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Quanyu Tang

Undergraduate student, Xi’an Jiaotong University
Email: tang_quanyu@163.com

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In fact, in my earlier comment at ''14:55 on 03 Oct 2025'' (''[Post deleted]''), I had already made essentially the same conjecture: I conjectured that when $n = 4k+2$, the maximum of $\Delta / n^n$ is attained at the $n$ points constructed as follows. Take a regular Reuleaux $(2k+1)$--gon of constant width $2$ obtained from a regular $(2k+1)$--gon, and choose the $2k+1$ vertices together with the $2k+1$ arc midpoints. In this case one can even compute the value of $\Delta$ in closed form:\[
\Delta_n
=\Bigl(\frac{n}{2}\Bigr)^{\!n}
\Bigl(\, \sec\!\frac{\pi}{n}\,\bigl(2-\sec\!\tfrac{\pi}{n}\bigr)\Bigr)^{\frac{n(n-2)}{4}}
\Bigl(\, \sec^{\,\frac{n}{2}}\!\frac{\pi}{n}
+\bigl(2-\sec\!\tfrac{\pi}{n}\bigr)^{\frac{n}{2}}\Bigr)^{\!n}.
\]However, Stijn and Yanni later found counterexamples; for instance, when $n=10$ this configuration is no longer optimal.

Thanks, that makes sense.

I have a question about one step in your argument. In the last part ("Suppose the contrary"), in Step 2) you write that\[
\text{as each element of } D'' \text{ lies in } \bigsqcup_i B_{p,i},
\text{ each element of } D'' \text{ lies in } [1,q/k].
\]I do not see why this holds. Since \(B_{p,i} = \{ p x + i : x \in \{1,2,\dots,\lfloor (q-1)/(pk) \rfloor\} \}\), the largest element of \(B_{p,i}\) is \(p\left\lfloor \frac{q-1}{pk} \right\rfloor + i\), and this number does not necessarily satisfy \[
p \left\lfloor \frac{q-1}{pk} \right\rfloor + i \le \frac{q}{k}.
\]Am I missing something here?

As a follow-up to my previous comment: in my note I included a table listing (Table 1), for each integer $k$, the OEIS sequence corresponding to\[
A(k) = \{ n \ge 1 : 2^n \equiv k \pmod n \}.
\]This table is essentially the same as the one collected on the OEIS wiki page 2^n mod n.

I am wondering whether we should include these sequences in the ''Related OEIS sequences'' part, since there are $38$ OEIS sequences in total, or instead just add a reference to the OEIS wiki page 2^n mod n, which already collects all $38$ of them.

(The site has been updated to address this comment.)

This website states that β€œErdΕ‘s and Graham report that Graham, Lehmer, and Lehmer have proved this for $k = 2^i$ for $i \ge 1$, or if $k = -1$, but I cannot find such a paper.” I have also tried to locate this manuscript, but without success.

Fortunately, I have just written a short note in which I give a complete proof of the case $k = 2^i$ for all $i \ge 1$, and survey the existing results for the set\[
A(k) := \{ n \ge 1 : 2^n \equiv k \pmod n \},
\]recording all integers $k$ for which $A(k)$ is currently known to be infinite. In short, this problem remains open for every fixed $k$ other than \[k \in \{0,-1,-2, 2^i : i \ge 1\}.\]This note is available on my GitHub page (here).

(The site has been updated to address this comment.)