1 Introduction

A numerical semigroup is an additively closed subset S of \({\mathbb {N}}\) with \(0\in S\) and only finitely many positive integers outside from S, the so-called gaps of S. The genus g of S is the number of its gaps. The set \(E=S^*\setminus (S^*+S^*)\), where \(S^*=S\setminus \{0\}\), is the (unique) minimal system of generators of S. Its elements are called the atoms of S; their number e is the embedding dimension of S. The multiplicity of S is the smallest element p of \(S^*\).

From now on we assume that \(S\ne {\mathbb {N}}\). Then the greatest gap f is the Frobenius number of S. Since \((f+1)+{\mathbb {N}}\subseteq S^*\) we have \((p+f+1)+{\mathbb {N}}\subseteq p+S^*\), hence the atoms of S are contained in the interval \([p,p+f]\).

For our investigation of certain numerical semigroups S generated by prime numbers, the fractions

$$\begin{aligned} \frac{f}{p},\frac{1+f}{p},\frac{g}{1+f}\,{\text {and}}\,\frac{e-1}{e} \end{aligned}$$

will play a role. For general S, what is known about these fractions?

First of all it is well known and easily seen that

$$\begin{aligned} \frac{1}{2}\le \frac{g}{1+f}\le \frac{p-1}{p}, \end{aligned}$$

and both bounds for \(\frac{g}{1+f}\) are attained.

However, the following is still open:

Wilf’s question [17]: Is it (even) true that

$$\begin{aligned} \frac{g}{1+f}\le \frac{e-1}{e} \end{aligned}$$
(1)

for every numerical semigroup?

A partial answer is given by the following result of Eliahou:

[4, Corollary 6.5] If \(\frac{1+f}{p}\le 3\), then \(\frac{g}{1+f}\le \frac{e-1}{e}\).

In [18], Zhai has shown that \(\frac{1+f}{p}\le 3\) holds for almost all numerical semigroups of genus g (as g goes to infinity).

Therefore, for randomly chosen S, one has \(\frac{g}{1+f}\le \frac{e-1}{e}\) almost surely.

We shall consider the following semigroups: Let \(p_1=2\), \(p_2=3\), \(p_3=5, \ldots\) be the sequence of prime numbers in natural order and let \(S_n\), for \(n\ge 1\), be the numerical semigroup generated by all prime numbers not less than \(p_n\); the multiplicity of \(S_n\) is \(p_n\) and we denote the aforementioned invariants of \(S_n\) by \(g_n\), \(f_n\), \(e_n\) and \(E_n\). Since \(S_{n+1}\) is a subsemigroup of \(S_n\) it is clear that \(f_n\le f_{n+1}\) for all \(n\ge 1\). The atoms of \(S_n\) are contained in the interval \([p_n,p_n+f_n]\); conversely, each odd integer from \(S_n\cap [p_n,3p_n[\) is an atom of \(S_n\).

As a major result we will see that Wilf’s question has a positive answer for \(S_n\). Further \({g_n}/{p_n}\) converges to 5/2 for \(n\rightarrow \infty\).

The prime number theorem suggests that there should be—like for the sequence \((p_n)\)—some asymptotic behavior of \((g_n)\), \((f_n)\) and \((e_n)\).

Based on the list \(f_1, f_2, \ldots , f_{2000}\) from [8], extensive calculations (cf. our table 1 in [9]) gave evidence for the following three conjectures:

  1. (C1)

    \(f_n\sim 3p_n\), i. e. \(\lim _{n\rightarrow \infty }\frac{f_n}{p_n}=3\),

as already observed by Kløve [12], see also the comments in [6, p. 56]; note that Kløve works with distinct primes, therefore his conjecture is formally stronger than ours, however see also [10, comment by user “Emil Jer̆ábek”, Apr 4 ’12].

In Proposition 1, we will show that

$$\begin{aligned} 3p_n-6\le f_n. \end{aligned}$$
(2)
  1. (C2)

    \(f_{n+1}<4p_n\) for all \(n\ge 1\).

and

  • \(3p_n<f_{n+1}\) for \(n\ge 3\).

