Extremely Abundant Numbers and the Riemann Hypothesis

Sadegh Nazardonyavi and Semyon Yakubovich Departamento de Matemática Faculdade de Ciências Universidade do Porto 4169-007 Porto Portugal

                   sdnazdi@yahoo.com
                   syakubov@fc.up.pt

Abstract

Robin’s theorem states that the Riemann hypothesis is equivalent to the inequality σ(n) < e^{γ}n log log n for all n > 5040, where σ(n) is the sum of divisors of n and γ is Euler’s constant. It is natural to seek the first integer, if it exists, that violates this inequality. We introduce the sequence of extremely abundant numbers, a subsequence of superabundant numbers, where one might look for this first violating integer. The Riemann hypothesis is true if and only if there are infinitely many extremely abundant numbers. These numbers have some connection to the colossally abundant numbers. We show the fragility of the Riemann hypothesis with respect to the terms of some supersets of extremely abundant numbers.

1 Introduction

There are several statements equivalent to the famous Riemann hypothesis (RH) [4]. Some of them are related to the asymptotic behavior of arithmetic functions. In particular, the well-known Robin theorem (inequality, criterion, etc.) deals with the upper bound of the sum-of-divisors function σ, which is defined by σ(n) := ∑_{d|n} d. Robin [24, Theorem 1] established an elegant connection between RH and the sum of divisors of n by proving that the RH is true if and only if

\frac{σ(n)}{n log log n} < e^{γ},  for all n > 5040,  (1)

where γ is Euler’s constant.

Throughout the paper, as in Robin [24], we define the function f by setting

f(n) = \frac{σ(n)}{n log log n}.  (2)

Gronwall, in his study of the asymptotic maximal size for the sum of divisors of n [11], found that the order of σ(n) is always “very nearly n” [12, Theorem 323]. More precisely, he proved the following theorem.

Theorem 1 (Gronwall). Let f be defined as in (2). Then

lim sup_{n→∞} f(n) = e^{γ}.  (3)

Let us call a positive integer n [2, 23]

(i) colossally abundant, if for some ε > 0,

\frac{σ(n)}{n^{1+ε}} ≥ \frac{σ(m)}{m^{1+ε}},  (m < n) and  \frac{σ(n)}{n^{1+ε}} > \frac{σ(m)}{m^{1+ε}},  (m > n);  (4)

(ii) generalized superior highly composite, if there is a positive number ε such that

\frac{σ_{−s}(n)}{n^{ε}} ≥ \frac{σ_{−s}(m)}{m^{ε}},  (m < n) and  \frac{σ_{−s}(n)}{n^{ε}} > \frac{σ_{−s}(m)}{m^{ε}},  (m > n),

where σ_{−s}(n) = ∑_{d|n} d^{−s}. The parameter s is assumed to be positive in [23]. In the case s = 1, (ii) becomes (i).

Ramanujan initiated the study of these classes of numbers in an unpublished part of his 1915 work on highly composite numbers ([21, 23] and [22, pp. 78–129, 338–339]). More precisely, he defined rather general classes of these numbers. For instance, he defined generalized highly composite numbers, containing as a subset superabundant numbers [21, Section 59], and he introduced the generalized superior highly composite numbers, including as a particular case colossally abundant numbers. For more details we refer the reader to [2, 10, 23].

We denote by CA the set of all colossally abundant numbers. We also use CA as an abbreviation for the term “colossally abundant”. Ramanujan [23] proved that if n is a generalized superior highly composite number, i.e., a CA number, then under the RH we have

lim inf_{n→∞} (\frac{σ(n)}{n} − e^{γ} log log n) \sqrt{log n} ≥ − e^{γ}(2\sqrt{2} + γ − log 4π) ≈ −1.558,

lim sup_{n→∞} (\frac{σ(n)}{n} − e^{γ} log log n) \sqrt{log n} ≤ − e^{γ}(2\sqrt{2} − 4 − γ + log 4π) ≈ −1.393.

Robin [24] also established (independent of the RH) the following inequality

f(n) ≤ e^{γ} + \frac{0.648214}{(log log n)^{2}} ,  (n ≥ 3),  (5)

where 0.648214 ≈ (\frac{7}{3} − e^{γ} log log 12) log log 12 and the left-hand side of (5) attains its maximum at n = 12. In the same spirit, Lagarias [15] proved the equivalence of the RH to the problem

σ(n) ≤ e^{H_{n}} log H_{n} + H_{n}, (n ≥ 1),

where H_{n} := ∑_{j=1}^{n} 1/j is the nth harmonic number.

Investigating upper and lower bounds for arithmetic functions, Landau [16, pp. 216–219] obtained the following limits:

lim inf_{n→∞} \frac{φ(n) log log n}{n} = e^{−γ} ,  lim sup_{n→∞} \frac{φ(n)}{n} = 1,

where φ(n) is the Euler totient function, which is defined as the number of positive integers not exceeding n that are relatively prime to n. It can also be expressed as a product extended over the distinct prime divisors of n [3, Theorem 2.4] by

φ(n) = n ∏_{p|n} (1 − \frac{1}{p}) .

Nicolas [18, 19] proved that if the RH is true, then we have for all k ≥ 2

\frac{N_{k}}{φ(N_{k}) log log N_{k}} > e^{γ},  (6)

where N_{k} = ∏_{j=1}^{k} p_{j} and p_{j} is the jth prime. On the other hand, if the RH is false, then for infinitely many k, inequality (6) is true and for infinitely many k, inequality (6) is false.

