Explicit Estimates Involving the Primorial Integers and Applications

Safia Aoudjit and Djamel Berkane University of Blida1 Department of Mathematics P. O. Box 270, Route de Soumaa Blida Algeria

                safiaoudjit9@gmail.com
                djaber72@univ-blida.dz

Pierre Dusart XLIM UMR 7252 Faculté des sciences et techniques Université de Limoges P. O. Box 123, avenue Albert Thomas 87060 Limoges Cedex

             France pierre.dusart@unilim.fr

Abstract

In this paper, we propose several explicit bounds for the function that counts the number of primorial integers less than or equal to a given positive real number. As applications, we obtain an effective version of Pósa’s inequality, and a method to estimate the maximal value of the sum over prime divisors ∑_{p|q} f(p) for a positive decreasing function f, when q ranges over all integers less than x. In particular, we improve the upper bound for the maximal value of ∑_{p|q} \frac{log p}{p−1}.

1 Introduction and statement of results

As usual, let (p_{k})_{k≥1} denote the increasing sequence of prime numbers, and let N_{k} be the primorial (prime-factorial) integer of index k, the product of its k first terms. The integers N_{k} are the terms of the sequence \underline{A002110} in the On-line Encyclopedia of Integer Sequences (OEIS) [?], and play an important role in number theory from Euclid’s proof of the infinity of primes to the remarkable equivalences of the Riemann hypothesis due to Nicolas [?] and Robin [?].

The approximations linked to N_{k} have found some unexpected applications in various areas of number theory; see the papers of Betts [?], Planat et al. [?] and Zhang [?]. However, the most frequently used results concern their logarithm θ(p_{k}) where θ denotes the Chebyshev function, here we cite the recent papers of Axler [?] and Dusart [?].

The principal object of this manuscript is to study the function K(x) which counts the number of primorial integers less than or equal to x; for instance K(1) = 0, K(2) = 1, K(3) = 1, K(5) = 1, K(6) = 2,… and a table of some large values of x is given in Section ??.

The function K(x) is very close to π(x) (the usual prime counting function) since it will be shown in what follows that K(x) ≈ π(log x). It often appears implicitly as an important key in the demonstration of several results; see, for instance, Balazard [?] and Hassani [?].

We let log_{i} denote the i-fold iterated logarithm. The asymptotic expansion of K(x) is easy to evaluate. Let us show this using less precise results than those in Balazard [?]:

Theorem 1 (Balazard). For every real number x ≥ 2 and every integer m ≥ 0, we have

K(x) = \frac{log x}{log_{2} x} (∑_{j=0}^{m} \frac{j!}{log_{2}^{j} x} + O (\frac{1}{log_{2}^{m+1} x})) .

Proof. Since for a real x ≥ 2, the integer K := K(x) is also given by the inequalities

θ(p_{K}) ≤ log x < θ(p_{K+1}),

using the prime number theorem in the forms

θ(t) = t + O_{m} (\frac{t}{log^{m+2} t})

and

π(t) = li(t) + O_{m} (\frac{t}{log^{m+2} t}) ,

where li(x) indicates the usual logarithmic integral function, one gets successively

p_{K} = log x + O_{m} (\frac{log x}{log_{2}^{m+2} x}) ,
K(x) = π(p_{K}) = li(log x) + O_{m} (\frac{log x}{log_{2}^{m+2} x}) .

Finally, by substituting log x in the classic asymptotic expansion of li(x) below

li(x) = ∫_{2}^{x} \frac{dt}{log t} = \frac{x}{log x} (0! + \frac{1!}{log x} + ⋯ + \frac{m!}{log^{m} x} + O (\frac{1}{log^{m+1} x})) ,

we establish the result.

