nk)−logNk (1)

M_k(q) = 1 + M(k) logq

If for some i6ω(n), we havea_i >M_ω(n)(qi), then the integer nsatisfies Robin’s inequality, regardless of the multiplicities of the other prime divisors.

For instance, if n=Q

i65p^a_iⁱ, M5(2) = 11.3367. . . All integers of the form 2^a1Q5

i=2p^a_iⁱ satisfy Robin’s inequality ifa1>12, anda2, a3, a4, a5>1.

Corollary 1.5 Let n be a positive integer. If f(Nω(n))−e^γlog logNω(n) 60, then n satisfies Robin’s inequality.

Proof: Iff(Nω(n))−e^γlog logNω(n)60, we haveM(k)60, and for alli6ω(n) and all prime numbers p,

a_i>1>M_ω(n)(p)

We conclude using theorem 1.4.

Theorem 1.6 Letc=

ω(n)

Q

i=1

q^c_iⁱ(c>5041)be the least number (if it exists) not satisfying Robin’s inequality.

The following properties hold:

1. c is superabundant, 2. ω(c)>969 672 728,

3. #{i6ω(c) ;ci6= 1}<ω(c) 14 , 4. e⁻

1

logpω(c) < p_ω(c) logc <1, 5. for all1< i6ω(c),

p^c_iⁱ <min

2^c1+2, p_ie^M(k)

2. Preliminary results

We will need some properties about the functions sand f to prove the above-mentioned theorems.

Lemma 2.1 Let n= Q

i6k

p^a_iⁱ,p, q be two prime numbers, andk, aandb positive integers.

1. the functions sandf are multiplicative 2. s(n) = Q

i6k

p_i−p^−a_i ⁱ

pi−1 ,f(n) = Q

i6k

p_i

pi−1 andf(n) =f(rad(n)).

3. 1 + 1

p6s(p^k)< p

p−1 and lim

k7→∞s(p^k) = p

p−1 =f(p) 4. If p < q,s(p^a)> s(q^b)for all positive integers a, b

5. s(p^k)increases as k >0 increases, decreases aspincreases, 6. s(p^k+1)

s(p^k) decreases ask increases, 7. If a > b >0, s(p^a)

s(p^b) decreases aspincresases,

8. Let pbe a prime number, andqany prime number, q < p(p+ 1). Then, for all a, b>1 s(p^a+1)

s(p^a) < s(q^b) For instance:s(p^a+1_n )< s(p^a_n)s(p^b_n+1)for alla, b>1.

9. If m, n>2,s(mn)6s(m)s(n).

IMPORTANT: note thats(p^a)> s(q^b)requires only to havep < q, and no condition on positive integers a, b.

Proof:

1. The multiplicativity ofsandf is a consequence of the multiplicativity ofσandϕ,

2. The proof is straightforward: recall that rad(p^k) = p,ϕ(p^k) =p^k−p^k−1, and hencef(p^k) =_p−1^p = f(rad(p^k)), and use the multiplicativity off and rad to conclude.

3. It is deduced from the propertiesσ(p^k) = ^pk+1_p−1⁻¹ andϕ(p^k) =p^k−p^k−1. To prove the asymptotic result, uses(p^k) = ^p−

1 pk

p−1 , 4. We have

1 +1 p > q

q−1 whenq > p, and we deduce

s(p^a)>1 + 1 p> q

q−1 > s(q^b) 5. Uses(p^k) =^p−

1 pk

p−1 , and we can conclude thats(p^k) increases askincreases. Use (2.1) witha=b=k to prove that ifp < q, s(p^k)> s(q^k).

6. We have

s(p^k+1)

s(p^k) =p^k+2−1 p^k+2−p which decreases asp^k increases.

Let us show that it also decreases aspincreases. The sign of the first derivative ofp7→ ^s(p_s(p^k+1k)⁾is the same as the sign ofp^k+1(k+2)−p^k+2(k+1)−1 =p^k+1(k+2−p(k+1))−16p^k+1(k+2−2(k+1))−1 =

−p^k+1(k−1)<0.

7. The equality

s(p^a) s(p^b) =

a−1

Y

k=b

s(p^k+1) s(p^k)

along with the monotonicity of each fraction of the product yields the desired result.

