In this post we denote the Dedekind psi function as $\psi(m)$ for integers $m\geq 1$. This is an important arithmetic fuction in several subjects of mathematics. As reference I add the Wikipedia Dedekind psi function. On the other hand I add the reference that Wikipedia has the article Fermat number, $F_l=2^{2^l}+1$ and that I was inspired in the results showed in page 101 of [1].

The Dedekind psi function can be represented for a positive integer $m>1$ as $$\psi(m)=m\prod_{\substack{p\mid m\\p\text{ prime}}}\left(1+\frac{1}{p}\right)$$ with the definition $\psi(1)=1$.

Question. I would like (what work can be done about it) if one can to deduce some claim about the behaviour of $$\frac{\psi(F_m)}{F_m}$$ as $m\to \infty.$ Many thanks.

If the question is in the literature please answer the question as a reference request and I try to search and read the results from the literature. If the question is very difficult I ask about the behaviour or heuristic for the quotient $\frac{\psi(F_m)}{F_m}$ for very large integers $m\geq 1$.


[1] Michal Krizek, Florian Luca, and Lawrence Somer, 17 Lectures on Fermat Numbers, CMS Books in Mathematics, Springer (2001).

  • $\begingroup$ This post is crossposted from Mathematics Stack Exchange, same title and identificator 4430576 asked two months ago. I've edited the question and few minutes after I got a downvote in Mathematics Stack Exchange. The guy who downvote the question isn't interested neither in respect the ownship of the post (that is the site MSE) nor in the question. Thus I'm going to ask the question here and delete the post from MSE. (My apologizes for this polemics, but it seems to me unfair) $\endgroup$
    – user142929
    Commented Jun 17, 2022 at 17:48
  • 1
    $\begingroup$ Unless the downvoter left a comment, you have no idea who the downvoter is, and you know nothing about why the user cast the downvote. It could be as simple and innocent as a slip of the finger. In any event, it's one downvote from one of thousands of users of that site and, in my opinion, a huge over-reaction on your part; moreover, the matter should be taken up on the site where it happened (or dropped entirely), not here. $\endgroup$ Commented Jun 18, 2022 at 3:01
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    $\begingroup$ In case anyone is interested (and has the necessary number of points), the deleted stackexchange post is math.stackexchange.com/questions/4430576/… $\endgroup$ Commented Jun 18, 2022 at 3:05
  • $\begingroup$ I'm sorry professor @GerryMyerson but my feeling is that some posts (some of them that I've flagged due the feeling that I got) are being downvoted in an unfair way (the content of my flags is weird in this case because the feeling that I have, including legal threats). I think that this is not the forum to add more about it, thus I done it, now. $\endgroup$
    – user142929
    Commented Jun 18, 2022 at 9:47

1 Answer 1


It is standard that all prime factors of Fermat number $F_m$ are of the form $2^{m+2}k+1$, in particular they are all at least $2^m$. It is then also clear that $F_m$ can only be divisible by at most $2^m/m$ such primes. Therefore $$\frac{\psi(F_m)}{F_m}=\prod_{p\mid m}\left(1+\frac{1}{p}\right)<\left(1+\frac{1}{2^m}\right)^{2^m/m}=\left(\left(1+\frac{1}{2^m}\right)^{2^m}\right)^{1/m}<e^{1/m}.$$ Certainly tighter estimates are possible, but this alone shows the limit of this quantity is equal to $1$.

  • $\begingroup$ Many thanks for your nice answer, I'm going to wait, before I'm accept your anwer, according your words that tighter estimates are possible if some of your colleagues (professors) wants to add some answer, but your work is excellent. I want to thank you for your contributions, many thanks. $\endgroup$
    – user142929
    Commented Jun 18, 2022 at 9:32
  • $\begingroup$ After a week I'm accepting your excellent answer. Many thanks. $\endgroup$
    – user142929
    Commented Jun 24, 2022 at 13:52

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