Consequently, the asymptotic expansion of the function xK(x) corresponds to a type of formula among a large class of expansions related to the iterated logarithms studied in Belbachir and Berkane [?] to which an asymptotic expansion for the sum of inverses is determined. The following proposition is a straightforward application of Belbachir and Berkane [?, Theorem 1]:

Proposition 2. For every real number x ≥ 2 and every integer m ≥ 2, we have

∑_{2≤n≤x} \frac{1}{nK(n)} = \frac{1}{2} log_{2}^{2} x − log_{2} x − log_{3} x + C

+ \frac{δ_{2}}{log_{2} x} + ⋯ + \frac{δ_{m}}{(m − 1) log_{2}^{m−1} x} + O (\frac{1}{log_{2}^{m} x}) ,

where C is an absolute constant, and \{δ_{j}\}_{j≥0} is the sequence \underline{A233824} in the OEIS [?] given by the recurrence relation δ_{n} + 1!δ_{n−1} + 2!δ_{n−2} + ⋯ + (n − 1)!δ_{1} = n · n!.

In this article, by comparing K(x) with π(log x) over large and special ranges, our first result (see Theorem 10) is an effective version of an inequality of Pósa [?], where he proved that for all n > 1 there is a k_{n} such that N_{k} > p_{k+1}^{n} for all n ≥ k_{n}. We describe an algorithm calculate k_{n} for every n.

The second result (see Theorem 19) is a good improvement of an approximation of the greatest sum over prime divisors 𝔏(q) = ∑_{p|q} \frac{log p}{p−1} when q ranges over all integers not exceeding x. The method employed to prove this theorem can also be applied to evaluate the maximum of sum of type 𝔏_{f}(q) = ∑_{p|q} f(p) for any positive decreasing function f on (1, ∞). In Section 5, we show that the maximal value of 𝔏_{f}(q) is 𝔏_{f}(N_{K(x)}).

These results are follow-ups of fully explicit bounds for the O-term in the formula of K(x), mainly Theorems 14–18.

Finally, throughout this paper, e represents Napier’s constant, p a prime number and the calculations are made by using Maple 17.

2 Technical lemmas

The sequence log p_{k} has the same asymptotic behavior as its Cesàro average \frac{log N_{k}}{k}, where the first few terms are

\frac{log N_{k}}{k} = log k + log_{2} k − 1 + \frac{log_{2} k − 2}{log k} − \frac{log_{2}^{2} k − 6 log_{2} k + 11}{2 log^{2} k} + O (\frac{log_{2}^{3} k}{log^{3} k} ) .

Robin [?] and Dusart [?] obtained fully explicit bounds for p_{k} and θ(p_{k}), we gathered the ones we need in the following lemma.

Lemma 3.

θ(p_{k}) > k log k, ∀k ≥ 3,  (1)
θ(p_{k}) ≥ k(log k + log_{2} k − a), ∀k ≥ 2, and a = 1.0769,  (2)
θ(p_{k}) ≥ k(log k + log_{2} k − 1 + \frac{log_{2} k − 2.1454}{log k}), ∀k ≥ 3,  (3)
θ(p_{k}) ≤ k(log k + log_{2} k − 0.9465), ∀k ≥ 14,  (4)
p_{k} ≤ k(log k + log_{2} k − \frac{1}{2}), ∀k ≥ 20,  (5)
p_{k} ≤ k log p_{k}, ∀k ≥ 4.  (6)

Similarly, the following lemma directly gives an upper bound of θ(p_{k+1}) in terms of k instead of k + 1. Note that c = 1 − log 2 is an absolute constant.

Lemma 4. We have, when k ≥ 2:

θ(p_{k+1}) ≤ k (log k + log_{2} k − c + \frac{log_{2} k + c}{log k}) .  (7)

Proof. From inequality (5), by taking the logarithm and using the fact that log(1 + x) ≤ x for all x > 0, we easily obtain

log p_{k} ≤ log k + log_{2} k + \frac{log_{2} k − 0.5}{log k};  (8)

this is valid even for k ≥ 18. In particular, replacing k by 2m − 1 and using the inequality log(2m − 1) < log(m) + log(2), one gets that for all m ≥ 2:

log p_{2m−1} ≤ log m + log_{2} m + log 2 + \frac{log_{2} m + log 2 − 0.5}{log m} + \frac{log 2}{log^{2} m}.

