# On values of $n\geq 1$ satisfying that for all primorial $N_k$ less than $n$ the difference $n-N_k$ is a prime number

In this post I present a similar question that shows section A19 of [1], I was inspired in it to define my sequence and question. I am asking about it since I think that the problem that arises from my question harmonizes with this.

For integers $$k\geq 1$$ we denote the primorial of order $$k$$ as $$N_k$$, that is

$$N_k=\prod_{r=1}^k p_r$$

where thus $$p_r$$ denotes the $$r-th$$ prime number. I add as reference that Wikipedia has the article Primorial.

In next paragraph $$n\geq 1$$ denotes the integer variable.

Definition. Now we consider the values of $$n$$ making $$n-N_k$$ prime for all those primorials $$N_k$$ satisfying $$2\leq N_k.

I've calculated the first few terms from the definition. Our sequence starts as $$n=4,9,25,43,49,73\ldots...$$

To illustrate our sequence we add the example about why $$25$$ is in our sequence: the only primorials less than $$n=25$$ are $$N_1=2$$ and $$N_2=6$$, and we get that the integers $$25-2$$ and $$25-6$$ are both prime numbers.

I have no idea/intutition if this sequence has only a finite number of terms.

Question. I don't know if it exists only a finite number of positive integers $$n's$$ being $$n-N_k$$ a prime number for all primorial $$2\leq N_k. What work can be done to study if the sequence of $$n's$$ from this definition should have a finite number of terms? Many thanks.

I am asking if it is possible to deduce a proposition, get some idea about it, or set some conjecture.

I emphasize that I am asking about what work or statements/heuristics we can get for the Question, since it is a hard question (see the comments) I think that I should accept an available and useful answer.

## References:

[1] Richard K. Guy, Unsolved Problems in Number Theory, Unsolved problems in Intuitive Mathematics Volume I, Second Edition, Springer-Verlag (1994).

• To prove that the sequence is infinite is at least as hard as to prove that both $n-2$ and $n-6$ are prime for infinitely many $n$ which seems out of reach. – Dmitry Krachun Aug 12 '19 at 3:47
• I was asking from the ignorance about the difficulty of the problem that I propose and therefore technical details like the one you have provided I did not know. If there is more activity for this post feel free to add your nice contribution and related thoughts as an answer. Many thanks @DmitryKrachun – user142929 Aug 12 '19 at 4:16
• I think this requires the $m$-th term of your sequence to be coprime with the first $m^{1+o(1)}$ primes. – Sylvain JULIEN Aug 12 '19 at 14:44
• Note also that the terms you listed are all primes or powers of primes. – Sylvain JULIEN Aug 12 '19 at 14:45
• Many thanks for your valuable comments, as I've said to the other user, feel free to add your remarks if you want as an answer @SylvainJULIEN when you consider it. On the other hand I've calculated only the previous few terms that were showed. – user142929 Aug 12 '19 at 15:31

Instead of looking at primorials, one can look at so-called admissible sets. We say that a set of integers $$S = \{h_1, \cdots, h_k\}$$ is admissible if for all primes $$p$$, $$S \pmod{p}$$ is not a complete set of residues. Note that for all $$p > |S|$$, $$S \pmod{p}$$ is automatically not a full set of residues.

For example, the set $$\{1,2\}$$ is not admissible since modulo 2 the set represents a complete set of residues, whereas $$\{0,2\}$$ is admissible.

The conjecture then (originally attributed to Hardy-Littlewood) that for any admissible set $$S$$, there exists infinitely many integers $$n$$ such that each of $$n + h_1, n + h_2, \cdots, n + h_k$$ is prime. In fact the Hardy-Littlewood $$k$$-tuple conjectures predicts an asymptotic formula for the number of such $$n \leq X$$.

This conjecture, of course, is totally out of reach. What is known is due to the seminal work of Yitang Zhang, James Maynard and the two polymath projects: that for for $$h_i \geq 0$$ for $$i = 1, \cdots, k$$ and $$h_i < h_j$$ for $$i < j$$, whenever $$h_k - h_1 \geq 246$$ there exists infinitely many $$n$$ such that there exists two indices $$i < j$$ such that $$n + h_i, n + h_j$$ are both prime.

Maynard also proved the following which is out of reach using Zhang's approach: for each positive integer $$m$$ there exists a number $$B(m)$$ such that whenever $$S$$ is an admissible set of size $$k$$ exceeding $$B(m)$$, there exists infinitely many $$n$$ such that $$n + h_i$$ is prime for at least $$m$$ indices $$i$$. In fact, one can take $$B(m) = O(m \log m)$$.

These results are known as bounded gaps between primes. They represent an extremely high achievement in prime number theory in recent times, perhaps the greatest results in a generation.

• However, the location of the admissible set is an important element of this problem. As fat as I know Maynard's work does not say how close the n are to each other, much less where they are. Gerhard "Where Is The Smallest One?" Paseman, 2020.05.01. – Gerhard Paseman May 2 at 0:22
• I am not sure what you mean by "location" of the admissible set. For each $k$ one can compute an admissible set having cardinality $k$; this is an easy exercise for small $k$ but non-trivial even for $k$ in the hundreds (indeed, the admissible set given in Maynard's original paper was computed by Andrew Sutherland). The goal is to find admissible sets of a given cardinality of the smallest diameter, i.e., minimizing $h_k - h_1$. It is true that none of the techniques that work actually tell you which indices are prime for a given $n$ which guarantee at least two indices will be prime. – Stanley Yao Xiao May 2 at 0:24
• @StanleyYaoXiao, the locations of admissible $k-$tuples are hard to find, please if you have more informations about this can you answer my question about this problem : mathoverflow.net/questions/358119/… – LAGRIDA May 2 at 1:38
• The admissible set in Maynard is a finite set independent of n. The admissible set in the problem depends on n, and is relatively close to the origin. Given an admissible set and an n which witnesses it, can you assure me there is no smaller witness? Gerhard "I Don't Think Maynard Can" Paseman, 2020.05.01. – Gerhard Paseman May 2 at 1:39
• It seems @GerhardPaseman , if I understand well the mentioned person, that he was awarded with other prestigious prize, see the article 2020 Frank Nelson Cole Prize in Number Theory, from the section From the Secretary of Notices Of The American Mathematical Society (April 2020). This is just as an invitation if you want to read this good journal, that I suppose that you know, and if you want to see my recent updated post on this MathOverflow What work can be done to study the solutions of $\varphi\left(x^{\sigma(x)}\sigma(x)^x\right)=2^{x-1} x^{3x-1}\varphi(x)$ ? Good day! – user142929 May 11 at 13:27

I will extend Stanley Yao Xiao definition of admissible k-tuple.

Your conjecture is not true if we arrive to a $$k-$$tuple not admissible!

Suppose that the $$m-$$tuple $$(0,h_1,\cdots,h_m)$$ not admissible, then for $$k>m$$ and $$a > 0$$ if you put values $$a, a+h_1,a+h_2,\cdots,a+h_{m-1}$$ in the $$k-$$tuple, it will never be admissible!

Your conjecture will stop with the first not admissible $$k-$$tuple:

$$n=9 : 3+(0, 4)$$

$$n=25 : 19+(0, 4)$$

$$n=43 : 13+(0, 24, 28)$$

$$n=49 : 19+(0, 24, 28)$$

For $$N_k < n \leq N_{k+1}$$, you have a unique $$k-$$tuple $$(0,h_1,\cdots,h_{k-1})$$ and the next $$(k+1)-$$tuple will be of the form $$(0,b,b+h_1,\cdots,b+h_{k-1})$$