It is immediate from (2) that at least

$$\begin{aligned}3p_n\le 3(p_{n+1}-2)\le f_{n+1}\,{\text {for}}\,n\ge 2.\end{aligned}$$

As already noticed in [12] and in [10, answer by user “Woett”, Apr 3 ’12], both conjectures (C1) and (C2) are closely related to Goldbach’s conjecture. As we will see in Proposition 4, (C1) would be a consequence of conjecture

  1. (C3)

    \(f_n\) is odd for \(n\ge 5\).

Notice again, that a conjecture similar to (C3) was already formulated in [12], however for the (related) notion ’threshold of completeness’ for the sequence of all prime numbers, in the sense of [6].

Figure 1 indicates, that \(\lim _{n\rightarrow \infty }\frac{f_n}{p_n}=3\) should be true.

As for (C2), by Figs. 1 and 2, evidently \(4p_n-f_{n+1}\) should stay positive for all time.

Fig. 1
figure 1

\(4p_n-f_{n+1}\) versus \(p_n\)

Full size image
Fig. 2
figure 2

\(4p_n-f_{n+1}\) versus \(p_n\)

Full size image

Observations Numerical experiments suggest that similiar conjectures can be made if one restricts the generating sequence to prime numbers in a fixed arithmetic progression \(a+kd\) for \((a,d)=1\). In such a case the limit of \(\frac{f_n}{p_n}\) would apparently be \(d+1\) (d even) or \(2d+1\) (d odd), see Fig. 3, and table 2 in [11].

Fig. 3
figure 3

f versus p for some series of semigroups as in the ’Observations’

Full size image

The following version of Vinogradov’s theorem is due to Matomäki, Maynard and Shao. It is fundamental for the considerations in this paper.

[13, Theorem 1.1] Let \(\theta >\frac{11}{20}\). Every sufficiently large odd integer n can be written as the sum \(n=q_1+q_2+q_3\) of three primes with the restriction

$$\begin{aligned} \left| q_i-\frac{n}{3}\right| \le n^\theta \,{\text {for}}\,i=1, 2, 3. \end{aligned}$$

Of course we could have used just as well one of the predecessors of this theorem, see the references in [13].

2 Variants of Goldbach’s conjecture

For \(x,y\in {\mathbb {Q}}\), \(x\le y\) we denote by [x, y] the ’integral interval’

$$\begin{aligned}\{n\in {\mathbb {Z}}|x\le n\le y\},\end{aligned}$$

accordingly we define [x, y[, ]x, y], ]x, y[, \([x,\infty [\).

For \(x\ge 2\) we define \(S_n^x\) to be the numerical semigroup generated by the primes in the interval \(I_n^x:=[p_n,x\cdot p_n[\) and \(f_n^x\) its Frobenius number.

A minor step towards a proof of conjecture (C1) is

Proposition 1

$$\begin{aligned}f_n\ge 3p_n-6.\end{aligned}$$

In particular for the null sequence \(r(n):=6/p_n\) we have

$$\begin{aligned}\frac{f_n}{p_n}\ge 3-r(n)\,{\text {for every}}\,n\ge 1. \end{aligned}$$

Proof

For \(n\ge 3\), obviously, the odd number \(3p_n-6\) is neither a prime nor the sum of primes greater than or equal to \(p_n\), hence \(3p_n-6\) is not contained in \(S_n\). \(\square\)

Remark 1

A final (major) step on the way to (C1) would be to find a null sequence l(n) such that

$$\begin{aligned}3+l(n)\ge \frac{f_n}{p_n}.\end{aligned}$$

Proposition 2

If (C1) is true then every sufficiently large even number x can be written as the sum \(x=p+q\) of prime numbers p, q.

Addendum The prime number p can be chosen from the interval \(]\frac{x}{4},\frac{x}{2}]\).

Proof

By the prime number theorem, we have \(p_{n+1}\sim p_n\). (C1) implies

$$\begin{aligned}f_{n+1}\sim 3p_{n+1}\sim 3p_n,\end{aligned}$$

i. e.

$$\begin{aligned}\lim _{n\rightarrow \infty }\frac{f_{n+1}}{p_n}=3.\end{aligned}$$

In particular, there exists \(n_0\ge 1\) such that \(\frac{f_{n+1}}{p_n}<4\) for all \(n\ge n_0\).

