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For $k \in \mathbb{N},\, k \geq 2$, define $S(k) := \operatorname{lcm}(2,\dots,k)$.

$S$ stands for "superprimorial", as it is a name I've seen come up once before and that I kinda like: it can be written as the (finite) product $S(k) = \prod_{p \in \mathbb{P}} p^{\max\{\alpha \in \mathbb{N} \,|\, p^\alpha \,\leq\, k\}} = \prod_{p \in \mathbb{P}} p^{\big\lfloor \frac{\ln(k)}{\ln(p)} \big\rfloor}$ where $\mathbb{P}$ is the set of positive prime numbers, and since the sequence of the exponents is decreasing when $p$ increases, it is a product of primorials, a "super" primorial so to speak.

Below are my two personal conjectures, based on manual realisation for smaller $k$ and a rudimentary Python program I made for bigger but still small $k$. I tried searching on this site among other $\operatorname{lcm}$ and number theory posts to see if this was answered or at least already talked about, to no avail. I'm rather new here though so maybe I missed some?

Conjectures:

  • $(1)$ For $k \neq 8$, $S(k) + 1$ and $S(k) - 1$ are squarefree, i.e. for all $p \in \mathbb{P}$, $p^2$ does not divide $S(k) \pm 1 $.
  • (EDIT: known to be false sadly) $(2)$ For all $k \geq 2$, $S(k) + 1$ and $S(k) - 1$ each have at most $\bigg\lfloor \frac{\ln(k)}{\ln(2)} \bigg\rfloor = \max\{\alpha \in \mathbb{N} \,|\, 2^\alpha \,\leq\, k\}$ prime divisors counted with multiplicity.

$(2)$ would imply, for example, that $S(2^{100}) + 1$, which is a pretty big number, would have at most $100$ prime divisors, which, while not that small for the purpose of finding prime candidates, is still really small. Plus, the fact that this concerns both $S(k) + 1$ AND $S(k) - 1$ could potentially lead towards the twin prime conjecture or something...?

$k = 8$ stands out since $S(8) + 1 = 841 = 29^2$ is not squarefree, and of course if $(1)$ holds then there's no need of talking about multiplicity in $(2)$ for $k \neq 8$. Do note that $(1)$ and $(2)$ still remain independent questions though, as in, unless the proof of either of them proves the other one, $(1)$ and $(2)$ should not imply each other (I don't think?).

Here are the questions I have in mind right now, having written those conjectures:

  • A) The kinda obvious question: do we know those assertions to be true? Or any partial version, like "true for big enough $k$", "an infinity of $k$", and so forth...? If not, maybe there are counterexamples other than $k = 8$ that I missed, which is very possible?
  • B) At the moment I'm not really knowledgeable on number theory proofs: any recommendations on results and/or techniques that might help me start? For $(2)$, I've tried upper bounding the number of divisors of $S(k) \pm 1$ by the number of numbers coprime to $S(k)$ ($S(k)$ being coprime to them, their divisors also have to be coprime to $S(k)$) due to the definition of $S(k)$ making it so that a "lot" of numbers are not coprime to $S(k)$ AND that it is "easy" to evaluate $\varphi(S(k))$ due to the multiplicativity of $\varphi$ (Euler's totient function for those unaware), and I've thought of removing the numbers between $S(k)$ and $S(k)/k$ since all of them are guaranteed to not be "divisor candidates", but besides that I'm sort of at a loss. As for $(1)$, I'm aware of the existence of the Möbius function and the fact there are quite a few results involving it, but I don't really know if that'll really help considering the challenge of the "$\pm 1$"...

(Hopefully my English is correct enough. Any advice on the form of the post itself is also welcome.)

EDIT: It would seem that conjecture $(2)$ is false by looking at $S(359) - 1$ which has $10$ prime factors (thanks a lot Peter)! Yet, it could still be interesting to know if such counterexamples are rare, or if a different upper bound could still apply, and so on, so I'll leave conjecture $(2)$'s false statement on the post.

Yajat Shamji
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Bruno B
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    We won't be able to factor $S(2^{100})+1$ , it is just too big :) But an interesting question (+1). The squarefree-conjecture holds for every prime factor $p\le 10^4$ , but there still might be well a counterexample. Also twin primes $S(k)\pm 1$ are probably very rare. Did you find some with $k>48$ ? – Peter Oct 09 '22 at 18:17
  • Apart from the freebies near the start, there's one twin prime pair at $k = 47$ it seems actually (somehow)! (Oh you just edited whoops) – Bruno B Oct 09 '22 at 18:38
  • Might take a look higher up at some point, but yeah, I'm not expecting twin primes to be any kind of common, it's probably just a happy coincidence for lower $k$ – Bruno B Oct 09 '22 at 18:39
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    $S(359)-1$ has $10$ distinct prime factors ! – Peter Oct 10 '22 at 08:53
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    If there is a counterexample for the squarefree conjecture , a prime $p$ doing the job for any $k$ must exceed $4\cdot 10^5$ , so this conjecture will be difficult to be refuted. – Peter Oct 10 '22 at 09:53
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    Extended to $10^6$ – Peter Oct 10 '22 at 12:28
  • @Peter ahhh bye-bye conjecture $(2)$ then, since $359 < 2^{10}$... Thanks though! Better to have a negative result than no result (I guess it could still hold for an infinity of $k$ or maybe for another upper bound...?) – Bruno B Oct 10 '22 at 14:14
  • As for the squarefree conjecture, if you don't mind, could you tell me how to check that? I'm not that much of a Python expert, so I just went with factorising with sympy.factorint and checking what it prints, which is of course most likely very unoptimised... – Bruno B Oct 10 '22 at 14:19
  • Well, there is only one more or less efficient way : I start with a prime $p$ , then I calculate $S(k)$ mod $p^2$ for $k=2,\cdots,p$ (note that a prime factor is always larger than $k$) by using that $S(k)$ only changes when $k$ is a prime power $q^l$ and then multiplies with $q$. If $S(k)^2\equiv 1\mod p^2$ holds , we found a solution. If $k\le 10^4$ , the conjecture even holds beyond $p=10^7$. – Peter Oct 10 '22 at 15:24
  • Let us continue this discussion in chat. – Peter Oct 10 '22 at 15:26

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