Compared to numbers N_{k} which are the smallest integers that maximize n/φ(n), there are integers which play this role for σ(n)/n and they are called superabundant numbers. A positive integer n is said to be superabundant [2, 23] if

\frac{σ(n)}{n} > \frac{σ(m)}{m}  for all m < n.  (7)

We will use the symbol SA to denote the set of superabundant numbers and also as an abbreviation for the term “superabundant”. Briggs [5] described a computational study of the successive maxima of the relative sum-of-divisors function σ(n)/n. He also studied the density of these numbers. Wójtowicz [26] showed that the values of the function f defined in (2) are close to 0 on a set of asymptotic density 1. Another study on Robin’s inequality (1) is due to Choie et al. [9]. They have shown that the RH holds true if and only if every natural number divisible by a fifth power greater than 1 satisfies Robin’s inequality (1).

Akbary and Friggstad [1] established the following interesting theorem which enables us to limit our attention to a narrow sequence of positive integers, in order to find a probable counterexample to inequality (1).

Theorem 2 ([1, Theorem 3]). If there is any counterexample to Robin’s inequality (1), then the least such counterexample is a superabundant number.

Unfortunately, to our knowledge, there is no known algorithm (except the formula (7) in the definition) to produce SA numbers. Alaoglu and Erdős [2] proved that

Q(x) > c \frac{log x log log x}{(log log log x)^{2}} ,

where Q(x) denotes the number of superabundant numbers not exceeding x. Later, Erdős and Nicolas [10] demonstrated a stronger result that for every δ < 5/48 we have

Q(x) > (log x)^{1+δ}, (x > x_{0}).

As a natural question in this direction, it is interesting to introduce and study a set of positive integers to which the first probable violation of inequality (1) belongs. Following this aim, we introduce the sequence of extremely abundant numbers. We will establish another criterion equivalent to the RH by proving that the RH is true if and only if there are infinitely many extremely abundant numbers. Also, we give a connection between extremely abundant numbers and CA numbers. Moreover, we present approximate formula for the prime factorization of (sufficiently large) extremely abundant numbers. Finally, we establish the fragility of the Riemann hypothesis with respect to the terms of certain subsets of superabundant numbers which are quite close to the set of extremely abundant numbers.

Before stating the main definition and results of this paper, we mention recent work of Caveney et al. [6]. They defined a positive integer n as an extraordinary number if n is composite and f(n) ≥ f(kn) for all

k ∈ ℕ ∪ {1/p : p is a prime factor of n}.

Under these conditions they showed that the smallest extraordinary number is n = 4. Then they proved that the RH is true if and only if 4 is the only extraordinary number. For more properties of these numbers and comparisons with SA and CA numbers, we refer the reader to [7].

2 Extremely abundant numbers

We define a new sequence of positive integers related to the RH. Our primary contribution and motivation of this definition are Theorems 6 and 7. Let us now state the main definition of this paper.

Definition 3. A positive integer n is extremely abundant if either n = 10080, or n > 10080 and

\frac{σ(n)}{n log log n} > \frac{σ(m)}{m log log m},  for all 10080 ≤ m < n.  (8)

Here 10080 has been chosen as the smallest SA number greater than 5040. In Table 1 we list the first 20 extremely abundant numbers. To find them, we used a list of SA numbers (see Proposition 5) provided by Kilminster [14] and Noe [20].

Remark 4. If we choose (instead of 10080) n_{1} such that 2520 < n_{1} ≤ 5040, and define n to be extremely abundant if either n = n_{1}, or n > n_{1} and

\frac{σ(n)}{n log log n} > \frac{σ(m)}{m log log m}, for all n_{1} ≤ m < n,

then we have a finite number of elements n ≤ 5040 that satisfy the above inequality. Using inequality (5), we have

f(n) < e^{γ} + \frac{0.648214}{(log log n)^{2}} < f(5040), for some s_{10308} < n ≤ s_{10309},

where s_{k} denotes the kth SA number listed in [20]. Checking by computer for values n between 5040 and s_{10309}, we derive a finite set with maximum 5040. Similarly, one can easily check for n_{1} < 2520, and get only sets with finite number of elements.

Let XA denote the set of all extremely abundant numbers. (We also use XA as an abbreviation for the term “extremely abundant”.) Clearly, XA ≠ CA (see Table 1). Indeed, we shall prove that infinitely many elements of CA are not in XA and that, if RH holds, then infinitely many elements of XA are in CA. As an elementary result from the definition of XA numbers we have the following proposition.

Proposition 5. The inclusion XA ⊂ SA holds.

Proof. First, 10080 ∈ SA. Next, if n > 10080 and n ∈ XA, then for 10080 ≤ m < n we have

\frac{σ(n)}{n} = f(n) log log n > f(m) log log m = \frac{σ(m)}{m}.

In particular, for m = 10080 we get

\frac{σ(n)}{n} > \frac{σ(10080)}{10080}.

So for m < 10080, we have

\frac{σ(n)}{n} > \frac{σ(10080)}{10080} > \frac{σ(m)}{m}

since 10080 ∈ SA. Therefore, n belongs to SA.