Since the right-hand side is a strictly increasing function M(m), the sum of all log p_{2m−1} until k − 1 is bounded above by ∫_{2}^{k} M(t)dt, and we have

∑_{m=1}^{k−1} log p_{2m−1} ≤ k (log k + log_{2} k − c + \frac{log_{2} k + c}{log k}) − D(k),

where

D(k) = (2 log 2 − \frac{3}{2}) li(k) + (\frac{3}{2} − 2 log 2) li(2) − 3 log 2 − 2 log_{2} 2 + 4 − 2\frac{log_{2} 2}{log 2} − \frac{2}{log 2}.

Finally, since the function D(x) is negative when x ≥ 11, by applying the logarithm to the inequality obtained in the lemma below, one gets the result for k ≥ 11. A computer check handles the cases 2 ≤ k ≤ 11.

Lemma 5. We have, when k ≥ 11:

N_{k+1} < ∏_{i=1}^{k−1} p_{2i−1}.

Proof. We proceed by induction on k. The inequality holds for k = 11. As k ≥ 3 implies 2k − 1 ≥ k + 2 and then p_{k+2} ≤ p_{2k−1}, the lemma follows from the fact that

N_{k+2} < p_{k+2} ∏_{i=1}^{k} p_{2i−1} ≤ ∏_{i=1}^{k+1} p_{2i−1}.

Lemma 6. For all k ≥ 1, we have

p_{k+1} ≤ \frac{5}{3}p_{k},

and the inequality is sharp at k = 2.

Proof. According to Dusart [?], the interval [x, \frac{1}{25 log^{2}(x)}] contains at least one prime for all x ≥ 396738. As 396833 is prime, we have that, for p_{k} ≥ 396833 = p_{33609},

p_{k+1} ≤ (1 + \frac{1}{25 log^{2}(p_{k})})p_{k} < \frac{4001}{4000}p_{k} < \frac{5}{3}p_{k}.

Finally, by computer, the last inequality is shown to be also valid for 2 ≤ p_{k} ≤ 396832.

Lemma 7. For every positive integer m, we have

∏_{k=1}^{m} N_{k} ≤ N_{\frac{m(m+1)}{2}} .

Proof. By setting s := ∑_{k=1}^{m} k, we want to show that ∏_{k=1}^{m} N_{k} is less than the primorial of index s. This is equivalent to

∑_{k=1}^{m} θ(p_{k}) ≤ θ(p_{s}).

According to the inequality(2), we have

θ(p_{s}) ≥ s(log s + log_{2} s − a)

≥ s (log m + log (\frac{m + 1}{2}) + log_{2} m − a) ,

which implies for m ≥ 14 that

θ(p_{s}) ≥ s (log m + log_{2} m + log \frac{15}{2} − a) .  (9)

On the other hand, according to the inequality (4), it follows that for m ≥ 14:

∑_{k=1}^{m} θ(p_{k}) ≤ s(log m + log_{2} m − 0.9465) + ∑_{k=1}^{13} θ(p_{k}).  (10)

However, the left-hand side of the subtraction of the inequality (10) from (9) is an increasing function on m, which is already positive for m ≥ 14. So, our inequality holds when m ≥ 14 and, we checked m ≤ 13 by computer to end the proof.

3 Effective version of Pósa’s inequality

Let us start by giving explicit formulations of the equivalence K(x) ≈ π (log x).

Lemma 8. For all ε > 0, there is a real number x_{0} such that

π (\frac{log x}{1 + ε}) ≤ K(x)

for all x ≥ x_{0}. In particular, ∀x ≥ 1 we have π (\frac{log x}{1.00000075}) ≤ K(x).

Proof. First, a form of the prime number theorem asserts that there exists a decreasing sequence to zero of positive real numbers (δ_{n})_{n≥0} and a sequence (u_{n})_{n≥0} such that |θ(x)−x| < δ_{n}x for all x ≥ u_{n}. Hence, for all ε > 0 there exists n_{0}(ε) such that |θ(x) − x| < εx for all x ≥ u_{n_{0}}. Then, for a given ε, θ(x) < (1 + ε)x we have

K(x) = max {k ∈ ℕ^{∗}, N_{k} ≤ x} = max {k ∈ ℕ^{∗}, θ(p_{k}) ≤ log x}
≥ max {k ∈ ℕ^{∗}, p_{k} ≥ u_{n_{0}}, (1 + ε)p_{k} ≤ log x}
≥ max {k ∈ ℕ^{∗}, (1 + ε)p_{k} ≤ log x} .

