The construction in cpcrt-claim-odd-reset-upper-bound proves \(\operatorname{rt}(A_n)\leq n(n-1)/2\) for every odd \(n\). A general fiber argument gives the rigorous lower bound \(\operatorname{rt}(A_n)\geq\lceil\log_2 n\rceil\): the letter \(a\) is a permutation, every fiber of \(b\) has size at most two, and a word containing \(k\) copies of \(b\) has fibers of size at most \(2^k\). A reset word has a fiber of size \(n\). Exact breadth-first search gives equality with the quadratic upper bound at every odd \(n\leq19\). Separately weighted searches find that every reset word in this tested range uses at least \(n-1\) copies of \(b\) and at least \((n-1)(n-2)/2\) copies of \(a\). These two sharper inequalities would prove the formula in general, but this record does not supply a universal proof of them.

Write \(n=2m+1\) and define \[ W_n=b\left(\prod_{j=0}^{m-2}a^{2j+1}b\,a^{n-2j-1}b\right)ab, \] where the product is read with increasing \(j\). After the first \(b\), the active set is \(\{0,2,\ldots,2m\}\). After the block \(a^{2j+1}b\) in pair \(j\), it is \(\{2j,2j+2,\ldots,2m-2\}\). After the following block \(a^{n-2j-1}b\), it is \[ \{0,2,\ldots,2(m-j-2)\}\cup\{2m\}. \] The formulas follow by reducing the displayed even states modulo \(2m+1\) and applying \(b\). The final \(ab\) sends \(\{0,2m\}\) to \(\{0\}\). The word contains \(2m=n-1\) copies of \(b\). Its powers of \(a\) sum to \((m-1)n+1=m(2m-1)\), so its length is \(m(2m+1)=n(n-1)/2\).

Assume \(n\) is even and follow the pair \(\{0,2\}\). The letter \(a\) rotates both states and preserves their cyclic difference. Any pair at cyclic difference two has endpoints of the same parity. The letter \(b\) fixes both endpoints when they are even and subtracts one from both when they are odd. It therefore preserves the difference two as well. Induction on the word length shows that the two images are always distinct. Hence \(A_n\) has no reset word for even \(n\).

The letter \(a\) is a transitive cycle, making this a one-cluster automaton, while \(b\) is an idempotent of rank \((n+1)/2\) at odd orders. Volkov's survey collects reset-threshold results for one-cluster automata, simple idempotents, and low-rank letters. Volkov's 2019 paper constructs different slowly synchronizing families whose letters are low-rank idempotents. A focused search by the transition formulas and the value \(n(n-1)/2\) did not locate the present binary family. Novelty therefore remains unverified.

Subsets are represented by bit masks. Ordinary breadth-first search either reaches a singleton at the claimed threshold or exhausts the reachable subsets. For each odd order, two zero-one breadth-first searches assign cost one to only \(a\), then only \(b\). They reproduce the lower-bound counts \((n-1)(n-2)/2\) and \(n-1\). A direct construction check applies \(W_n\) for every odd \(n\leq101\).