## 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.

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].

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 [xy] the ’integral interval’

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

accordingly we define [xy[, ]xy], ]xy[, $$[x,\infty [$$.

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 pq.

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).

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.