It remains to show:

Lemma 1

If \(n_0\ge 1\) is such that \(\frac{f_{n+1}}{p_n}<4\) for all \(n\ge n_0\) then every even number \(x>2\) with \(x>f_{n_0}\) can be written as the sum

$$\begin{aligned} x=p+q\text { with prime numbers }p\le q \text { and such that }\frac{x}{4}<p\le \frac{x}{2}. \end{aligned}$$
(1)

Proof

By our hypothesis,

$$\begin{aligned}f_n\le f_{n+1}<4p_n<4p_{n+1}\text { for all }n\ge n_0\end{aligned}$$

and hence, for \(I_n:=[1+f_n,4p_n[\) (\(n\ge n_0\)),

$$\begin{aligned} {}[1+f_{n_0},\infty [=\bigcup _{n\ge n_0}I_n. \end{aligned}$$

Therefore it suffices to prove (1) for all even numbers \(x>2\) from the interval \(I_n\), for \(n\ge n_0\).

By definition of \(f_n\), every \(x\in I_n\) can be written as the sum of primes \(p\ge p_n\).

If in addition \(x>2\) is even, then, because of \(f_n<x<4p_n\), the number x is the sum of precisely two prime numbers \(p\le q\) with

$$\begin{aligned}p_n\le p\le q=x-p<4p_n-p\le 3p,\end{aligned}$$

hence

$$\begin{aligned}\frac{x}{4}<p\le \frac{x}{2}.\end{aligned}$$

\(\square\)

The special case \(n_0=1\) of Lemma 1 gives

Proposition 3

If (C2) is true then every even number \(x>2\) can be written as the sum \(x=p+q\) of prime numbers \(p\le q\) as described in the Addendum above. In particular for each \(n\ge 1\), \(4p_n=p+q\) with primes \(p_{n+1}\le p\le q\), implying Bertrand’s postulate. \(\square\)

Lemma 2

Let \(\varepsilon >0\). For odd N large enough, there are prime numbers \(q_1\), \(q_2\), \(q_3\) with

$$\begin{aligned}N=q_1+q_2+q_3\end{aligned}$$

and such that

$$\begin{aligned} \frac{1}{3+\varepsilon }\cdot N<q_i<\frac{3+2\varepsilon }{9+3\varepsilon }\cdot N\text {, i. }\,\text {e. }\left| q_i-\frac{N}{3}\right| <\frac{\varepsilon }{9+3\varepsilon }\cdot N\,{\text {for}}\,i=1,2,3. \end{aligned}$$

Proof

The claim follows immediately from [13, Theorem 1.1], since \(\theta :=\frac{3}{5}>\frac{11}{20}\) and, for large N, \(N^\frac{3}{5}<\frac{\varepsilon }{9+3\varepsilon }\cdot N\). \(\square\)

Lemma 3

Let \(\varepsilon >0\). Then for large n, each odd integer \(N\ge (3+\varepsilon )p_n\) is contained in \(S_{n+1}\). In particular, for large n

$$\begin{aligned} f_{n+1}< & {} (3+\varepsilon )p_n\text { if }f_{n+1}\text { is odd, and}\\ f_{n+1}< & {} (3+\varepsilon )p_n+p_{n+1}\text { if }f_{n+1}\text { is even,} \end{aligned}$$

since then \(f_{n+1}-p_{n+1}\) is odd and not in \(S_{n+1}\).

Proof

Since N is odd and large for large n, by Lemma 2 there exist prime numbers \(q_1\), \(q_2\), \(q_3\) with

$$\begin{aligned}N=q_1+q_2+q_3\end{aligned}$$

and such that

$$\begin{aligned}\frac{N}{3+\varepsilon }<q_i\,{\text {for}}\,i=1,2,3.\end{aligned}$$

By assumption, \(\frac{N}{3+\varepsilon }\ge p_n\), hence

$$\begin{aligned}q_i>p_n\text {, i. }\,\text {e. }q_i\ge p_{n+1}\end{aligned}$$

for the prime numbers \(q_i\). This implies \(N=q_1+q_2+q_3\in S_{n+1}.\) \(\square\)

Proposition 4

If the Frobenius number \(f_n\) is odd for all large n, then \(f_n\sim 3p_n\). In particular, conjecture (C3) implies conjecture (C1).

Proof

This is immediate from Proposition 1 and Lemma 3. \(\square\)

For a similar argument, see [10, answer by user “Anonymous”, Apr 5’12].