Next, motivating our construction of XA numbers, we will establish the first interesting result of the paper.

Theorem 6. If there is any counterexample to Robin’s inequality (1), then the least one is an XA number.

Proof. By doing some computer calculations we observe that there is no counterexample to Robin’s inequality (1) for 5040 < n ≤ 10080. Now let n > 10080 be the least counterexample to inequality (1). For m satisfying 10080 ≤ m < n we have

f(m) < e^{γ} ≤ f(n).

Therefore n is an XA number.

As we mentioned in Section 1 we will prove an equivalent criterion to the RH for which the proof is based on Robin’s inequality (1) and Gronwall’s theorem. Let #A denote the cardinality of the set A. The second stimulus result is the following theorem. This result also has its own interest that will be discussed in Section 5.

Theorem 7. The RH is true if and only if #XA = ∞.

Proof. Sufficiency. Assume that RH is not true. Then by Theorem 6 we have f(m) ≥ e^{γ} for some m ≥ 10080. From Gronwall’s theorem, we know that M = sup_{n≥10080} f(n) is finite and there exists n_{0} such that f(n_{0}) = M ≥ e^{γ} (if M = e^{γ} then set n_{0} = m). An integer n > n_{0} satisfies f(n) ≤ M = f(n_{0}) and n can not be in XA, so #XA ≤ n_{0}.

Necessity. On the other hand, if RH is true, then Robin’s inequality (1) holds. If #XA is finite, then let m be its largest element. For every n > m the inequality f(n) ≤ f(m) holds and therefore

lim sup_{n→∞} f(n) ≤ f(m) < e^{γ},

which is a contradiction to Gronwall’s theorem.

There are some primes which cannot be the largest prime factor of any XA number. For example, referring to Table 1, there is no XA number with the largest prime factor p(n) = 149. Do there exist infinitely many such primes?

3 Auxiliary lemmas and inequalities

Chebyshev’s functions ϑ(x) and ψ(x) are defined by

ϑ(x) = ∑_{p≤x} log p, ψ(x) = ∑_{p^{m}≤x} log p = ∑_{p≤x} ⌊\frac{log x}{log p}⌋ log p,

where ⌊x⌋ denotes the largest integer not exceeding x. It is known that the prime number theorem (PNT) ([12, Theorem 434] and [13, Theorem 3, 12]) is equivalent to

ψ(x) ∼ x.  (9)

In his mémoir, Chebyshev proved the following lemma that we call Chebyshev’s result.

Lemma 8 ([8, p. 379]). For all x > 1

ϑ(x) < \frac{6}{5}Ax − Ax^{\frac{1}{2}} + \frac{5}{4 log 6} log^{2} x + \frac{5}{2} log x + 2
ϑ(x) > Ax − \frac{12}{5}Ax^{\frac{1}{2}} − \frac{5}{8 log 6} log^{2} x − \frac{15}{4} log x − 3,

where

A = log \frac{2^{\frac{1}{2}} 3^{\frac{1}{3}} 5^{\frac{1}{5}}}{30^{\frac{1}{30}}} ≈ 0.92129202.

We will use the following corollary in the proof of Theorem 26.

Corollary 9. We have

ϑ(x) > \frac{x}{3}, (x ≥ 3).

The next lemma here provides Littlewood’s result for oscillation of Chebyshev’s ϑ function.

Lemma 10 ([7, Lemma 4]). There exists a constant c > 0 such that for infinitely many primes p we have

ϑ(p) < p − c\sqrt{p} log log log p,  (10)

and for infinitely many other primes p we have

ϑ(p) > p + c\sqrt{p} log log log p.

In what follows, we shall frequently use the following elementary inequalities:

\frac{t}{1 + t} < log(1 + t) < t,  (t > 0),  (11)

and

\frac{2t}{2 + t} < log(1 + t),  (t > 0).  (12)

4 Some properties of SA, CA and XA numbers

We divide this section into three subsections, for which we shall exhibit several properties of SA, CA and XA numbers, respectively. We denote by p(n) the largest prime factor of n or, when there is no ambiguity, simply by p.

4.1 SA Numbers

Proposition 11. Let n < n′ be two consecutive SA numbers. Then

\frac{n′}{n} ≤ 2.

Proof. Let n = 2^{k_{2}} ⋯ p. We compare n with 2n. In fact,

\frac{σ(2n)/(2n)}{σ(n)/n} = \frac{2^{k_{2}+2} − 1}{2^{k_{2}+2} − 2} > 1.

Hence n′ ≤ 2n.

Corollary 12. For any positive real number x ≥ 1 there exists at least one SA number n such that x ≤ n < 2x.

Alaoglu and Erdős [2] have shown that if n = 2^{k_{2}} · 3^{k_{3}} ⋯ p^{k_{p}} is an SA number, then k_{2} ≥ k_{3} ≥ ⋯ ≥ k_{p} and k_{p} = 1, except for n = 4, 36.

Proposition 13 ([2, Theorem 2]). Let n ∈ SA, and let q < r be prime factors of n with corresponding exponents k_{q} and k_{r}. Set

β := ⌊\frac{k_{q} log q}{log r} ⌋ .

Then k_{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 an SA number with error of at most 1 in terms of a smaller prime factor of that number. In the next lemma we prove a relation between the lower bound of an exponent of a prime factor of n and its largest prime factor p.

