Finding joker numbers

Question

Let $n$ be an integer, $q(n)$ be the smallest prime number which divides $n$ and $r(n)$ be the biggest prime number less than or equal to $\sqrt{n}$. We say that $n$ is joker if $q(n)=r(n)$.

Except $8$, joker numbers are of the form $p_n\cdot p_{n+k}$, with $k\geq 0$. The converse is obviously false. If $(p_n)$ is the prime numbers sequence, it is easy to prove that a number of the form $p_n\cdot p_{n+k}$ is joker if and only if $p_{n+1}^2 > p_n\cdot p_{n+k}$. In particular, if a number is not joker for $k$, it is not joker for $k+1$ (with the same $n$). So, all the numbers of the form $p_n^2$ and $p_n\cdot p_{n+1}$ are jokers. Using a spreadsheet, we can see that for $n\leq 1000$, the percentage of numbers of the form $p_n\cdot p_{n+k}$ which are jokers, with $k=2, 3, 4, 5, 6$ and $7$ is respectively $52 \%, 19.2 \%, 7.8 \%, 2 \%, 0.5 \%$ and $0.1 \%$.

Question : for any $k$, does there exist $n$ such that the number $p_n\cdot p_{n+k}$ is joker ?

Answer 1

With PARI/GP, I could find examples upto $k=22$ , listed in the following output :

gp > for(s=2,25,z=primes(s);[a,b,c]=[z[1],nextprime(z[1]+1),z[s]];q=1;while(b^2<=a*c,q=q+1;a=b;b=nextprime(b+1);c=nextprime(c+1));print(s-1,"   ",q))
1   1
2   2
3   4
4   30
5   180
6   462
7   890
8   1532
9   3385
10   19871
11   29040
12   31545
13   31545
14   31545
15   597311
16   1293698
17   1293698
18   1293698
19   2279181
20   2279181
21   118374763
22   118374763

If the gap to the next prime is denoted with $a$ and the difference $p_{n+k}-p_n$ is denoted with $b$ , then $2a>b$ is sufficient for an example.

Considering that the merit of a prime gap defined as $$\frac{p_{n+1}-p_n}{\ln(p_n)}$$ can be arbitarily large, such a solution should exist for arbitarily large $k$ , but this is of course no proof.

Note that a solution immediately implies that there is a solution for all smaller $k$ as well. So, chances are very good that the answer to your question is "yes".

Answer 2

NOT AN ANSWER. An extended discussion to restate the question being asked, which won't fit as a comment

OP uses $n$ both as the name for joker numbers and as the primary index of primes. For clarity, I will avoid using $n$ as the primary index of primes.

A number $n$ with $3$ or more prime factors can have at most one prime factor larger than $\sqrt{n}$; if it had more than one such prime factor, their product alone would be greater than $n$. Thus if $n$ has $3$ or more prime factors, at least two of them must be smaller than $\sqrt{n}$. The only way that such a number might be a joker number is if the two prime factors $p_a$ smaller than $\sqrt{n}$ are equal: $n=(p_a)^2\cdot c$. Then $\sqrt{n}=p_a\sqrt{c}$ where $c$ has prime factors $\ge p_a$.

Specific cases: $p_a=c=2$ affords the noted exception $n=8$. $p_a=2,\ c=3 \Rightarrow n=12$, and $12$ is not a joker number by examination. $p_a\ge 3 \Rightarrow n\ge 27$, and since $27>25=p_{a+1}^2$, this affords no joker numbers. $c\ge 4 \Rightarrow p_a\sqrt{c}\ge 2p_a$ and by Bertrand's postulate, $p_a$ is not the smallest prime $\le \sqrt{n}$. Since $n$ has at least $2$, and (excepting $8$) cannot have $3$ or more prime factors, $n$ is a semiprime.

For any particular $p_a$, the joker numbers containing it as a prime factor lie within the interval $(p_a)^2$ to $(p_{a+1})^2$. OP's question can be stated: Let $k$ be the count of numbers in that interval that are products $p_ap_b$ such that $p_a<p_b<\frac{(p_{a+1})^2}{p_a}$; does there always exist $p_a$ such that $k$ can be arbitrarily large?

Let $p_{a+1}=p_a+g_a$, where $g_a$ is the gap between $p_a$ and $p_{a+1}$. Then acceptable $p_b$ to be counted are defined as $p_a<p_b<\frac{(p_{a}+g_a)^2}{p_a}=p_a+2g_a+\frac{g_a^2}{p_a}$. For sufficiently large $p_a$ we can consider $\frac{g_a^2}{p_a}$ to be an error term. We also know that there is only one prime to be counted within the interval from $p_a$ to $p_{a+1}$. So the question becomes: In the interval from $p_{a+1}$ to $p_{a+1}+g_a$, can there be arbitrarily many prime numbers?

At this time, I don't have an answer to that question.