Remark 2

  1. (a)

     It is immediate from Lemma 3 that

    $$\begin{aligned}\limsup _{n\rightarrow \infty }\frac{f_n}{p_n}\le 4.\end{aligned}$$

    As a consequence, a proof of \(\limsup _{n\rightarrow \infty }\frac{f_n}{p_n}\ne 4\) would imply the binary Goldbach conjecture for large x with the Addendum from above—see Lemma 1 and the proof of Proposition 2.

  2. (b)

    The estimate \(\limsup _{n\rightarrow \infty }\frac{f_n}{p_n}\le 4\) together with a sketch of proof was already formulated in [10, comment by user “François Brunault” (Apr 6 ’12) to answer by user “Anonymous” (Apr 5 ’12)]. Our proof is essentially an elaboration of this sketch.

  3. (c)

    Lemma 3 shows that

    $$\begin{aligned}f_{n+1}<5p_{n+1}\text { for large }n.\end{aligned}$$

    Because of \(p_{n+1}<2p_n\) (Bertrand’s postulate) this implies also that there exists a constant C with

    $$\begin{aligned} f_{n+1}<Cp_n\text { for all }n. \end{aligned}$$
    (2)

    Conjecture (C2) says that in (2) one can actually take \(C=4\).

    Notice that (2) already follows from [1, Lemma 1].

Problem Find an explicit pair \((n_0,C_0)\) of numbers such that

$$\begin{aligned}f_{n+1}<C_0\cdot p_n\text { for every }n\ge n_0.\end{aligned}$$

Next we shall study the asymptotic behavior of the set of atoms of \(S_n\).

Lemma 2 implies

Lemma 4

Let \(\varepsilon >0\). Then \(S_n=S_n^{3+\varepsilon }\) for large n.

Proof

It suffices to prove the claim for arbitrarily small values of \(\varepsilon\):

First we show that, if \(\varepsilon <3\), then

$$\begin{aligned}S_{n+1}^{3+\varepsilon }\subseteq S_n^{3+\varepsilon }\end{aligned}$$

for large n. For this it suffices to show that every prime number p on the interval \([p_{n+1},(3+\varepsilon )p_{n+1}[\) is in \(S_n^{3+\varepsilon }\):

Firstly, \(p\ge p_{n+1}>p_n\).

Now we distinguish two cases:

  1. I

    \(p<(3+\varepsilon )p_n\): Then \(p\in I_n^{3+\varepsilon }\), hence \(p\in S_n^{3+\varepsilon }\).

  2. II

    \(p\ge (3+\varepsilon )p_n\): For n large enough, by Lemma 2 there exist prime numbers \(q_1,q_2,q_3\) with

    $$\begin{aligned}p=q_1+q_2+q_3\end{aligned}$$

    and such that

    $$\begin{aligned}p_n\buildrel \text {II}\over \le \frac{p}{3+\varepsilon }<q_i<\frac{3+2\varepsilon }{9+3\varepsilon }p\,{\text {for}}\,i=1,2,3.\end{aligned}$$

    By Chebyshev, Bertrand’s postulate \(p_{n+1}<2p_n\) holds. Therefore,

    $$\begin{aligned}p\buildrel \text {hypothesis}\over<(3+\varepsilon )p_{n+1}<(6+2\varepsilon )p_n\end{aligned}$$

    and hence

    $$\begin{aligned}q_i<\frac{3+2\varepsilon }{9+3\varepsilon }p<\frac{3+2\varepsilon }{9+3\varepsilon }(6+2\varepsilon )p_n<(3+\varepsilon )p_n,\end{aligned}$$

    if \(\varepsilon <3\). It follows that

    $$\begin{aligned}q_i\in [p_n,(3+\varepsilon )p_n[\,{\text {for}}\,i=1,2,3\text { and hence}\\p=q_1+q_2+q_3\in S_n^{3+\varepsilon },\end{aligned}$$

    which proves the above claim.