Lemma 14. Let n ∈ SA, and let q be a prime factor of n. Then

⌊\frac{log p}{log q}⌋ ≤ k_{q}.

Proof. If q = p (= p(n)), then the result is trivial. Let q < p and k_{q} = k. Suppose that k ≤ ⌊log p/ log q⌋ − 1. Then

q^{k+1} < p.  (13)

Now we compare values of σ(ν)/ν, taking ν = n and ν = m = nq^{k+1}/p. Since σ(ν)/ν is multiplicative, we restrict our attention to different factors. But n is an SA number and m < n. Hence

1 < \frac{σ(n)/n}{σ(m)/m} = \frac{q^{2k+2} − q^{k+1}}{q^{2k+2} − 1} (1 + \frac{1}{p}) = \frac{1}{1 + 1/q^{k+1}} (1 + \frac{1}{p}) .

Consequently, p < q^{k+1}, which contradicts inequality (13).

Proposition 15 ([2, Theorem 5]). Let n ∈ SA. If k_{q} = k and q < (log p)^{α}, where α is a constant, then

log \frac{q^{k+1} − 1}{q^{k+1} − q} > \frac{log q}{p log p} (1 + O (\frac{(log log p)^{2}}{log p log q} )) ,  (14)

log \frac{q^{k+2} − 1}{q^{k+2} − q} < \frac{log q}{p log p} (1 + O (\frac{(log log p)^{2}}{log p log q} )) .  (15)

Lemma 16. Let n ∈ SA, and let q be a fixed prime factor of n. Then there exist two positive constants c and c′ (depending on q) such that

c p\frac{log p}{log q} < q^{k_{q}} < c′p\frac{log p}{log q}.

Proof. By inequality (11)

log \frac{q^{k+1} − 1}{q^{k+1} − q} = log (1 + \frac{q − 1}{q^{k+1} − q} ) < \frac{q − 1}{q^{k+1} − q} ≤ \frac{1}{q^{k}}

and (14), there exists a c′ > 0 such that

q^{k} < c′ \frac{p log p}{log q}.

On the other hand, again by inequality (11)

log \frac{q^{k+2} − 1}{q^{k+2} − q} = log (1 + \frac{q − 1}{q^{k+2} − q} ) > \frac{q − 1}{q^{k+2} − 1} > \frac{1}{2q^{k+1}}

and (15), there exists a c > 0 such that

q^{k} > c \frac{p log p}{log q}.

Corollary 17. Let n = 2^{k} ⋯ p be an SA number. Then for n sufficiently large we have

⌊\frac{k log 2}{log p}⌋ = 1.

We showed [17] that for sufficiently large n ∈ SA

log n < p (1 + \frac{1}{2 log p}) .  (16)

Our computation on the list of SA numbers in [20] suggests that a weaker inequality

log n < p (1 + \frac{2}{3 log p})  (17)

holds for all n ≥ s_{365}. The product of exponent of a prime factor and the logarithm of the corresponding prime factor of an SA number can be controlled, on average, by the logarithm of the largest prime factor of that number. More precisely,

Proposition 18 ([2, Theorem 7]). If n ∈ SA, then

p(n) ∼ log n.

The next proposition gives a lower bound of n ∈ SA in terms of Chebyshev’s ψ function compared to the above-mentioned asymptotic relation.

Proposition 19. Let n ∈ SA. Then

ψ(p(n)) ≤ log n.

Moreover,

lim_{\begin{matrix}n→∞ \\ n∈SA \end{matrix}} \frac{ψ(p(n))}{log n} = 1.  (18)

Proof. In fact, by Lemma 14

ψ(p(n)) = ∑_{q≤p(n)} ⌊\frac{log p(n)}{log q}⌋ log q ≤ ∑_{q≤p(n)} k_{q} log q = log n.

To prove (18) we appeal to (the equivalent of) the PNT (9) and Proposition 18.

4.2 CA Numbers

By the definition of CA numbers (4) it is easily seen that CA ⊂ SA. Here we give a concise description of the algorithm (essentially borrowed from [7, 10, 24]) to produce CA numbers. For more details on this introduction we refer the reader to [2, 7, 10, 24].

Let F be defined by

F (x, k) = \frac{log(1 + 1/(x + ⋯ + x^{k}))}{log x}.

For ε > 0 we define x_{1} = x_{1}(ε) to be the only number such that

F (x_{1}, 1) = ε,  (19)

and x_{k} = x_{k}(ε) (for k > 1) to be the only number such that

F (x_{k}, k) = ε.

Let

E_{p} = {F (p, α) : α ≥ 1}, p is a prime

and

E = ⋃_{p} E_{p} = {ε_{1}, ε_{2}, …}.

If ε ∉ E, then the function σ(n)/n^{1+ε} attains its maximum at a single point N_{ε} whose prime decomposition is

N_{ε} = ∏ p^{α_{p}(ε)}, α_{p}(ε) = ⌊\frac{log \frac{p^{1+ε}−1}{p^{ε}−1}}{log p} ⌋ − 1,

or if one prefers

α_{p}(ε) = \begin{cases}k, & if x_{k+1} < p < x_{k}, k ≥ 1; \\ 0, & if p > x = x_{1}. \end{cases}

If ε ∈ E, then at most two x_{k}’s are prime. Hence, there are either two or four CA numbers of parameter ε, defined by

N_{ε} = ∏_{k=1}^{K} ∏_{\begin{matrix}p<x_{k} \\ or \\ p≤x_{k} \end{matrix}} p,  (20)

where K is the largest integer such that x_{K} ≥ 2. In particular, if N is the largest CA number of parameter ε, then

F (p, 1) = ε ⇒ p(N) = p,  (21)

where p(N) is the largest prime factor of N. Therefore for any ε, formula (20) gives all possible values of a CA number N of parameter ε [7].

