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    $\begingroup$ Nice! I think you mean "not highly abundant" in the first sentence. The desired claim should follow from a combination of the prime number theorem in arithmetic progressions and the Brun-Titchmarsh inequality en.wikipedia.org/wiki/Brun%E2%80%93Titchmarsh_theorem to subtract off multiples of $p^2$, perhaps after passing to a finer congruence class to dodge multiples of very small squares such as $2^2$. By the way, this opens up the possibility of a computer assisted proof that $L_n$ is not highly abundant for all $n \geq 71$. $\endgroup$ Commented Oct 4, 2025 at 19:17
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    $\begingroup$ @TerryTao: In fact, $L_{148}$ ($=L_{139}$ if we restrict $n$ to ptime powers) is highly abundant. I can suggest it as a candidate for the largest one. (I'm searching for a larger one now with no success.) $\endgroup$ Commented Oct 4, 2025 at 22:13
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    $\begingroup$ Ops. $L_{169}$ is also highly abundant. $\endgroup$ Commented Oct 4, 2025 at 22:18
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    $\begingroup$ @TerryTao We do not need Brun-Titchmarsh. In fact by elaborating on Saúl RM's nice argument, I prove in my "Third response" that $L_n$ is not highly abundant if there exists a prime $p$ such that $p-1$ is divisible by $2310$ and lies in the interval $[(120/121)n,n-1]$. This will quickly give an effective upper bound for the largest $n$ such that $L_n$ is highly abundant. $\endgroup$ Commented Oct 4, 2025 at 23:15
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    $\begingroup$ After thinking more about it, I am not sure how to use Brun-Titchmarsh to prove the claim (probably just my lack of experience with this kind of argument), so for now I'll return the answer to its previous state. Hopefully it won't matter thanks to @GHfromMO's new arguments. If anyone wants to edit the answer proving the claim anyways, that would be great though, it seems like an interesting claim on its own $\endgroup$ Commented Oct 5, 2025 at 0:11