However, the maximum in the last inequality is just π(log x/(1 + ε)).

In particular, according to Platt and Trudgian [?] we have θ(x) < 1.00000075x for all x > 0. Then, for all x ≥ 8 and further by computer for x ≥ 1 we deduce that

K(x) ≥ π (\frac{log x}{1.00000075}) .

Lemma 9. For all ε > 0, there is a real number x_{0} such that

∀x ≥ x_{0}, K(x) ≤ π (\frac{log x}{1 − ε}) .

In particular, for x ≥ x_{0} we have

K(x) ≤ π (\frac{log x}{log α}) ,

for the values in Table 1.

\begin{matrix}α & 1.7 & 2 & 2.1 & 2.2 & 2.5 \\ x_{0} & 3 & 32 & 3503 & N_{11} & N_{50} \end{matrix}

Table 1: Some values of x_{0}.

Proof. Similarly to the first part of the above proof, for all ε > 0 there exists n_{0}(ε) such that |θ(x) − x| < εx for all x ≥ u_{n_{0}}. Thus, for a given ε, θ(x) < (1 − ε)x implies readily that

K(x) ≤ max {k ∈ ℕ^{∗}, p_{k} ≥ u_{n_{0}}, (1 − ε)p_{k} ≤ log x}
≤ max {k ∈ ℕ^{∗}, (1 − ε)p_{k} ≤ log x} .

However, the maximum in the last inequality is just π(log x/(1 − ε)).

For the effective upper bounds, the idea is to look at the best practical values of α > 1 and k_{0}(α) such as N_{k_{0}} > α^{p_{k_{0}}}, since this would result in K(x) is not being greater than π(log x/ log α) for all x ≥ N_{k_{0}}. Using inequality (3) combined with inequality (8), the problem is reduced to finding α > 1 and k_{0} such that

(log k + log_{2} k + \frac{log_{2} k}{log k} ) (1 − log α) ≥ 1 + \frac{2.1454}{log k} − \frac{0.5 log α}{log k}, ∀k ≥ k_{0}.

Starting from a certain k_{0}(α), the function on the right is greater than the one on the left as long as 1 < α < e. After checking the small values, we obtain the results in Table 2.

\begin{matrix}α & 1.7 & 2 & 2.1 & 2.2 & 2.5 \\ k_{0} & 2 & 4 & 6 & 11 & 50 \end{matrix}

Table 2: Some values of k_{0}.

The values of x_{0} given in Table 1 are the possible minimums except for α = 2.2 and α = 2.5, where the calculations showed that K(x) is smaller than π(log x/ log α) for x = N_{10} and x = N_{49}.

Through an elementary method, this last proposition permits us to get an effective version of a result of Pósa [?] which proved that for all n > 1 there is a k_{n} such that N_{k} > p_{k+1}^{n} for all n ≥ k_{n}.

Theorem 10. For all integer n ≥ 1, there exists x_{0}(n) ≥ 1 such that for all x ≥ x_{0}(n),

(∀k ∈ ℕ^{∗}, K(x) < k < π(\sqrt[n]{x})) ⇒ N_{k} > p_{k+1}^{n}.

Furthermore, if k verifies p_{k+1} ≥ (\frac{5}{3})^{n}:

∀m ≥ k, N_{m} > p_{m+1}^{n}.

Proof. For n ≥ 1 and x_{0}(n) = \frac{n^{2n}}{(log 1.7)^{2n}}, we can show easily that \frac{log x}{log 1.7} ≤ \sqrt[n]{x} as long as x ≥ x_{0}(n). Thus, we obtain that N_{k} > x ≥ p_{k+1}^{n} for all k satisfying the inequalities

K(x) < k < π(\sqrt[n]{x}), x ≥ x_{0}(n).