    Recursively, we get from \(S_{n+1}^{3+\varepsilon }\subseteq S_n^{3+\varepsilon }\) that

    $$\begin{aligned}p_k\in S_k^{3+\varepsilon }\subseteq S_n^{3+\varepsilon }\text { for all }k\ge n.\end{aligned}$$

    Therefore,

    $$\begin{aligned}S_n=S_n^{3+\varepsilon }.\end{aligned}$$

    \(\square\)

For \(x\ge 0\) let \(\pi (x)\) be the number of primes less than or equal to x. Applying Lemma 4, the prime number theorem (PNT) yields

Theorem 1

Let \(u_n\) be the largest atom of \(S_n\). Then

$$\begin{aligned} \pi (u_n)\sim 3n, u_n\sim 3p_n\,{\text {and}}\,e_n\sim 2n. \end{aligned}$$

Proof

By Lemma 4, for each \(\varepsilon >0\) there is an \(n(\varepsilon )>0\) such that \(E_n\subseteq [p_n,(3+\varepsilon )p_n]\) for all \(n\ge n(\varepsilon )\). On the other hand, the primes from \([p_n,3p_n]\) are atoms of \(S_n\). Hence

$$\begin{aligned} \pi (3p_n)\le \pi (u_n)\le \pi ((3+\varepsilon )p_n) \,{\text {for}}\,n\ge n(\varepsilon ). \end{aligned}$$
(3)

By the PNT, \(\pi (3p_n)\sim 3n\) and \(\pi ((3+\varepsilon )p_n)\sim (3+\varepsilon )n\text { for }\varepsilon >0\). Hence (3) implies \(\pi (u_n)\sim 3n\).

From \(p_n\le u_n\le (3+\varepsilon )p_n\) we get \(\log u_n\sim \log p_n\), hence again by the PNT,

$$\begin{aligned}u_n\sim \pi (u_n)\cdot \log u_n\sim 3n\cdot \log p_n\sim 3p_n.\end{aligned}$$

Finally, \(e_n=\pi (u_n)-n+1\sim 3n-n+1\sim 2n\). \(\square\)

By [4, Cor. 6.5], for arbitrary numerical semigroups S, Wilf’s inequality \(\frac{g}{1+f}\le \frac{e-1}{e}\) holds, whenever \(f<3\cdot p\). Further by [18], the latter is true for almost every numerical semigroup of genus g (as g goes to infinity).

In contrast, according to table 1 in [9], for the semigroups \(S_n\), the relation \(f_n<3\cdot p_n\) seems to occur extremely seldom, but over and over again (see Fig. 4).

Fig. 4
figure 4

\(f_n-3p_n\) versus n

Full size image

The following considerations are related to [10, answer by user “Aaron Meyerowitz”, Apr 3 ’12]:

Let \(f_n<3\cdot p_n\). Then the odd number \(3\cdot p_n+6\) is in \(S_n\), but not a prime; hence \(p_{n+1}\le p_n+6\).

  1. 1.

    If \(p_{n+1}=p_n+4\), since \(3\cdot p_n+6\in S_n\) is not a prime, \(p_n+6\) must be prime.

  2. 2.

    If \(p_{n+1}=p_n+6\), then the odd numbers \(3p_n+2\) and \(3p_n+4\) must be atoms in \(S_n\), hence primes.

In any case:

Nota bene If \(f_n<3p_n\), then there is a twin prime pair within \([p_n,3p_n+4]\).

So we cannot expect to prove, that \(f_n<3p_n\) happens infinitely often, since this would prove the twin prime conjecture, that there are infinitely many twin prime pairs. Another consequence would be that

$$\begin{aligned}\liminf _{n\rightarrow \infty }\frac{f_n}{p_n}=3,\end{aligned}$$

since one always has that this limit inferior is \(\ge 3\), by Proposition 1.

The next section is attended to Wilf’s question mentioned above.

3 The question of Wilf for the semigroups \(\mathbf {S_n}\)

Proposition 5

For the semigroups \(S_n\), Wilf’s (proposed) inequality

$$\begin{aligned} \frac{g_n}{1+f_n}\le \frac{e_n-1}{e_n} \end{aligned}$$
(1)

holds.

Proof

For \(n<429\), have a look at table 1 in [9]. Now let \(n\ge 429\).