8. The functiona7→ s(p^a+1)

s(p^a) is decreasing, and we have s(p^a+1)

s(p^a) −s(q)< s(p²)

s(p) −1−1

q = 1

p(p+ 1) −1 q Hence

s(p^a+1)

s(p^a) < s(q)6s(q^b) 9. Letn=Q

p|n

p^ap andm= Q

p|m

p^bp. We have m n=Y

p|n p∤m

p^apY

p∤n p|m

p^bpY

p|n p|m

p^ap+bp

yielding

s(m n) =Y

p|n p∤m

s(p^ap)Y

p∤n p|m

s(p^bp)Y

p|n p|m

s(p^ap+bp)

Ifpis a prime number, anda, btwo positive integers, then

s(p^a+b)−s(p^a)s(p^b) =−(p^a−1)(p^b−1) p^a+b−1(p−1)² <0 Now we have

s(m n)6Y

p|n p∤m

s(p^ap)Y

p∤n p|m

s(p^bp)Y

p|n p|m

(s(p^ap)s(p^bp)) =s(m)s(n)

Lemma 2.2 For all positive integersn>2, we haves(n)< f(Nω(n)) Proof: Letn=

k

Q

i=1

q_i^ai whereq1, q2, q3, ...is a finite increasing sequence of prime numbers. Therefore, for alli,p_i6q_i. Setm=

k

Q

i=1

p^a_iⁱ. Clearly,m6nand s(p^a_iⁱ) s(q^a_iⁱ) >1 yielding

s(n)6s(m)< f(m) =f(Nω(n))

Theorem 2.3 (Multiplicity Permutation) Letnbe a positive integer,pandqbe two prime divisors ofn (p < q), andaandb their multiplicity respectively.

Consider the integer n^⋆ obtained by switching aandb. Ifa < b n^⋆< n and s(n^⋆)> s(n) and if a > b

n^⋆> n and s(n^⋆)< s(n) Proof: Switchingaandb means thatn^⋆=n

p q

b−a

. Hence, if a < b, we haven^⋆< nand s(n)

s(n^⋆) = s(p^a)s(q^b)

s(p^b)s(q^a) < s(p^a)s(p^b) s(p^b)s(p^a) = 1 Indeed, since a < b, we know from lemma 2.1 that s(q^b)

s(q^a) decreases as q increases, and sinceq > p, we have

s(q^b)

s(q^a) < s(p^b) s(p^a)

The proof for the casea > bfollows the same logic.

Letn= Q^k

i=1

q^a_iⁱ, andb1, b2, . . . , bkbe a reordering ofa1, a2, . . . , ak such thatbi>bi+1. Define the functions AandH as follows:

A



 Y

i6k

q^a_iⁱ



=Y

i6k

q_i^bi (2)

H



 Y

i6k

q^a_iⁱ



=Y

i6k

p^b_iⁱ (3)

We claim thatA(n)6nands(n)6s(A(n)): letn1be the integer obtained by switchinga1andb1. Using theorem 2.3, we haven16nands(n1)>s(n). Continue by switchingb2and the multiplicity ofp2in the prime factorization ofn1. The result is an integern2 such thatn26n16nand s(n2)>s(n1)>s(n).

Repeat the process a total ofk−1 times, and you obtain a positive integernk−1 such that nk−1=Y

i6k

q_i^bi =A(n) with bi>bi+1

nk−16n and s(nk−1)>s(n)

This proves that A(n) 6 n and s(A(n)) > s(n). The same inequalities hold for the function H. The function H replaces all the prime divisors by the first prime divisors, and reorganize the multiplici- ties in a decreasing order. It is easy to check that H(n) 6 n and s(H(n)) > s(n). Indeed, note that H(n)6A(n)6nands(H(n))>s(A(n))>s(n).

Letn>3. IfH(n) satisfies Robin’s inequality, so does n. Indeed,

s(n)6s(H(n))< e^γlog logH(n)< e^γlog logn thus proving:

Theorem 2.4 If Robin’s inequality is valid for all Hardy-Ramanujan numbers greater than5040, then it is valid for all positive integers greater than5040.

Note: compare this result to proposition 5.1 in [6]: ”if Robins inequality holds for all Hardy-Ramanujan integers 50416n6x, then it holds for all integers50416n6x”.

3. Numbers satisfying Robin’s inequality

Some categories of numbers satisfy Robin’s inequality. Most of them have already been identified in previous works[6], and we report these results below. Note that the proofs are not the original ones. In- stead, we used simple methods to provide shorter and simpler proofs. To do so, we will need the following inequalities ([3], th. 2).