Furthermore, if k fulfils previous inequalities together with p_{k+1} ≥ (5/3)^{n}, it implies by induction on k that N_{m} > p_{m+1}^{n} for all m > k. Indeed, according to Lemma 6, we have

p_{m+2}^{n} ≤ (5/3)^{n}p_{m+1}^{n} < (5/3)^{n}N_{m} ≤ p_{m+1}N_{m} = N_{m+1}.

In practice, using the Maple algorithm below, the calculations of the values of k_{n} listed in Table ??, show that the first choice x_{0}(n) for n ≤ 30 is sufficient.

// Algorithm1 Computation of $K(x)$.
restart; with(numtheory);
K := proc (L) local s, k;
s := 2; for k from 2 do
if s <= L then s := s*ithprime(k)
else return k-2 end
if end do end proc
// Algorithm2 Computation of $k_n$.
posa := proc (n) local x0, x, k, R, m, s, i, t;
x0 := floor((n/ln(1.7))^(4*n)); R := (5/3)^n;
for x from x0 to x0+5 do
for k from K(x)+1 to pi(floor(x^(1/n)))-1 do
if R <= ithprime(k+1) then m := k end if end do
end do; for s to m do
if ithprime(s+1)^n < product(ithprime(i), i = 1 .. s) then t := s;
return t end if
end do end proc

We thus regain the results of Euclid: N_{k} > p_{k+1} for k ≥ 1 and those of Bonse [?]: N_{k} > p_{k+1}^{2} for k ≥ 4 and N_{k} > p_{k+1}^{3} for k ≥ 5. The sequence (k_{n})_{n≥1} is sequence \underline{A056127} in the OEIS [?].

Pósa’s inequality has been studied extensively. The reader can find similar forms in in papers of Hassani [?] and Sándor [?].

Let us end this section by giving some special approximations for limited ranges.

Theorem 11. (i) For 1 < x ≤ exp(10^{19}), we have π(log x) ≤ K(x).

(‘ii) For 2310 ≤ x ≤ exp(10^{19}), we have \frac{log x}{log_{2} x}(1 + \frac{1}{4 log_{2} x}) ≤ K(x).

(iii) For 210 ≤ x ≤ exp(10^{19}), we have \frac{log x}{log_{2} x} ≤ K(x).

Proof. Let 0 < log x ≤ 10^{19}. According to Büthe [?], the primes p_{k} smaller than log x verify θ(p_{k}) < p_{k}. So,

π(log x) = max {k ∈ ℕ^{∗}, p_{k} ≤ log x}
≤ max {k ∈ ℕ^{∗}, θ(p_{k}) ≤ log x} = K(x).

After a direct computation of the small values, we derive the last two lower bounds from the first result together with the lower bounds for π(x), namely (see Dusart [?]):

\frac{x}{log x} ≤ π(x), for x ≥ 17 and \frac{x}{log x} (1 + \frac{1}{log x}) ≤ π(x), for x ≥ 599.

4 Explicit bounds for the primorial counting function

First, we give some upper bounds for K(x) by using a classical method. We will use appropriate bounds for θ(p_{k}) (essentially formulas (1) and (2)) while taking advantage of the fact that K(x) is constant over the intervals [N_{k}, N_{k+1}). Thankfully, with the following lemma, we rediscover the estimates given by Robin [?] of the large values of ω(n). Here ω(n) denotes the number of prime distinct divisors of n.

Lemma 12. For all real numbers x ≥ 1, we have K(x) = max_{1≤n≤x} ω(n). Furthermore, if

K(x) = K, then for all integers n ≤ x with ω(n) = K, we have n ≥ N_{K}.

In other words, N_{K(x)} is the smallest integer less than x whose decomposition into prime numbers is the longest. Thus, (K(n))_{n≥1} is the sequence \underline{A111972} in the OEIS [?].

Proof. As N_{k} ≤ n ≤ x < N_{k+1} means that ω(n) ≤ k and K(n) = k, one sees that ω(n) ≤ K(n) in every interval [N_{k}, N_{k+1}), which implies that

max_{1≤n≤x} ω(n) = max_{1≤n<N_{K+1}} ω(n) = K.