Robin [24, Proposition 1] proved that the maximum order of the function f defined in (2) is attained by CA numbers. More precisely, if 3 ≤ N < n < N′, where N and N′ are two successive CA numbers, then

f(n) ≤ max{f(N), f(N ′)}.

In the next proposition we improve the above inequality to a strict one.

Proposition 20. Let 3 ≤ N < n < N′, where N and N′ are two successive CA numbers. Then

f(n) < max{f(N), f(N′)}.  (22)

Proof. In fact, due to the strict convexity of the function t ↦ εt − log log t, Robin’s proof extends to the strict inequality (22).

Proposition 20 shows that if there is a counterexample to inequality (1), then there exists at least one CA number that violates it.

Lemma 21. Let N < N′ be two consecutive CA numbers. If there exists an XA number n > 10080 satisfying N < n < N′, then N′ is also an XA.

Proof. Let us set

B = {m ∈ XA : N < m < N′}.

By assumption n ∈ XA, we have B ≠ ∅. Let n′ = max B. Since n′ ∈ XA and n′ > N, it follows that f(n′) > f(N). From inequality (22) we must have f(n′) < f(N′). Hence N′ ∈ XA.

Remark 22. If n = 10080, then we have N = 5040 and N′ = 55440, and

f(N) ≈ 1.7909,  f(n) ≈ 1.7558,  f(N′) ≈ 1.7512.

Hence, f(N′) < f(n) < f(N) and inequality (22) is satisfied with

f(n) < f(N) = max{f(N), f(N′)}.

Therefore N′ ∉ XA. The point here is that n = 10080 is the initial XA number, so that it misses the property (8) of the definition of XA numbers which is used in the proof of Lemma 21.

Theorem 23. If RH holds, then there exist infinitely many CA numbers that are also XA.

Proof. If RH holds, then #XA = ∞ by Theorem 7. Let n ∈ XA. Since #CA = ∞ [2, 10], there exist two successive CA numbers N, N′ such that N < n ≤ N′. If N′ = n then it is already in XA, otherwise N′ belongs to XA via Lemma 21.

It can be seen that there exist infinitely many CA numbers N for which the largest prime factor p (= p(N)) is greater than log N. For this purpose we use the following lemma.

Lemma 24 ([7, Lemma 3]). Let N be a CA number of parameter ε with

ε < F (2, 1) = log(3/2)/ log 2

and define x = x(ε) by (19). Then

(i) for some constant c > 0

log N ≤ ϑ(x) + c\sqrt{x}.

(ii) Moreover, if N is the largest CA number of parameter ε, then

ϑ(x) ≤ log N ≤ ϑ(x) + c\sqrt{x}.

Theorem 25. There are infinitely many CA numbers N_{ε} such that log N_{ε} < p(N_{ε}). Proof. Let p be sufficiently large satisfying the inequality (10), and let N_{ε} be the largest CA number of parameter

ε = F (p, 1).

Then from (21) it follows that p(N_{ε}) = p. By part (ii) of Lemma 24 we have

log N_{ε} − ϑ(p) < c\sqrt{p}, (for some c > 0).

On the other hand, by Lemma 10 there exists a constant c′ > 0 such that

ϑ(p) − p < −c′\sqrt{p} log log log p, (c′ > 0).

Hence

log N_{ε} − p < \{c − c′ log log log p\}\sqrt{p} < 0,

which is the desired conclusion.

4.3 XA Numbers

We return to XA numbers and present some of their properties. We begin by the first interesting property of the XA numbers whose proof is essentially an application of the definition of XA numbers. Theorem 26. Let n ∈ XA. Then

p(n) < 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(n). Then m > 10080, since for all primes p we have ϑ(p) > p/3 (Corollary 9). Thus for n ∈ SA we have

log n ≥ ψ(p(n)) ≥ ϑ(p(n)) > \frac{1}{3}p(n)

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 we obtain

1 + \frac{1}{p(n)} = \frac{σ(n)/n}{σ(m)/m} > \frac{log log n}{log log m}.

Therefore,

\frac{1}{p(n)} > \frac{log log n}{log log m} − 1 = \frac{log(1 + log p(n)/ log m)}{log log m}.

Using inequality (11) we have

\frac{1}{p(n)} > \frac{log p(n)}{log n log log m} > \frac{log p(n)}{log n log log n} ⇒ p(n) < log n.

We mention here a similar result proved by Choie et al. [9].

Proposition 27 ([9, Lemma 6.1]). Let t ≥ 2 be fixed. Suppose that there exists a t-free integer exceeding 5040 that does not satisfy Robin’s inequality (1). Let n be the smallest such integer. Then p(n) < log n.

In Theorem 23 we showed that if RH holds, then there exist infinitely many CA numbers that are also XA. Next theorem is a conclusion of Theorems 25 and 26 which is independent of the RH.

