Yet another conjecture about primes

Let $\mathcal{N}(n)$ be the next prime greater than $n$. Conjecture:

$\mathcal{N}(n!)-n!\:$ is either $1$ or a prime.

It holds for n=1 to 99 and the expression is 1 for 3,11,27,37,41,73,77 and primes for all other $n<100$.

I have no idea how to proceed, but will go on testing for $n>99$.

Solution 1:

This is similar to Fortune's conjecture, with factorials replacing primorials. The relevant sequence in the OEIS is A033932 where you can find your question as a 2004 conjecture by Amarnath Murthy. I expect others have asked the same question earlier. You can see that the conjecture has been verified for the first 2000 terms.

Solution 2:

I kept in my favorites your question, and finally this week I have a computer! so I did also some tests. I think that the rule could be generalized to something like this (1):

(1) $\mathcal{N}(E(n))-E(n)\ \in \{0,1,\Bbb P\}$.

Where $E(n)$ is an expression based on $n$, and as @Charles mentioned, the known ones are: Fortunate numbers (using primorials), and the $n!$ version by Amarnath Murthy that you also found by yourself.

I have been trying other expressions and the following ones seem to comply as well (maybe they are known, but I just did some trial-error). I tested both over the interval $[1,300]$ with Python:

$E_1(n)=2^n\cdot n!$

I wondered if it was possible to find an expression $E(n) \lt n!$, and the following one seems to work as well! (at least up to $n=300$).

$E_2(n)=\frac{n!}{2^{\lfloor \frac{n}{2} \rfloor}}$

I found other expressions (*) $E(n) \lt E_2(n)$ that seem also to comply with the original rule, but I am still testing them. The topic is really amazing (and must confess I did not know anything about it before).

(*) I tried also for instance Catalan numbers, but they do not comply.

In my humble opinion, one interesting point about your research would be:

Which one is the minimum expression $E_{min}(n)$?

$E_{min}(n)\ /$

$\ \mathcal{N}(E_{min}(n))-E_{min}(n) \in \{0,1,\Bbb P\} \land$

$\not \exists\ E(n)\ ,\ E(n) \lt E_{min}(n) \land \ \mathcal{N}(E(n))-E(n) \in \{0,1,\Bbb P\}$

UPDATE (2015/05/14):

The following expression seems to comply as well, tested as far as I could with Pytyon in the interval $n \in [1..700]$

$E_3(n) = (\frac{n}{2})^{(n-1-\lfloor\frac{n}{2}\rfloor)}$

Where $(x)^k = x*(x+1)*...*(x+k-1)$ is the Pochhammer symbol.

The origin of the expression was the product of the even numbers $e \in [p,2p-2]$, where $p$ is a prime. Such expression can be written as $2^{k_1}(\frac{p}{2})^{k_2}$. That expression failed the test for few $p$'s. For that reason I tested separately $2^{k_1}$ and $(\frac{p}{2})^{k_2}$.

While the $2^{k_1}$ side failed, the Pochhammer side complied with the rule for all the primes up to $n=700$. Then I applied that same calculation for not only primes but any odd number, and also worked in the interval. And finally applied the same for any even number. So finally the candidate expression that works in the whole interval $[1,700]$ is the Pochhammer side of the initial expression I tested.

It would be cool to be able to test in a bigger interval (I cannot test it due to my laptop) because $E(n) = (\frac{n}{2})^{(n-1-\lfloor\frac{n}{2}\rfloor)}$ is much smaller than $n!$ so the computational cost is smaller (anyway they are semi factorials, so it takes time)

What I was not able to find yet is a power-based only expression. All of them require factorials...