Let q_{1}q_{2} ⋯ q_{K} be an integer less than x such that q_{1} < q_{2} < … < q_{K} are prime numbers. For K = 1 it is obvious that q_{1} ≥ p_{1}. Now, assuming q_{i} ≥ p_{i} for i < K, it is necessary that q_{K} ≥ p_{K} otherwise q_{K} < q_{K−1}.

For to prove our first upper bound, we need the following lemma.

Lemma 13. For a large real number A > 0, if κ(A) is a root of the equation t log t = A, then

κ(A) = \frac{A}{log A}(1 + o(1)),

furthermore, for A > e:

\frac{A}{log A} < κ(A) ≤ (1 + \frac{1}{e})\frac{A}{log A}.

Proof. See Olver [?, Theorem 5. 1, Ex. 5. 7].

Theorem 14. We have, when x ≥ N_{13}:

K(x) ≤ (1 + \frac{1}{e}) \frac{log x}{log_{2} x},

and

K(x) ≤ 1.3841\frac{log x}{log_{2} x}, for x ≥ 3.

Proof. From inequality (1), one easily deduces that, for x ≥ N_{13}:

K(x) ≤ Max {k ∈ ℕ^{∗}, k log k ≤ log x} ,

however this last set is only a part of the set of roots of inequality t log t ≤ log x. So, according to Lemma 13, one gets

K(x) ≤ κ(log x) ≤ (1 + \frac{1}{e}) \frac{log x}{log_{2} x}, ∀x ≥ N_{13}.

Now, as the function F_{0}(x) = \frac{K(x) log_{2} x}{log x} is decreasing over every interval [N_{k}, N_{k+1}) once k ≥ 3, then, a computer verification can be done only on N_{k} where 3 ≤ k ≤ 12. This verification shows that the maximum is reached on N_{9} and it is also true for all real numbers x < N_{3} with F_{0}(N_{9}) ≤ 1.3841, which concludes the proof.

Theorem 15. We have, when x ≥ 3

K(x) ≤ \frac{log x}{log_{2} x}(1 + \frac{1.4575}{log_{2} x}).

Proof. For x ≥ 2, we consider the function

F_{1}(x) = \frac{K(x)(log_{2} x)^{2}}{log x} − log_{2} x.

The function F_{1} is decreasing over every interval [N_{k}, N_{k+1}) as long as k ≥ 5 since − log_{2} x is decreasing and the function \frac{K(x)(log_{2} x)^{2}}{log x} decreases when x ≥ 1619. So, F_{1} reaches its maximum at an integer N_{k_{0}} where k_{0} ≥ 5. On the other hand, for k ≥ 6, using inequality (2), invoking the decrease of \frac{(log(x))^{2}}{x} (valid for x ≥ 8) together with the fact that log(1 + x) < x and \frac{1}{1+x} ≤ 1, we obtain after a long expansion that

F_{1}(N_{k}) = \frac{k(log θ(p_{k}))^{2}}{θ(k)} − log θ(p_{k})) ≤ G(k),

with

G(k) = a + \frac{log_{2} k − a}{log k} + \frac{1}{log k} (\frac{log_{2} k − a}{log k})^{2} + \frac{2a}{log k} (\frac{log_{2} k − a}{log k}) + \frac{a^{2}}{log k} .

The function G is decreasing and smaller than 1.3832444 for k ≥ exp(exp(a + 1)) ≃ 2922. Then, F_{1}(N_{k}) ≤ 1.3833 for k ≥ 2922.

Finally, for 5 ≤ k ≤ 2921, we conclude by computer verification over intervals [N_{k}, N_{k+1}) that the maximum of F_{1}(x) is reached at N_{47} with F_{1}(N_{47}) ≤ 1.4575 and our upper bound is valid for x < N_{5} as well.

With the techniques used above, the inequality in the following theorem requires a wider estimate of θ(p_{k}). Nevertheless, for this last round, we will use a strong relation between π(x) and θ(x) were given by Robin [?], namely

∀x ≥ 2, \frac{π(x) log θ(x)}{θ(x)} ≤ 1 + \frac{1}{log θ(x)} + \frac{2.89726}{log^{2} θ(x)}.  (11)

Theorem 16. We have, when x ≥ 3

K(x) ≤ \frac{log x}{log_{2} x} (1 + \frac{1}{log_{2} x} + \frac{2.89726}{log_{2}^{2} x}) .