Theorem 28. There exist infinitely many CA numbers that are not XA.

We conclude this subsection with a result describing the structure of (sufficiently large) XA numbers. More precisely, the next theorem will determine the exponents of the prime factors of a (sufficiently large) XA number with an error at most 1.

Theorem 29. Let n = 2^{k_{2}} ⋯ q^{k_{q}} ⋯ p ∈ XA. Set

α_{q}(p) = ⌊log_{q} (1 + (q − 1)\frac{p log p}{q log q})⌋ .

Then for sufficiently large n ∈ XA we have |k_{q} − α_{q}(p)| ≤ 1.

Proof. Assume that k_{q} = k and k − α_{q}(p) ≥ 2. Then we have

q^{k} ≥ q^{α_{q}(p)+2} > q (1 + (q − 1)\frac{p log p}{q log q}) .  (23)

Now let us compare f(n) with f(m), where m = n/q. Since n ∈ XA we must have

\frac{σ(n)/n}{σ(m)/m} = \frac{q^{k+1} − 1}{q^{k+1} − q} > \frac{log log n}{log log m},

or using inequality (11),

q^{k} < 1 + (q − 1)\frac{log n log log m}{q log q}.  (24)

Comparison of (23) and (24) gives that

log n log log m − qp log p > q log q.

This contradicts inequality (16). Now we assume that k − α_{q}(p) ≤ −2. Then

\frac{q^{k+2} − 1}{q − 1} ≤ \frac{p log p}{q log q}.  (25)

Choose m = nq/p. Comparing f(n) with f(m) we have

\frac{σ(n)/n}{σ(m)/m} = (1 − \frac{q − 1}{q^{k+2} − 1}) (1 + \frac{1}{p}) > \frac{log log n}{log log m} > 1 + \frac{log p/q}{log n log log m}.

or simply by (25)

−\frac{q log q}{log p} (1 + \frac{1}{p}) > \frac{p log p/q}{log n log log m} − 1.

Hence by (16) we have

−\frac{q log q}{log p} (1 + \frac{1}{p}) > \frac{log p/q}{(log p) (1 + \frac{1}{2 log p}) (1 + \frac{1}{2 log^{2} p})} − 1,

which is false for all 2 ≤ q ≤ p.

Remark 30. Note that if inequality (17) holds for XA numbers n ≥ s_{365}, then by performing computations for two smaller values of XA numbers, i.e., s_{20} and s_{356} (cf. Table 1) we see that the above theorem holds true for all n ∈ XA.

5 Fragility of the RH and certain supersets of XA numbers

In Theorem 7 we proved that under the RH the cardinality of the set of XA numbers is infinite. Here we present some interesting theorems which demonstrate the fragility of the RH showing the infinitude of some supersets of XA numbers independent of the RH. These sets are defined by inequalities quite close to that in (8). The basic inequalities used here to define these sets are (11) and (12).

Lemma 31. If m ≥ 3, then there exists n > m such that

\frac{σ(n)/n}{σ(m)/m} > 1 + \frac{log n/m}{log n log log m}.

Proof. Let m ≥ 3. Then by inequality (5)

\frac{σ(m)}{m} ≤ (e^{γ} + \frac{0.648214}{(log log m)^{2}} ) log log m.  (26)

Since for m′ > m

\frac{log log m}{log log m′} (1 + \frac{log m′/m}{log m′ log log m}) < 1

and the left-hand side is decreasing with respect to m′ and tends to zero as m′ → ∞, then for some m′ > m we have

\frac{log log m}{log log m′} (1 + \frac{log m′/m}{log m′ log log m}) (e^{γ} + \frac{0.648214}{(log log m)^{2}} ) = e^{γ} − ε,  (27)

where ε > 0 is arbitrarily small and fixed. Hence by Gronwall’s theorem there exists n ≥ m′ such that

\frac{σ(n)}{n} > (e^{γ} − ε) log log n

= \frac{log log m}{log log m′} (1 + \frac{log m′/m}{log m′ log log m}) (e^{γ} + \frac{0.648214}{(log log m)^{2}} ) log log n

≥ (1 + \frac{log n/m}{log n log log m}) \frac{σ(m)}{m},

where the last inequality holds by (26) and (27).

Definition 32. Let n_{1} = 10080, and let n_{k+1} be the least integer greater than n_{k} such that

\frac{σ(n_{k+1})/n_{k+1}}{σ(n_{k})/n_{k}} > 1 + \frac{log n_{k+1}/n_{k}}{log n_{k+1} log log n_{k}} , (k = 1, 2, …).

We define X′ to be the set of all n_{1}, n_{2}, n_{3}, ….

One can easily show that

XA ⊂ X′ ⊂ SA.  (28)

Our first result towards the fragility of the RH is the following theorem. Theorem 33. The set X′ has an infinite number of elements. Proof. If the RH is true, then the cardinality of X′ is infinite by (28). If RH is not true, then by Theorem 7 there exists m_{0} ≥ 10080 such that

\frac{σ(m_{0})/m_{0}}{σ(m)/m} ≥ \frac{log log m_{0}}{log log m} , for all m ≥ 10080.

By Lemma 31 there exists m′ > m_{0} such that m′ satisfies

\frac{σ(m′)/m′}{σ(m_{0})/m_{0}} > 1 + \frac{log m′/m_{0}}{log m′ log log m_{0}} .

