Hi,
I took the partial draft Zach Hunter worked on and shared with this group in 2019, fixed some issues and filled in the gaps using AI. To be clear - I did not do any mathematics, just directed the search for the proof. The result is a Lean-formalized proof of a lower bound.
S(k) ≥ k! + (k−1)! + (k−2)! + ((k−2)! − (k−2))/(k² − 3k + 1) + k − 3 for every k ≥ 3,
an improvement over the 2018 HPV bound of order (k−4)!, roughly twice the coefficient-two improvement I posted earlier.
Kernel-checked consequences: S(6) ≥ 869, S(7) ≥ 5888, S(8) ≥ 46103.
There is a PDF write-up and the Lean proof at
https://github.com/urdvr/superpermutations-hunter[AI summary]
The proof works in the weighted overlap graph on the k! permutations, where S(k) = k + L(k) for L(k) the minimum Hamiltonian-path weight. A transformation F rewires any Hamiltonian path into a covering by "exitless" paths — paths that never leave a rotation class while its weight-1 edge is still fresh — with the rewiring cost tracked by an exact ledger: one bound counts components and cheapest re-entries, a second converts this to a cost-per-vertex ratio. Everything reduces to i_k(ℓ), the minimum weight of an exitless path on ℓ vertices.
The heart is the inequality i_k(ℓ) ≥ j_k(ℓ) for an explicit comparison function j_k. An exitless path is a chain of full rotation blocks; its weight-2 continuations follow a unique door map of period k−1, and of the six weight-3 continuations after a completed run exactly one sustains another full run (period k−2). A minimum-excess path through (k−1)(k−2) blocks is thereby forced into a single deterministic chart with excess exactly k−3, and from the chart's final exit every successor of weight ≤ 3 — all nine of them — lands in a visited class, so a fresh continuation costs at least 4. A strong induction on excess (opened blocks ≤ (E+1)(k−1) − ⌊E/(k−2)⌋) then gives i_k ≥ j_k at every supported length, partial terminal blocks included. Minimizing j_k(ℓ) − q_k·ℓ, whose ceiling remainders vanish exactly at ℓ = k(k−1)(k−2), yields the coefficient q_k.
(The formalization also proves the relaxation is tight — i_k = j_k on the whole base region, attained by the greedy chart — so further improvement must come from the reduction, not from exitless paths; and that the graph model is exact: L*(k) = S(k).)
[/AI summary]
Zach saw the work, agreed to be listed as a co-author and consented to sharing it.
I believe there is slack left - an integer-capacity argument on top of this bound appears to give roughly an additional (k−5)!-order term, which I'm still verifying.
As somewhat of a meta comment - I don't want to spam the group. I'm (obviously?) not used to communicating mathematics, I don't know what is generally interesting and to what detail. So if I'm overstepping somehow or violating some community rules, please just let me know.