The asymptotic generating-triple count conjecture for prefix reversals

About 1 year old · traced to

For nn and integers m,km,k with 2⩽k<m<n2\leqslant k<m<n, let f(n)f(n) be the number of pairs (m,k)(m,k) for which the prefix reversals rn,rm,rkr_n,r_m,r_k generate the symmetric group Sym⁡n\operatorname{Sym}_n. Generating-triple count conjecture. As nn tends to infinity,

f(n)={113n2+O(n)n≡0(mod4),19n2+O(n)n≡1(mod4),110n2+O(n)n≡2(mod4),17n2+O(n)n≡3(mod4).f(n)=\begin{cases} \frac{1}{13}n^2+O(n)&n\equiv0\pmod{4},\\[6pt] \frac{1}{9}n^2+O(n)&n\equiv1\pmod{4},\\[6pt] \frac{1}{10}n^2+O(n)&n\equiv2\pmod{4},\\[6pt] \frac{1}{7}n^2+O(n)&n\equiv3\pmod{4}. \end{cases}

This is based on computational experiments for 6⩽n⩽706\leqslant n\leqslant70; the source provides no proof or resolution.

References

Primary source

Saúl A. Blanco, Mikhail P. Golubyatnikov, Elena V. Konstantinova, Natalia V. Maslova and Luka A. Nikiforov, “Generating the symmetric group by three prefix reversals”, arXiv:2511.16959 (2025).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.