Timeline for answer to Is the least common multiple sequence $\text{lcm}(1, 2, \dots, n)$ a subset of the highly abundant numbers? by Eduardo Rodríguez Golvano
Current License: CC BY-SA 4.0
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| Oct 15, 2025 at 17:29 | history | edited | Eduardo Rodríguez Golvano | CC BY-SA 4.0 |
added 19 characters in body
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| Oct 15, 2025 at 17:16 | comment | added | Eduardo Rodríguez Golvano | @GHfromMO The proof is complete now. As Terry Tao pointed out, it does not make use of the prime number theorem. | |
| Oct 15, 2025 at 17:05 | history | edited | Eduardo Rodríguez Golvano | CC BY-SA 4.0 |
Provided the details for big n. The proof is complete now.
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| Oct 10, 2025 at 14:25 | comment | added | Eduardo Rodríguez Golvano | @GHfromMO absolutely, the proof is not complete without the last part. I am trying to remember what I had thought about this. In the meantime I have edited my answer to add an alternative for big n (based on your proof) | |
| Oct 10, 2025 at 14:23 | history | edited | Eduardo Rodríguez Golvano | CC BY-SA 4.0 |
I added an alternative for big n
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| Oct 10, 2025 at 0:38 | comment | added | GH from MO | Before seeing the proof above, based on Terry's ideas, I worked out my own proof that $n!$ is not highly abundant for $n\geq 168$. In this range, by Theorem 1 in Rosser-Schoenfeld (1962), we have primes $n/3<p_1<p_2<p_3\leq n/2$. Let us also take a prime $n<q<2n$, and define the contender $M:=\frac{qrn!}{p_1p_2p_3}$, where $qr<p_1p_2p_3<q(r+1)$. The contender wins: $$\frac{\sigma(M)}{n!}>\frac{\sigma(q)}{q}\left(\prod_j\frac{p_j\sigma(p_j)}{\sigma(p_j^2)}\right)\left(1-\frac{q}{p_1p_2p_3}\right)>\left(1+\frac{1}{2n}\right)\left(\frac{n^2+3n}{n^2+3n+9}\right)^3\left(1-\frac{54}{n^2}\right)>1.$$ | |
| Oct 10, 2025 at 0:25 | comment | added | GH from MO | By the way, if one wants a quick proof for $n\geq n_0$ based on the results of Erdős-Alaoglu (1944), I recommend to look at their Corollary below Theorem 16. By this Corollary, if $n!$ is highly abundant, then every prime $q<c_4\log(n!)$ divides $n!$. So for such $n$, there is no prime strictly between $n$ and $c_4\log(n!)\sim c_4 n\log n$, which means that $n$ is bounded. The proof of this Corollary contains an annoying typo. In the second display of the proof, $\sigma(qn)/nq^{k+1}$ should be $\sigma(n)/nq\sigma(q^k)$. | |
| Oct 10, 2025 at 0:19 | comment | added | GH from MO | @EduardoRodríguezGolvano Your proof looks nice, but can you provide the details for $n>1320$? How to choose $k$ precisely, and why does it really work? Thanks in advance. | |
| Oct 9, 2025 at 18:55 | comment | added | Terry Tao | Nice! In particular one is able to handle the asymptotic regime without needing any form of the prime number theorem (as opposed to the Alaoglu-Erdos analysis, which needs variants of PNT at many places in their paper). | |
| Oct 9, 2025 at 15:54 | comment | added | Eduardo Rodríguez Golvano | @MaxAlekseyev yes, I mentioned Mersenne primes to mean the requirement could obviously be satisfied | |
| Oct 9, 2025 at 15:29 | comment | added | Matthew Bolan | This together with the OEIS data means $n!$ is highly abundant if and only if $n \le 8$. | |
| Oct 9, 2025 at 15:27 | comment | added | Max Alekseyev | "$2^k-1$ is not a Mersenne prime" can be easily enforced by taking composite $k$. | |
| S Oct 9, 2025 at 12:45 | history | suggested | G. Melfi | CC BY-SA 4.0 |
improved formatting
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| Oct 9, 2025 at 12:19 | review | Suggested edits | |||
| S Oct 9, 2025 at 12:45 | |||||
| Oct 9, 2025 at 11:36 | comment | added | Sam Hopkins♦ | As Terry Tao pointed out, this was also mentioned in the Alaoglu-Erdos paper. | |
| Oct 9, 2025 at 11:35 | comment | added | Saúl RM | Eduardo decided to create an account (this is the same person that told me that $L_{2310}$ is not highly abundant, which then led to the proof that $L_{n}$ is not highly abundant for big $n$) | |
| S Oct 9, 2025 at 11:31 | review | First answers | |||
| Oct 9, 2025 at 12:40 | |||||
| S Oct 9, 2025 at 11:31 | history | answered | Eduardo Rodríguez Golvano | CC BY-SA 4.0 |