Theorem 3.1 For alln>3,

s(n)6e^γlog logn+ 0.6483 log logn and [2](th. 15)

Theorem 3.2 For alln>3,

f(n)6e^γlog logn+ 2.51 log logn

We will start with primorials. We will then investigate odd positive integers, and integers n such that ω(n) 6 4 (all known integers not satisfying Robin’s inequality are such that ω(n) 64), and we will conclude with square-free and square-full integers.

3.1. Primorial numbers

Theorem 3.3 All primorialsN_k (k>4) satisfy Robin’s inequality:

s(Nk) =

k

Y

i=1

1 + 1

p_i

< e^γlog logN_k The following inequality is a bit sharper (fork>2):

k

Y

i=1

1 + 1

pi

63

4

e^γlog logNk+ 2.51 log logNk

Proof: Letk>2 be an integer. We have s(Nk) =s(2)s

Nk

2

6s(2)f Nk

2

= s(2)

f(2)f(Nk) =3 4f(Nk) Using theorem 3.2, we have

s(Nk)−e^γlog logNk63 4

e^γlog logNk+ 2.51 log logNk

−e^γlog logNk

6−0.25e^γlog logNk+ 1.8825 log logNk

<0

ifk>6. Fork65, only the numbersN1= 2, N2= 6 andN3= 30 do not satisfy Robin’s inequality.

3.2. Odd integers

The following theorem can be found in [6], but the proof is not the original one. A very simple proof is given instead.

Theorem 3.4 Any odd positive integern distinct from3,5and9 satisfies Robin’s inequality.

Proof: Letn>3 be a positive odd integer. Using theorem 3.1, we have s(n) = s(2n)

s(2) <2 3

e^γlog log(2n) + 0.6483 log log(2n)

< e^γlog logn for alln>17. Indeed, set

g(x) = 2 3

e^γlog log(2x) + 0.6483 log log(2x)

−e^γlog logx

We have, forx>2,

g^′(x)<−1.23(log log(2x))²+ (0.43 + 0.6(log log(2x))²) logx x(logx)(log(2x))(log log(2x))² <0

Whenx> 17, g(x)6g(17)<0. Therefore, all odd integers greater than 15 satisfy Robin’s inequality.

For odd integers up to 15, only 3,5 and 9 do not satisfy Robin’s inequality.

3.3. Integersn such thatω(n)≤4

Theorem 3.5 Let n such that ω(n)≤4. The exceptions to Robin’s inequality such thatω(n) =k form the setsCk where

C1={3,4,5,8,9,16}

C2={6,10,12,18,20,24,36,48,72}

C3={30,60,84,120,180,240,360,720}

C4={840,2520,5040}

Proof: Numbersnunder 5041 that do not satisfy Robin’s inequality are well known, and the number of prime divisorsω(n) of these counterexamples does not exceed 4. We will now show that ifω(n)64, the only counterexamples are the elements ofC.

Letnbe an integer such thatω(n)64. We have (lemma 2.2) using the monotonicity ofk7→f(Nk) s(n)< f(N_ω(n))6f(N4) = 4.375

Therefore, ifn >116144, log logn>e^−γ4.375,

e^γlog logn>4.375> s(n)

and Robin’s inequality is satisfied if n >116144 and ω(n) 6 4. Numerical computations confirm that there are no counterexamples to Robin’s inequality in the range 5041< n6116144.

3.4. Square-free integers

A positive integernis calledsquare-freeif for every prime numberp, p² is not a factor ofn. Hencen has the form

n=Y

i6k

q_i

The following two theorems can be found in [6], but not the proofs (we provide simpler proofs).

Theorem 3.6 Any square-free positive integer distinct from2,3,5,6,10,30satisfy Robin’s inequality.

Proof: Ifnis square-free and even, the prime number 2 has multiplicitya1= 1. This implies s(n)

s(2n) =s(2) s(4) =6

7

and using theorem 3.1

s(n)< 6 7

e^γlog log(2n) + 0.6483 log log(2n)

< e^γlog logn

ifn >418. Ifn6418, numerical computations show that only the square-free numbers 2,6,10,30 do not satisfy Robin’s inequality.

Ifnis odd, see theorem 3.4.

3.5. Square-full integers

Letnbe an integer. If for every prime divisorpofn, we havep²|n, the integernis said to besquare-full.

This result can also be found in [6]. The proof we give here is shorter and simpler.

