Timeline for Is the least common multiple sequence $\text{lcm}(1, 2, \dots, n)$ a subset of the highly abundant numbers?
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| Jan 4 at 8:35 | history | edited | Martin Sleziak |
Removed the tag (erdos) baes on the discussion on meta: https://meta.mathoverflow.net/q/5477
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| Oct 9, 2025 at 11:31 | answer | added | Eduardo Rodríguez Golvano | timeline score: 8 | |
| S Oct 6, 2025 at 19:02 | vote | accept | José Damián Espinosa | ||
| Oct 6, 2025 at 15:16 | comment | added | Terry Tao | I actually got around to reading the Alaoglu-Erdos paper at users.renyi.hu/~p_erdos/1944-03.pdf . On pages 466-467 they write "it would be easy to see that neither n! nor the multiple of the first n integers can be highly abundant infinitely often", but offer no proof. (I found this claim through an AI deep research tool, though I initially believed it to be a hallucination.) Strangely, though, I am almost glad that we didn't find this reference earlier, given the outcome! | |
| Oct 6, 2025 at 6:09 | comment | added | José Damián Espinosa | I had been thinking about it for weeks, and I never imagined that when I asked, you would be the one who answer and solve it. The most epic experience I'll ever have, thank you very much! | |
| Oct 6, 2025 at 6:00 | comment | added | José Damián Espinosa | @TerryTao, Thank you very much for taking my question seriously, you are my inspiration as you are to many many others!! | |
| Oct 6, 2025 at 5:53 | comment | added | José Damián Espinosa | Finally, it was a true honor for me to have been able to follow and witness firsthand how the most brilliant minds such as Terence Tao, Alekseyev, GH from MO, Bolan, and the entire mathematical community that was involved solved a beautiful question. I am grateful to have been part of this epic adventure and to be able to see everything that a beautiful question generates, I love mathematics more than anything! This occurred to me as an idea that suggested a certain beauty, my question indeed has a long, long story behind it. It was the most beautiful experience of my life... | |
| Oct 6, 2025 at 5:44 | comment | added | José Damián Espinosa | @RaviVakil I've asked myself other interesting questions, which I haven't been able to prove or refute on my own! But I've discovered that this is the perfect place to ask them! I would love to learn so much more and everything necessary, all the tools required to be able to refute or prove so many questions I ask myself every day, to be able to solve them on my own! | |
| Oct 6, 2025 at 5:38 | comment | added | José Damián Espinosa | Thank you very much for your congratulations AMS @RaviVakil ! I've been fascinated by every moment of this story like nothing else ever has been! My biggest dream has always been to be a mathematician! I can honestly tell you that it was the most exciting and thrilling experience I've ever had in my life! The most incredible and joyful experience I've ever had! Mathoverflow is incredible! The best site that can exist to do mathematics! I loved this story like no other! I'm so glad I dared to ask this conjecture, which, although it turned out to be false, led to this! | |
| Oct 6, 2025 at 4:22 | comment | added | Terry Tao | A followup account: mathstodon.xyz/@tao/115325228243131134 | |
| Oct 6, 2025 at 1:55 | history | protected | Sam Hopkins♦ | ||
| Oct 6, 2025 at 1:54 | answer | added | B. Mehta | timeline score: 21 | |
| Oct 6, 2025 at 1:51 | history | unprotected | David Roberts♦ | ||
| Oct 6, 2025 at 0:49 | history | protected | Sam Hopkins♦ | ||
| Oct 5, 2025 at 22:23 | comment | added | Terry Tao | @RaviVakil I have an initial account of this project at mathstodon.xyz/@tao/115316787727719049 | |
| Oct 5, 2025 at 20:49 | answer | added | Terry Tao | timeline score: 39 | |
| Oct 5, 2025 at 19:12 | comment | added | Ravi Vakil | Everything about this story, including how it is unfolding, is wonderful. I would like to congratulate everyone involved, and in particular @283212123 for the bravery of the initial post (as a new user!) which has brought joy to so many people. I really hope this story is published (ideally with the backstory) so it can be shared by a wider audience for posterity. | |
| Oct 5, 2025 at 19:01 | history | edited | Sam Hopkins♦ | CC BY-SA 4.0 |
added 76 characters in body
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| S Oct 5, 2025 at 18:59 | history | suggested | Geoffrey Sangston |
The divisors tag says to use divisors-multiples for questions related to divisors in number theory. The divisors-multiples tag mentions sums of divisors.