Instead of (1), we would rather prove the equivalent relation

$$\begin{aligned} e_n(1+f_n-g_n)\ge 1+f_n. \end{aligned}$$
(2)

According to [4, Cor. 6.5] we may assume, that \(3p_n<1+f_n\). Hence the primes in the interval \([p_n,3p_n[\) are elements of \(S_n\) lying below \(1+f_n\), and in fact, they are atoms of \(S_n\) as well. This implies for the prime-counting function \(\pi\)

$$\begin{aligned} e_n(1+f_n-g_n)\ge (\pi (3p_n)-n+1)^2. \end{aligned}$$
(3)

By Rosser and Schoenfeld [15, Theorem 2] we have

$$\begin{aligned} \pi (x)< & {} \frac{x}{\log x-\frac{3}{2}} \text { for }x>e^{\frac{3}{2}}\text {, and} \end{aligned}$$
(4)
$$\begin{aligned} \pi (x)> & {} \frac{x}{\log x-\frac{1}{2}}\,{\text {for}}\, x\ge 67. \end{aligned}$$
(5)

From (4) and (5) we will get in a moment:

$$\begin{aligned} 2n<\pi (3p_n)<3n\text { for }n\ge 429. \end{aligned}$$
(6)

Proof of (6) Since the function \(\lambda (x):=3\cdot \frac{\log x-\frac{3}{2}}{\log (3x)-\frac{1}{2}}\) is strictly increasing for \(x>1\), we get for \(n\ge 429\), i. e. \(p_n\ge 2971\)

$$\begin{aligned}&\pi (3p_n)\buildrel \text {(5)}\over>\frac{3p_n}{\log (3p_n)-\frac{1}{2}}\buildrel \text {(4)}\over>\pi (p_n)\cdot \lambda (p_n) \ge n\cdot \lambda (2971)>2n\text {, and}\\&\qquad \pi (3p_n)\buildrel \text {(4)}\over<\frac{3p_n}{\log p_n+\log 3-\frac{3}{2}}<\frac{3p_n}{\log p_n-\frac{1}{2}}\buildrel \text {(5)}\over <3n. \end{aligned}$$

In particular, by (3) and (6)

$$\begin{aligned}e_n(1+f_n-g_n)\buildrel \text {(3)} \over \ge (\pi (3p_n)-n+1)^2\buildrel \text {(6)}\over \ge (n+2)^2. \end{aligned}$$

It remains to prove

Lemma 5

If \(n\ge 429\), then

$$\begin{aligned}f_n<n^2.\end{aligned}$$

Proof

Let \(N\le a_1<\cdots <a_N\) be positive integers with \((a_1,\ldots ,a_N)=1\), \(S=\langle a_1,\ldots , a_N\rangle\) the numerical semigroup generated by these numbers and f its Frobenius number. Then, by Selmer [16] we have the following theorem (see the book [14] of Ramírez Alfonsín). It is an improvement of a former result [5, Theorem 1] of Erdős and Graham.

[14, Theorem 3.1.11]

$$\begin{aligned} f\le 2\cdot a_N\left\lfloor \frac{a_1}{N}\right\rfloor -a_1. \end{aligned}$$
(7)

We will apply this to the semigroup \(S_n^3\subseteq S_n\) generated by the primes

$$\begin{aligned}p_n=a_1<p_{n+1}=a_2<\ldots <p_{N+n-1}=a_N\end{aligned}$$

in the interval \(I_n^3=[p_n,3p_n[\), with Frobenius number \(f_n^3\), hence

$$\begin{aligned} N=\pi (3p_n)-n+1, a_N=p_{\pi (3p_n)}=\text {the largest prime in }I_n^3. \end{aligned}$$

By [15, Theorem 3, Corollary, (3.12)] we have

$$\begin{aligned} p_n>n\log n\ge n\log 429>6n\buildrel \text {(6)}\over >N, \end{aligned}$$

hence the above theorem can be applied.

By (6) and (7), \(p_{\pi (3p_n)}\buildrel \text {(6)}\over <p_{3n}\) and

$$\begin{aligned} f_n\le f_n^3\buildrel \text {(7)}\over<2\cdot p_{\pi (3p_n)} \cdot \frac{p_n}{\pi (3p_n)-n+1}\buildrel \text {(6)}\over <2 \cdot p_{3n}\cdot \frac{p_n}{n+2}. \end{aligned}$$

It remains to show, that \(2\cdot p_{3n}\cdot \frac{p_n}{n+2}<n^2\,{\text {for}}\,n\ge 429:\)

By [15, Theorem 3, Corollary, (3.13)], we have

$$\begin{aligned} p_k<k(\log k+\log \log k)\,{\text {for}}\,k\ge 6. \end{aligned}$$
(8)

We consider the function

$$\begin{aligned} \lambda _2(x):=6\cdot (\log (3x)+\log \log (3x))\cdot (\log x+\log \log x). \end{aligned}$$