Theorem 3.7 The only square-full integers not satisfying Robin’s inequality are 4, 8, 9, 16 and 36.

Proof:

Case 1:ω(n)64. From theorem 3.5, the only square-full counterexamples are 4,8,9,16,36.

Case 2:ω(n)>5. Letn= Q^k

i=1

q^a_iⁱ. Sincenis square-full, we haven>N_k²ands(n)< f(Nk) (lemma 2.2).

We can now write, using theorem 3.2,

s(n)−e^γlog logn < f(Nk)−e^γlog logN_k²< 2.51

log logN_k −1.2345<−0.0083

4. Proof of theorem 1.3

Lemma 4.1 Let M(k) =e^e−γf(Nk)−logNk

1. For all k>39, we haveM(k)6logN⌊^k3⌋ 2. For all k>18, we haveM(k)6logN⌊^k2⌋

3. For all k>969672728, we haveM(k)6logN⌊14^k⌋ Proof: Let us prove property 1.

We claim that for allk >38, we haveM(3k+j)6logN(k) for all 06j <3. Using theorem 3.2, we have M(k)6

e^{log log1.41Nk} −1 logN_k Define

ǫ_k=e^{log(3k1.41log(3k))} −1− klogk

(3k+ 2)(log(3k+ 2) + log log(3k+ 2)) Using the inequalities [4]

klogk <logNk < k(logk+ log logk) fork>13

we see that

ǫ_k>e^{log log1.41N3k} −1− logN_k logN3k+2

We claim thatǫk is decreasing, and thatǫk <0 fork >109. Indeed, rewrite the term klogk

(3k+ 2)(log(3k+ 2) + log log(3k+ 2)) = k

3k+ 2 × logk

log(3k+ 2) × log(3k+ 2)

log(3k+ 2) + log log(3k+ 2) as a product of three increasing functions: to prove they are increasing, you can use the monotonicity of x7→ ^log_x^x (x>e), and the monotonicity ofx7→ _log(x+a)^logx , whereais a positive real number .

We haveǫ109<−0.0003 implying for all 06i <3 e

1.41

log logN3k+i −1< e^{log log1.41N3k} −16 logNk

logN3k+2

6 logNk

logN3k+i

yielding

M(3k+i)6 e

1.41

log logN3k+i −1

logN3k+i6logNk

or simply

M(k)6logN_⌊k 3⌋

fork>109, and was confirmed for 396k <109 with the help of numerical computations.

•Proof of property 2:This result is deduced the same way: consider the function

ǫ_k=e^{log(2k1.41log(2k))} −1− klogk

(2k+ 1)(log(2k+ 1) + log log(2k+ 1)) We have

ǫk>e^{log log1.41N2k} −1− logNk

logN2k+1

andǫ_k is a decreasing function withǫ28<−0.003 yielding M(k)6logN_⌊k

2⌋

fork >27, and was confirmed for 186k <28 with the help of numerical computations.

•Proof of property 3:we proceed the same way. Consider the function

ǫk =elog(14klog(14k))^1.41 −1− klogk

(14k+ 13)(log(14k+ 13) + log log(14k+ 13)) We have

ǫ_k>e^{log log1.41N14k} −1− logN_k logN14k+13

andǫk is a decreasing function withǫ969672728<−0.001 yielding M(k)6logN_⌊k

14⌋

fork>969672728.

Proof of theorem 1.3

We will need the functionH defined in (3).

H(n)6n and s(H(n))>s(n)

The integerH(n) is a Hardy-Ramanujan number. In addition, ifH(n) satisfies Robin’s inequality, so does n. Indeed,

s(n)6s(H(n))< e^γlog logH(n)6e^γlog logn

Note that the value of #{i6ω(n) ; ai6= 1}is the same if we replace nwithH(n), and thatω(H(n)) = ω(n). So without loss of generality, we will prove the theorem for Hardy-Ramanujan numbers only.

Let n be a Hardy-Ramanujan number. Define j(n) := min{1 6 j 6 ω(n);M(ω(n)) 6 logN_j} and i(n) := #{i 6 ω(n);a_i 6= 1}. A sufficient condition for n to satisfy Robin’s inequality is i(n) > j(n).

Sincenis a Hardy-Ramanujan number, we havea_i>2 fori6i(n), anda_i= 1 otherwise, implying that n>N_i(n)N_ω(n)and yielding

s(n)−e^γlog logn < f(Nω(n))−e^γlog log(Ni(n)N_ω(n))<0 ifi(n)>j(n). Hence, nsatisfies Robin’s inequality.