Proof. Similarly, by studying F_{2}(N_{k}) which corresponds to the function

F_{2}(x) = \frac{K(x)(log_{2} x)^{3}}{log x} − (log_{2} x)^{2} − log_{2} x, when k ≥ 10.

Since π(p_{k}) = K(N_{k}) = k, formula (11) guarantees that F_{2}(N_{k}) ≤ 2.89726, ∀k ≥ 10. Hence, we must now verify that our inequality is valid for x < N_{10}.

For x ≥ 863, let M be the strictly increasing function that appears on the right side of our upper bound. A verification by computer over the intervals [N_{k}, N_{k+1}), with 5 ≤ k ≤ 9, shows that M(x) is always greater than K(x), by a difference of at least 1.5. The calculations also show that the upper bound is also true for x < N_{5}, which concludes the proof.

Theorem 17. If the Riemann hypothesis holds, we have

K(x) ≤ li^{−1}(log x) + 0.12\sqrt{log x}, ∀x ≥ 42,  (12)

where li^{−1} is the inverse of the logarithmic integral function.

Proof. According to Robin [?], for k ≥ 5

θ(p_{k}) > li^{−1}(k) − 0.12\sqrt{k log^{3} k},

and as li(x) is an increasing function, one gets successively

k ≤ li(θ(p_{k})) + ∫_{θ(p_{k})}^{θ(p_{k})+0.12\sqrt{k log^{3} k}} \frac{dx}{log x}
≤ li(θ(p_{k})) + 0.12\frac{\sqrt{k log^{3} k}}{log θ(p_{k})} .

Now, applying inequality (1), one gets

k ≤ li(θ(p_{k})) + 0.12\frac{\sqrt{θ(p_{k})} log k}{log θ(p_{k})} ≤ li(θ(p_{k})) + 0.12\sqrt{θ(p_{k})}, ∀k ≥ 5,

which is equivalent to

K(N_{k}) ≤ li(log N_{k}) + 0.12\sqrt{log N_{k}}, ∀k ≥ 5.

Simply, the term on the right side is an increasing function of k, which yields that

K(x) ≤ li^{−1}(log x) + 0.12\sqrt{log x}, ∀x ≥ N_{5}.

Finally, with computer verifications, we extend the result to x ≥ 42. Theorem 18. We have, when x ≥ 2310

K(x) ≥ \frac{log x}{log_{2} x} (1 + \frac{1}{4 log_{2} x}) .

Moreover, for x ≥ 210

K(x) ≥ \frac{log x}{log_{2} x}.

Proof. Let κ(x) = \frac{log x}{log_{2} x}(1 + \frac{1}{4 log_{2} x}) and f_{a}(x) = k(log k + log_{2} k − a). As the function \frac{log_{2} x−c}{log x} is decreasing for x ≥ 8, inequality (7) gives that

θ(p_{k+1}) ≤ f(k) = k(log k + log_{2} k − c + 0.006389), when x ≥ 10^{500}.

Hence, for x ≥ 10^{500}, the function K(x) is greater than the minimum of the set of positive integer solutions of the inequality f_{a}(k) > log x, where a = c − 0.006389 ≤ 0.30046382.

Now, let us show that the function κ(x) = \frac{log x}{log_{2} x}(1+ \frac{1}{4 log_{2} x}) is not in the above set. Indeed, for x ≥ 10^{500}

log κ(x) ≤ log_{2} x − log_{3} x + \frac{0.25}{log_{2} x}, log_{2} κ(x) ≤ log_{3} x,
log κ(x) + log_{2} κ(x) ≤ log_{2} x + \frac{0.25}{log_{2} x} ≤ log_{2} x + \frac{0.25}{log_{2} x},

so, we obtain the following

f_{a}(κ(x)) ≤ log x(1 + \frac{1}{4 log_{2} x})(1 + \frac{1}{4 log_{2}^{2} x} − \frac{a}{log_{2} x}) < log x.