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| Oct 5, 2025 at 16:17 | review | Suggested edits | |||
| S Oct 5, 2025 at 18:59 | |||||
| Oct 5, 2025 at 1:24 | history | edited | GH from MO |
edited tags
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| Oct 4, 2025 at 22:33 | answer | added | Terry Tao | timeline score: 34 | |
| Oct 4, 2025 at 19:43 | answer | added | Matthew Bolan | timeline score: 17 | |
| Oct 4, 2025 at 19:00 | answer | added | Saúl RM | timeline score: 17 | |
| Oct 3, 2025 at 19:44 | answer | added | Max Alekseyev | timeline score: 27 | |
| Oct 3, 2025 at 9:56 | answer | added | Der Schutz | timeline score: 21 | |
| Oct 2, 2025 at 23:07 | answer | added | GH from MO | timeline score: 42 | |
| Oct 2, 2025 at 23:03 | vote | accept | José Damián Espinosa | ||
| S Oct 6, 2025 at 19:02 | |||||
| Oct 2, 2025 at 23:02 | vote | accept | José Damián Espinosa | ||
| Oct 2, 2025 at 23:03 | |||||
| Oct 2, 2025 at 22:38 | comment | added | José Damián Espinosa | @TerryTao, Thank you very much!! I had been thinking about the conjecture for a long time, I really believed it was true, I am surprised that it wasn't, and your impressive and elegant way of seeing and solving it. After reading your comments, I made the program that would search for the primes that met the conditions you mentioned to find the counterexample, but it was slow; an impressive honor and pleasure read you, again, thank you very much!!! | |
| Oct 2, 2025 at 20:08 | answer | added | Terry Tao | timeline score: 78 | |
| Oct 2, 2025 at 15:08 | comment | added | Terry Tao | In fact one could probably make this rigorous with some Fourier analysis and known bounds on the zeta function. In any case we have an explicit criterion for a counterexample: find distinct primes $n^{1/2} < p_j, q_m < n$ such that $0 < \sum_m \log q_m - \sum_j \log p_j < \sum_j \log(1+\frac{1}{p_j^2+p_j}) - \sum_m \log(1+\frac{1}{q_m})$. I think a computer search using primes $p_j$ near $n^{1/2}$ and $q_m$ near $n$ perhaps combined with some shortest lattice vector type methods can produce a very small positive value of $\sum_m \log q_m - \sum_j \log p_j$ that would work. | |
| Oct 2, 2025 at 14:59 | comment | added | Terry Tao | (The $(1+O(\varepsilon))^n$ should have been $(1+O(\varepsilon))^k$.) To win one needs to find $k$ primes $p_j \in (n^{1/2}, (1+\varepsilon) n^{1/2})$ and $\ell$ primes $q_m \in ((1-\varepsilon) n, n)$ such that $\sum_m \log q_m - \sum_j \log p_j$ is between $0$ and $\frac{k}{3n}$ say. Standard equidistribution heuristics suggest this becomes possible for $n$ large and $k$ a bit larger than $\log n$, and $\varepsilon$ a small absolute constant. | |
| Oct 2, 2025 at 14:46 | comment | added | Terry Tao | Sorry, I wasn't able to specify all the parameters within the comment. The $p_j$ will range between $n^{1/2}$ and $(1+\varepsilon) n^{1/2}$ for some not too small $\varepsilon$, and the $q_m$ between $(1-\varepsilon) n$ and $n$, introducing factors of $(1\pm O(\varepsilon))^n$ that can counteract the gap you mention. It's easier to see things multiplicatively rather than additively: the ratio between the products over $j$ is like $1 + (1+O(\varepsilon)) k/n$ and the ratio between the products over $m$ is like $1 + (1+O(\varepsilon)) \ell/n$, revealing the favorable gap. | |
| Oct 2, 2025 at 6:35 | comment | added | Terry Tao | I predict that this proposition in fact fails for all sufficiently large $n$. If one adds $k$ primes slightly larger than $n^{1/2}$ to $\mathrm{lcm}(1,2,\dots,n)$, and then takes away slightly more than $k/2$ primes slightly less than $n$, in such a way that the product of the primes added is very slightly smaller than the product of the primes taken away, one should end up with a quantity a little smaller than $\mathrm{lcm}(1,2,\dots,n)$ with a larger sum of divisors. It may be possible to locate a numerical example of this type by computer. | |
| Oct 1, 2025 at 22:15 | comment | added | Conrad | @Andrei you can find the first 10000 HA numbers oeis.org/A002093/b002093.txt and the first 501 LCM oeis.org/A003418/b003418.txt and indeed there is no missing LCM in the HA data; there is no simple criterion to disprove the fact that a number is HA the way there is for the SA numbers, so it may be quite difficult to settle this conjecture | |
| Oct 1, 2025 at 22:07 | comment | added | Andrei Smolensky | Somewhat surprisingly, OEIS entry for highly abundant numbers only gets up to 2100, while it is relatively easy to calculate further. I ran it up to the current maximum 2329089562800 of the LCM sequence (producing 1265 HA numbers), and the conjecture cannot be refuted by this data. Can be pushed further, though. | |
| Oct 1, 2025 at 20:06 | history | edited | GH from MO | CC BY-SA 4.0 |
fixed OEIS identifier
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| Oct 1, 2025 at 16:33 | comment | added | Will Jagy | Not likely to be easily settled. The LCM sequence is a reasonable guess here, in the long run the exponent of a prime factor $p,$ over the exponent of a prime $q,$ is roughly $\frac{\log q}{\log p}.$ the same holds for the Colossally Abundant, not so strict for the superabundant numbers, less still for the highly abundant. So: $420$ is an LCM number but not superabundant or colossally | |
| Oct 1, 2025 at 8:47 | comment | added | joro | Related question: mathoverflow.net/questions/79927/… | |
| Oct 1, 2025 at 5:00 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Oct 1, 2025 at 4:04 | comment | added | Will Jagy | @283 What is your background in mathematics? | |
| Oct 1, 2025 at 3:43 | comment | added | José Damián Espinosa | Because there is an article by Erdos and Alaoglu on a similar subject called "On highly composite and similar numbers" | |
| Oct 1, 2025 at 3:36 | comment | added | Daniel Weber | Why did you add the Erdős tag? | |
| Oct 1, 2025 at 3:24 | history | asked | José Damián Espinosa | CC BY-SA 4.0 |