If ω(n) >18, and if #{i 6 ω(n);ai 6= 1} > ^ω(n)

2 , n satisfies Robin’s inequality. Indeed, if ω(n) >18, lemma 4.1 tells us thatj(n)6^ω(n)₂ . Using the assumption #{i6ω(n);ai6= 1}> ^ω(n)₂ yieldsi(n)>j(n).

Proceed the same way with the two cases:ω(n)>39 andω(n)>969672728

5. Proof of theorem 1.4

According to lemma 2.2, we haves(n)< f(Nω(n)) and n=

ω(n)

Y

j=1

q_j^aj >q_i^ai−1



q_iY

j6=i

q^a_j^j



>q^a_iⁱ⁻¹



p_iY p_j

j6=i



=q_i^ai−1N_ω(n) implying

s(n)−e^γlog logn < f(Nω(n))−e^γlog log(q^a_iⁱ⁻¹N_ω(n)) Ifa_i>M_ω(n)(qi) with

M_k(q) := 1 + e^e−γf(Nk)−logNk

logq then

f(Nω(n))−e^γlog log(q^a_iⁱ⁻¹Nω(n))60 and therefore

s(n)< e^γlog logn

6. Proof of theorem 1.6

Proof of 1. The superabundance of the numberc is proved in [7].

Proof of 2. In [5], numerical computations have confirmed thatc >10¹⁰¹⁰. Using lemma 2.2, we have

e^γlog logc6s(c)< f(Nω(c))< e^γ

logp_ω(c)+ 1 logp_ω(c)

yielding

logpω(c)>1 2

log logc+p

log logc−4 Since log logc >10 log 10 + log log 10>23.85988, we obtain

logp_ω(c)>23.81789 yielding

ω(c)>969 672 728

Proof of 3. ⋆Part 1:p_ω(c)<logc

By definition, the counterexamplecdoes not satisfy Robin’s inequality:

s(c)>e^γlog logc and since c > 5040 is the least one, the integer c^′ = c

p_ω(c) satisfy Robin’s inequality, because ω(c^′) =ω(c)−1>4. Therefore, we have

s c

p_ω(c)

< e^γlog log c p_ω(c) yielding

s(c) s

c p_ω(c)

> log logc

log log_pω(c)^c (4)

Now use s(c) s

c pω(c)

=s(pω(c)) = 1 + 1

p_ω(c), and using inequality log(1−x)< x(x6= 0), we establish that

log log c pω(c)

= log logc+ log

1−logp_ω(c) logc

<log logc−logp_ω(c)

logc (5)

Inequality (4) becomes 1

p_ω(c) > log logc log log_p^c

ω(c)

−1> log logc

log logc−^log_log^pω(c)_c −1 = logp_ω(c)

logclog logc−logp_ω(c) and

pω(c)logpω(c)<logclog logc−logpω(c)<logclog logc

implyingpω(c)<logc.

⋆ Part 2:logc < p_ω(c)e

1 logpω(c)

The counterexample inequalitys(c)>e^γlog logcyields e^γlog logc6s(c)< f(N_ω(c))< e^γ

logp_ω(c)+ 1 logp_ω(c)

implying logc < p_ω(c)e

1 logpω(c).

Proof of 4. See theorem 1.3.

Proof of 5. The proof of the inequalityp^a_iⁱ <2^a1+2 can be found in [1] (lemma 1). The other inequality p^c_iⁱ< pie^M(ω(c))

is deduced from theorem 1.4.

7. Sharper bounds

We can use some of the ideas of this article to improve upper bounds for the arithmetic functionss andf.

7.1. Primorials

For instance, when dealing with primorialsN_k, we can write s(Nk) = s(2)

s(4)s(2Nk)

and then use an upper bound fors(2Nk) (see theorem 3.1) to sharpen Robin’s bound fors(Nk). Or we can use another method :

Theorem 7.1 For alli>2, for alln>i,

n

Y

k=1

1 + 1

p_k

< αi

e^γlog logNn+ 2.51 log logN_n

with

αi=

i

Y

j=1

1− 1 p²_j

!