Let n be the least number greater than m_{0} for which

\frac{σ(n)/n}{σ(m_{0})/m_{0}} > 1 + \frac{log n/m_{0}}{log n log log m_{0}} .

Hence n ∈ X′.

The following lemma can be proved in the same manner as Lemma 31. Lemma 34. If m ≥ 3, then there exists n > m such that

\frac{σ(n)/n}{σ(m)/m} > 1 + \frac{2 log n/m}{(log m + log n) log log m}.

We continue our approach towards the fragility of the RH via (the stronger) inequality (12) defining a smaller superset of XA numbers as follows. Definition 35. Let n_{1} = 10080, and let n_{k+1} be the least integer greater than n_{k}, such that

\frac{σ(n_{k+1})/n_{k+1}}{σ(n_{k})/n_{k}} > 1 + \frac{2 log n_{k+1}/n_{k}}{(log n_{k} + log n_{k+1}) log log n_{k}} , (k = 1, 2, …).

We define X′′ to be the set of all n_{1}, n_{2}, n_{3}, ….

By elementary inequality (12) and

\frac{t}{1 + t} < \frac{2t}{2 + t}, (t > 0)

one can easily show the inclusion XA ⊂ X′′ ⊂ X′. The following theorem is a refinement of Theorem 33 with a similar proof.

Theorem 36. The set X′′ has an infinite number of elements.

We calculated the number of elements in XA, X′ and X′′ up to the 300, 000th element of SA in [20]. Note that

#XA = 9240,  #X′ = 9535,  #X′′ = 9279

and

#(X′ − XA) = 295, #(X′′ − XA) = 39.

It might be interesting to look at the list of elements of X′′ − XA up to s_{300,000}:

X′′ − XA = {s_{55}, s_{62}, s_{91}, s_{106}, s_{116}, s_{127}, s_{128}, s_{137}, s_{138}, s_{149}, s_{181}, s_{196}, s_{212}, s_{219},
s_{224}, s_{231}, s_{232}, s_{246}, s_{247}, s_{259}, s_{260}, s_{263}, s_{272}, s_{273}, s_{276}, s_{288}, s_{294},
s_{299}, s_{305}, s_{311}, s_{317}, s_{330}, s_{340}, s_{341}, s_{343}, s_{354}, s_{65343}, s_{271143}, s_{271151}}

Note that the second XA number is s_{356} (see Table 1) and only three out of 39 elements in the set X′′ − XA up to s_{300,000}, namely s_{65343}, s_{271143} and s_{271151}, are greater than s_{356}.

6 Numerical experiments

We present here some numerical results, mainly for the set of XA numbers (sorted in increasing order) up to 13770th element, which is less than C_{1} := s_{500,000}, based on the list provided by Noe [20].

Property 37. Let n = 2^{k_{2}} ⋯ q^{k_{q}} ⋯ r^{k_{r}} ⋯ p ∈ XA, where 2 ≤ q < r ≤ p. Then for 10080 < n ≤ C_{1}

(i) log n < q^{k_{q}+1},

(ii) r^{k_{r}} < q^{k_{q}+1} < r^{k_{r}+2},

(iii) q^{k_{q}} < k_{q}p,

(iv) q^{k_{q}} log q < log n log log n < q^{k_{q}+2}.

Property 38. Let n = 2^{k_{2}} ⋯ x_{k}^{k} ⋯ p ∈ XA, where 2 < x_{k} < p is the greatest prime factor of exponent k. Then

\sqrt{p} < x_{2} < \sqrt{2p}, for 10080 < n ≤ C_{1}.

Property 39. Let n = 2^{k_{2}} ⋯ q^{k_{q}} ⋯ p and n′ = 2^{k′_{2}} ⋯ q^{k′_{q}} ⋯ p′ be two consecutive XA numbers. Then for 10080 < n < n′ ≤ C_{1}

|k_{q} − k′_{q}| ≤ 1, for all 2 ≤ q ≤ p′.

Property 40. If m, n are XA numbers, then for 10080 ≤ m < n ≤ C_{1} (i) p(m) ≤ p(n), (ii) d(m) ≤ d(n). Remark 41. We note that Property 40 is not true for SA numbers. For instance,

s_{47} = (19♯)(3♯)^{2}2, s_{48} = (17♯)(5♯)(3♯)2^{3},

p(s_{47}) = 19 > 17 = p(s_{48}).

and

s_{173} = (59♯)(7♯)(5♯)(3♯)^{2}2^{3}, s_{174} = (61♯)(7♯)(3♯)^{2}2^{2},
d(s_{173}) = 5308416 > 5160960 = d(s_{174}).

Property 42. If n, n′ ∈ XA are consecutive, then for 10080 ≤ n < n′ < C_{1}

\frac{n′}{n} > 1 + c \frac{(log log n)^{2}}{log n} , (0 < c ≤ 4),

\frac{n′}{n} > 1 + c \frac{(log log n)^{2}}{\sqrt{log n}} , (0 < c ≤ 0.195).

Property 43. If n, n′ ∈ XA are consecutive, then for 10080 ≤ n < n′ < C_{1}

\frac{f(n′)}{f(n)} < 1 + \frac{1}{p′} ,

where p′ is the largest prime factor of n′.

