Abstract
Let \(p_1=2, p_2=3, p_3=5, \ldots\) be the consecutive prime numbers, \(S_n\) the numerical semigroup generated by the primes not less than \(p_n\) and \(u_n\) the largest irredundant generator of \(S_n\). We will show, that \(u_n\sim 3p_n\). Similarly, for the largest integer \(f_n\) not contained in \(S_n\), by computational evidence (https://www.uniregensburg.de/Fakultaeten/nat_Fak_I/Hellus/table_1.pdf) we suspect that (1) \(f_n\) is an odd number for \(n\ge 5\) and (2) \(f_n\sim 3p_n\); further (3) \(4p_n>f_{n+1}\) for \(n\ge 1\). If \(f_n\) is odd for large n, then \(f_n\sim 3p_n\). In case \(f_n\sim 3p_n\) every large even integer x is the sum of two primes. If \(4p_n>f_{n+1}\) for \(n\ge 1\), then the Goldbach conjecture holds true. Further, Wilf’s question in Wilf (Am Math Mon 85:562–565, 1978) has a positive answer for the semigroups \(S_n\).
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 socalled 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
will play a role. For general S, what is known about these fractions?
First of all it is well known and easily seen that
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
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 \(\frac{g}{1+f}\le \frac{e1}{e}\).
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 \(\frac{g}{1+f}\le \frac{e1}{e}\) 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:

(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

(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
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

(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 \(4p_nf_{n+1}\) 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
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’
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
In particular for the null sequence \(r(n):=6/p_n\) we have
Proof
For \(n\ge 3\), obviously, the odd number \(3p_n6\) is neither a prime nor the sum of primes greater than or equal to \(p_n\), hence \(3p_n6\) 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
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
i. e.
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
Proof
By our hypothesis,
and hence, for \(I_n:=[1+f_n,4p_n[\) (\(n\ge n_0\)),
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
hence
\(\square\)
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
and such that
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
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
and such that
By assumption, \(\frac{N}{3+\varepsilon }\ge p_n\), hence
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

(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.

(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.

(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
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
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:

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

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
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
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,
Finally, \(e_n=\pi (u_n)n+1\sim 3nn+1\sim 2n\). \(\square\)
By [4, Cor. 6.5], for arbitrary numerical semigroups S, Wilf’s inequality \(\frac{g}{1+f}\le \frac{e1}{e}\) 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.
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.
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
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
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
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 primecounting function \(\pi\)
By Rosser and Schoenfeld [15, Theorem 2] we have
From (4) and (5) we will get in a moment:
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. \(p_n\ge 2971\)
It remains to prove
Lemma 5
If \(n\ge 429\), then
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]
We will apply this to the semigroup \(S_n^3\subseteq S_n\) generated by the primes
in the interval \(I_n^3=[p_n,3p_n[\), with Frobenius number \(f_n^3\), hence
By [15, Theorem 3, Corollary, (3.12)] we have
hence the above theorem can be applied.
By (6) and (7), \(p_{\pi (3p_n)}\buildrel \text {(6)}\over <p_{3n}\) and
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
We consider the function
Since \(\frac{\log (3x)}{x}\) is decreasing and \(\lambda _2^{\prime }(x)<48\cdot \frac{\log (3x)}{x}\) for \(x\ge 3\), we obtain
Elementary calculus yields
Hence
\(\square\)
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
(which may be found elsewhere), and
At last we will see that, apparently, the quotient \(\frac{g_n}{1+f_n}\) should converge to \(\frac{5}{6}\) (whereas \(\lim _{n\rightarrow \infty }\frac{e_n1}{e_n}=1\), since \(e_n\sim 2n\) by our Theorem).
Proposition 6
The quotient \(\frac{g_n}{p_n}\) converges, and
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
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:
Hence
(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,
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.
Change history
10 June 2021
A Correction to this paper has been published: https://doi.org/10.1007/s00233021101987
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Open Access funding enabled and organized by Projekt DEAL. We thank F. Brunault, O. Forster and K. Matomäki for valuable hints.
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Hellus, M., Rechenauer, A. & Waldi, R. Numerical semigroups generated by primes. Semigroup Forum 101, 690–703 (2020). https://doi.org/10.1007/s00233020101029
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DOI: https://doi.org/10.1007/s00233020101029