Ex: for alln>4,

n

Y

k=1

1 + 1

p_k

<0.627

e^γlog logN_n+ 2.51 log logN_n

< e^γlog logN_n

Proof: Leti>2, n>i. We have s(Nn) =s(Ni)s

N_n N_i

< s(Ni)f N_n

N_i

= s(Ni)

f(Ni)f(Nn) =α_if(Nn)

Use theorem 3.2 to conclude.

7.2. Odd numbers

Lemma 7.2 Ifn>17 is odd,

σ(2n)<2e^γnlog log(2n) Proof: Write

s(2n) =s(2)

s(4)s(4n)6 6 7

e^γlog log(4n) + 0.6483 log log(4n)

< e^γlog log(2n) To prove the last inequality, define

g(x) =6 7

e^γlog log(4x) + 0.6483 log log(4x)

−e^γlog log(2x) We have

g^′(x)<−0.38 + 1.41(log log(4x))²+ (0.55 + 0.25(log log(4x))²) logx xlog(2x) log(4x)(log log(4x))² <0

and g(210) < 0 < g(209). The inequality s(2n) < e^γlog log(2n) is valid for n > 209, and numerical

computations show that this is also true when 176n6209.

Corollary 7.3 If n>17is odd,

σ(n)<2

3e^γnlog log(2n) The following result is an improvement of theorem 1.2 in [6].

Proof: Use the previous result to get

s(n) =s(2n) s(2) < 2

3e^γlog log(2n)

The following result is an improvement of theorem 2.1 in [6].

Theorem 7.4 Ifn>3 is odd, n ϕ(n) 61

2

e^γlog log(2n) + 2.51 log log(2n)

< e^γlog logn Proof: Write

f(n) = f(2n) f(2) ans use theorem 3.2:

f(n)6 1 2

e^γlog log(2n) + 2.51 log log(2n)

7.3. The general case

In previous sections, we have mentioned superabundant numbers and pointed out some of their prop- erties, thus confirming that all superabundant numbers are Hardy-Ramanujan numbers.

Robin’s inequality can be enhanced for non superabundant numbers. Indeed, ifnis not a superabundant number, there exists m < n such thats(m)>s(n). Let I(n) ={m < n;s(n)6s(m)}. The set I(n) is not empty, and we can defineB(n) = minI(n). We have

B(n)< n and s(n)6s(B(n) yielding

s(n)6s(B(n))< e^γlog logB(n)

ifB(n) satisfies Robin’s inequality. Hence, the upper bound is better because B(n)< n.

But this method to sharpen Robin’s bound does not say much about B(n). In order to find an up- per bound with exploitable informations, we can use the functionH (see (3)). This method concerns non Hardy-Ramanujan numbers, whereas the previous method concerns the larger set of non superabundant numbers.

In (3), we defined the functionH which transforms any number into a Hardy-Ramanujan number, leaves the multiplicities unchanged (but not teh way they are ordered) and is such that H(n)6nand s(n)6 s(H(n)). IfH(n) satisfies Robin’s inequality, then

s(n)6s(H(n))< e^γlog logH(n)

The property H(n) 6 n improves Robin’s inequality, and unlike the function B, there is an explicit method to determine its value for all positive integers.

References

[1] Alaoglu, L. and Erdos, P. On highly composite and similar numbers. Trans. Amer. Math. Soc., 56:448469, 1944.

[2] Rosser, J. Barkley and Schoenfeld, Lowell. Approximate formulas for some functions of prime numbers. Illinois J. Math.

6 (1): 6494, 1962.

[3] Robin, Guy. Grandes valeurs de la fonction somme des diviseurs et hypothse de Riemann, Journal de Mathmatiques Pures et Appliques, Neuvime Srie, 63 (2): 187213, ISSN 0021-7824, MR 0774171, 1984.

[4] Massias, Jean-Pierre and Robin, Guy. Bornes effectives pour certaines fonctions concernant les nombres premiers.

Journal de thorie des nombres de Bordeaux, Tome 8, no. 1, p. 215-242 , 1996.

[5] Briggs, K. Abundant numbers and the Riemann hypothesis. Experiment. Math., 15(2):251256, 2006.

[6] Choie, YoungJu and Lichiardopol, Nicolas and Moree, Pieter and Sole, Patrick. On Robins criterion for the Riemann hypothesis, Journal de Theorie des Nombres de Bordeaux 19, 357372, 2007.

[7] Akbary, Amir and Friggstad, Zachary. Superabundant numbers and the Riemann hypothesis. Amer. Math. Monthly, 116(3):273275, 2009.