We have checked the following properties up to C_{2} = s_{250,000} and up to 8150th element of XA numbers which is less than C_{2}. Property 44. If n, n′ ∈ SA are consecutive, then

\frac{σ(n′)/n′}{σ(n)/n} < 1 + \frac{1}{p′} , (n′ < C_{2}),

where p′ is the largest prime factor of n′.

The number of distinct prime factors of a number n is denoted by ω(n) (see [25]). Property 45. If n, n′ ∈ XA are consecutive, then for 10080 ≤ n < n′ < C_{2}, then

\frac{n}{ω(n)} ≤ \frac{n′}{ω(n′)}.

The comparison of the sets CA and XA is given. We calculated them up to C = s_{1,000,000} from the list of SA in [20].

#{n ∈ XA : n < C} = 24875,
#{n ∈ CA : n < C} = 21187,
#{n ∈ CA ∩ XA : n < C} = 20468,
#{n ∈ XA \ CA : n < C} = 4407,
#{n ∈ CA \ XA : n < C} = 719,

We conclude this paper with another remark on choosing the first element of XA as 10080.

Remark 46. If we replace n_{1} = 10080, the initial number in the definition of XA numbers, by n_{1} = 665280, we do not need to pose the condition (i.e., n > n_{1}) in Lemma 21. Indeed, if we choose the initial number n_{1} = 665280, then N = 55440 < 665280 < 720720 = N′, where in this case N′ is also an XA number. Therefore we do not need Remark 22. Moreover, if we choose n_{1} = 665280, then there are 37 more XA numbers.

n  Type f(n) p(n) log n k_{2}
1 s_{20} = (7♯)(3♯)2^{3} = 10080 1.75581 7 9.21831 5
2 s_{356} = (113♯)(13♯)(5♯)(3♯)^{2}2^{3} c 1.75718 113 126.444 8
3 s_{368} = (127♯)(13♯)(5♯)(3♯)^{2}2^{3} c 1.75737 127 131.288 8
4 s_{380} = (131♯)(13♯)(5♯)(3♯)^{2}2^{3} c 1.75764 131 136.163 8
5 s_{394} = (137♯)(13♯)(5♯)(3♯)^{2}2^{3} c 1.75778 137 141.083 8
6 s_{408} = (139♯)(13♯)(5♯)(3♯)^{2}2^{3} c 1.75821 139 146.018 8
7 s_{409} = (139♯)(13♯)(5♯)(3♯)^{2}2^{4} c 1.75826 139 146.711 9
8 s_{438} = (151♯)(13♯)(5♯)(3♯)^{2}2^{3}  1.75831 151 156.039 8
9 s_{440} = (151♯)(13♯)(5♯)(3♯)^{2}2^{4} c 1.75849 151 156.732 9
10 s_{444} = (151♯)(13♯)(7♯)(3♯)^{2}2^{4} c 1.75860 151 158.678 9
11 s_{455} = (157♯)(13♯)(5♯)(3♯)^{2}2^{4}  1.75864 157 161.788 9
12 s_{458} = (157♯)(13♯)(7♯)(3♯)^{2}2^{3}  1.75866 157 163.041 8
13 s_{459} = (157♯)(13♯)(7♯)(3♯)^{2}2^{4} c 1.75892 157 163.734 9
14 s_{476} = (163♯)(13♯)(7♯)(3♯)^{2}2^{4} c 1.75914 163 168.828 9
15 s_{486} = (163♯)(17♯)(7♯)(3♯)^{2}2^{4}  1.75918 163 171.661 9
16 s_{493} = (167♯)(13♯)(7♯)(3♯)^{2}2^{4} c 1.75943 167 173.946 9
17 s_{502} = (167♯)(17♯)(7♯)(3♯)^{2}2^{4} c 1.75966 167 176.779 9
18 s_{519} = (173♯)(17♯)(7♯)(3♯)^{2}2^{4} c 1.76006 173 181.933 9
19 s_{537} = (179♯)(17♯)(7♯)(3♯)^{2}2^{4} c 1.76038 179 187.120 9
20 s_{555} = (181♯)(17♯)(7♯)(3♯)^{2}2^{4} c 1.76089 181 192.318 9

Table 1: First 20 extremely abundant numbers (p_{k}♯ := ∏_{j=1}^{k} p_{j} and c represents a colossally abundant number).

7 Acknowledgments

We thank J. C. Lagarias, C. Calderon, J. Stopple, and M. Wolf for useful discussion and sending us some relevant references. Our sincere thanks to J.-L. Nicolas for careful reading of the manuscript, helpful comments, and worthwhile suggestions which improved the presentation of the paper. The work of the first author is supported by the Calouste Gulbenkian Foundation, under Ph.D. grant number CB/C02/2009/32. Research partially funded by the European Regional Development Fund through the programme COMPETE and by the Portuguese Government through the FCT under the project PEst-C/MAT/UI0144/2011.

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2010 Mathematics Subject Classification: Primary 11A25; Secondary 11N37, 11Y70, 11K31. Keywords: extremely abundant number, superabundant number, colossally abundant number, Robin’s theorem, Chebyshev’s function.

(Concerned with sequences \underline{A004394}, \underline{A004490}, and \underline{A217867}.) Received April 1 2013; revised versions received August 16 2013; January 15 2014. Published in Journal of Integer Sequences, February 7 2014. Return to Journal of Integer Sequences home page.