Since \(\frac{\log (3x)}{x}\) is decreasing and \(\lambda _2^{\prime }(x)<48\cdot \frac{\log (3x)}{x}\) for \(x\ge 3\), we obtain

$$\begin{aligned} \lambda _2^{\prime }(x)<48\cdot \frac{\log 1287}{429}<1=(x+2)^{\prime }\,{\text {for}}\,x \ge 429;\,{\text {further}}\,\lambda _2(429)<431. \end{aligned}$$

Elementary calculus yields

$$\begin{aligned} \lambda _2(x)<x+2\,{\text {for}}\,x\ge 429. \end{aligned}$$
(9)

Hence

$$\begin{aligned}2\cdot p_{3n}\cdot p_n\buildrel \text {(8)}\over<n^2\cdot \lambda _2(n)\buildrel \text {(9)}\over <n^2\cdot (n+2)\,{\text {for}}\,n\ge 429. \end{aligned}$$

\(\square\)

See also Dusart’s thèse [3] for more estimates like (4), (5) and (8).

Remark 3

Looking at table 3 in [7] we see, that even

$$\begin{aligned}\pi (3p_n)>2n\,{\text {for}}\,n> 8\,{\text {and}}\,\pi (3p_n)<3n\text { for n > 1}\end{aligned}$$

(which may be found elsewhere), and

$$\begin{aligned}f_n\le n^2\,{\text {for}}\,n\ne 5.\end{aligned}$$

At last we will see that, apparently, the quotient \(\frac{g_n}{1+f_n}\) should converge to \(\frac{5}{6}\) (whereas \(\lim _{n\rightarrow \infty }\frac{e_n-1}{e_n}=1\), since \(e_n\sim 2n\) by our Theorem).

Proposition 6

The quotient \(\frac{g_n}{p_n}\) converges, and

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{g_n}{p_n}=\frac{5}{2}. \end{aligned}$$

Proof

For that, we consider the proportion \(\alpha _k(n)\) of gaps of \(S_n\) among the integers in \([k\cdot p_n,(k+1)\cdot p_n]\), (\(k,n\ge 1\)). Besides [13, Theorem 1.1], we shall need the following similar result about the representation of even numbers as the sum of two primes:

[2, Theorem 1, Corollary] Let \(\varepsilon >0\) and \(A>0\) be real constants. For \(N>0\) let E(N) be the set of even numbers \(2m\in [N,2N]\), which cannot be written as the sum \(2m = q_1 + q_2\) of primes \(q_1\) and \(q_2\) with the restriction

$$\begin{aligned}|q_j-m|\le m^{\frac{5}{8}+\varepsilon }\,{\text {for}}\,j=1,2.\end{aligned}$$

Then there is a constant \(D>0\) such that \(\#E(N)<D\cdot N/(\log N)^A\).

From these two facts together with the prime number theorem, we conclude the following asymptotic behavior of the numbers \(\alpha _k(n)\), as n goes to infinity:

$$\begin{aligned}\alpha _0(n)\rightarrow 1, \alpha _1(n)\rightarrow 1, \alpha _2(n)\rightarrow \frac{1}{2}\,{\text {and}}\,\alpha _k(n)\rightarrow 0\,{\text {for}}\,k\ge 3.\end{aligned}$$

Hence

$$\begin{aligned}\lim _{n\rightarrow \infty }\frac{g_n}{p_n}=1+1+\frac{1}{2}=\frac{5}{2}.\end{aligned}$$

(Notice that for large n, by Lemma 3 we have \(f_n<5p_n\), hence \(\alpha _k(n)=0\) for \(k\ge 5\).) \(\square\)

Remark 4

Under the assumption \(\lim _{n\rightarrow \infty }\frac{p_n}{f_n}=\frac{1}{3}\) (C1) (which should be true by computational evidence), by Proposition 6,

$$\begin{aligned}\lim _{n\rightarrow \infty }\frac{g_n}{1+f_n}=\frac{5}{6}.\end{aligned}$$

Remark 5

Let \(f_{n,e}\) be the largest even gap of \(S_n\). Our computations (see table 1 in [9]) suggest that \(f_{n,e}\sim 2p_n\). In this case, by Proposition 1 and Proposition 4, \(f_n\) is odd for large n and conjecture (C1) holds.