Groups of Odd Order with Conjugate Deficiency 16m (Kourovka Notebook Problem 20.52)

For a finite group \(G\) let \(k(G)\) be its number of conjugacy classes and \(r(G)=|G|-k(G)\). Burnside proved that \(16\mid r(G)\) when \(|G|\) is odd, and D. MacHale asked (Kourovka Notebook, Problem 20.52) whether every positive multiple \(16m\) is \(r(G)\) for a non-abelian group \(G\) of odd order. Since \(r(G\times C_c)=c\,r(G)\), the question reduces to \(m=2^j\); this is Problem 5 of Buckley and MacHale (2013), who settled \(j\le 8\). We give explicit groups for every \(j\le 13\), so every \(m\lt 16384\) occurs, and, by a computer-assisted search with exact certificates, for every \(j\le 120\) except \(j\in\{14,34,64\}\). Besides direct products with Frobenius groups, the new examples are semidirect products \((M_1\times M_2)\rtimes H\) of abelian groups by groups \(H\) of order at most \(2187\), mostly \(3\)-groups, acting through two cyclic quotients; their class numbers are given by a closed formula that we prove: if \(H\) acts on \(M_i\) through \(H/K_i\), every element outside \(K_i\) acting without non-trivial fixed points, then \(k(G)=\sum_{S}\prod_{i\in S}(|M_i|-1)\,k(K_S)/|H:K_S|\) with \(K_S=\bigcap_{i\in S}K_i\). For example \((C_{13}\times C_{337})\rtimes 3^{1+4}_+\) has \(r(G)=2^{20}\), which gives \(j=16\). For \(j=14\), that is \(m=16384\), we show that no group in several natural families works, among them \(P\times F\) for every group \(P\) of odd order and every Frobenius group \(F\) with abelian kernel and complement of prime order. The question remains open, first for \(m=2^{14}\). This is an unrefereed note.