Now suppose that f(k) > log x. Then x → ∞ implies k → ∞, since f is bounded on every bounded subset of the set of all positive integers. Therefore, for x ≥ 10^{500}, we have f(k) ≤ f_{a}(k) and, since f_{a}(k) is increasing in k, we also have f_{a}(k) ≤ f_{a}(κ(x)) if k < κ(x), so that

f(k) ≤ f_{a}(k) ≤ f_{a}(κ(x)) < log x,

which means that, if x ≥ 10^{500} and f(k) ≥ log x, then k ≥ κ(x) and so K(x) ≥ κ(x).

For the values of x < 10^{500}, Theorem 11 is used as 10^{500} < exp 10^{19}. The last lower bound is partly due to the fact that \frac{log x}{4 log_{2}^{2} x} ≥ 1 when x ≥ 1.7·10^{30} and again to Theorem 11-(iii).

5 Maximal value of sums over prime divisors

Sums over primes and their evaluations are one of the main subjects of multiplicative number theory. Through a special case, we are concerned in this section to give an idea allowing to estimate the maximal value of sum of type 𝔏_{f}(q) = ∑_{p|q} f(p) for a positive decreasing function f on (1, ∞), when q ranges over all integers not exceeding x.

This type of sum appears without estimates in several papers, as in the work of Gordon and Rogers [?] where they derive refined forms of the sums of the divisor function; and in the work of Lehmer [?] where he studied a generalization of the Euler constant.

Our idea can be summarized as follows. As f is a positive decreasing function, then max_{1<q≤x} 𝔏_{f}(q) is reached on the smallest integer q(x) less than x among those having the

longest decomposition in prime numbers. However, according to Lemma 12, we can clearly specify that q(x) is nothing other than N_{K(x)}.

Finally, for a given real x ≥ 1, if K(x) = K, we deduce for all positive decreasing function f on (1, ∞) the following

max_{1<q≤x} 𝔏_{f}(q) = 𝔏_{f}(N_{K}) = ∑_{p≤p_{K}} f(p).

In particular, there is an approximation of the maximal value of 𝔏(q) = ∑_{p|q} \frac{log p}{p−1}. For a constant C, Hassani [?] proved the following result

max_{1<q≤x} 𝔏(q) ≤ log_{2} x + C and 𝔏(q) ≥ \frac{log q}{q − 1} for q ≥ 2.

We propose the following improvements Theorem 19. We have, when x ≥ N_{7}

max_{1<q≤x} 𝔏(q) ≤ log K(x) + log_{2} K(x).

Proof. As in Hassani [?], we start the estimation of the maximum as follows

max_{1<q≤x} 𝔏(q) = ∑_{p≤p_{K}} \frac{log p}{p} + ∑_{p≤p_{K}} \frac{log p}{p(p − 1)}

= ∑_{p≤p_{K}} \frac{log p}{p} + ∑_{p>1} \frac{log p}{p(p − 1)} − ∑_{p>p_{K}} \frac{log p}{p(p − 1)}.

However, Rosser and Shoenfeld [?] proved the following:

∀t ≥ 32, ∑_{p≤t} \frac{log p}{p} ≤ log t + E + \frac{1}{log t} and ∑_{p>1} \frac{log p}{p(p − 1)} + E ≈ −0.58.

So, recalling that

p_{k} ≤ k(log k + log_{2} k),

once k ≥ 6, we have for K ≥ 331 (p_{K} ≥ 32 and log_{2} K/ log K < 0.3) then after checking by hand for K ≥ 7 that

∑_{p≤p_{K}} \frac{log p}{p} + ∑_{p>1} \frac{log p}{p(p − 1)} ≤ log K + log_{2} K + \frac{log_{2} K}{log K} + \frac{1}{log p_{12}} − 0.58
≤ log K + log_{2} K.

Hence, for x ≥ N_{7} we reach

max_{1<q≤x} 𝔏(q) ≤ log K(x) + log_{2} K(x).  (13)

Finally, an computer verifications show that inequality 13 is not true for integers smaller than N_{7}. Corollary 20. We have, when x ≥ 3

max_{1<q≤x} 𝔏(q) ≤ log_{2} x + \frac{1.4575}{log_{2} x}.