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 divides 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 × C_c) = c·r(G), the question reduces to m = 2^j; this is Problem 5 of Buckley and MacHale (2013), who settled j ≤ 8. We give explicit groups for every j ≤ 13, so every m < 16384 occurs, and, by a computer-assisted search with exact certificates, for every j ≤ 120 except j ∈ {14, 34, 64}. Besides direct products with Frobenius groups, the new examples are semidirect products (M₁ × M₂) ⋊ 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 Mᵢ through H/Kᵢ, every element outside Kᵢ acting without non-trivial fixed points, then k(G) = Σ_S ∏_{i∈S} (|Mᵢ| − 1)·k(K_S)/|H : K_S| with K_S = ⋂_{i∈S} Kᵢ. For example (C₁₃ × C₃₃₇) ⋊ 3^{1+4}₊ has r(G) = 2²⁰, which gives j = 16. For j = 14, that is m = 16384, we show that no group in several natural families works, among them P × 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 = 2¹⁴. This is an unrefereed note. Unrefereed preprint released for independent mathematical scrutiny. Publication on Zenodo does not constitute peer review. AI-assisted tools supported research, computation, proof development, and manuscript preparation. The author remains responsible for all claims and the final text. Corpus identifier: KOU-20.52 (Kourovka Notebook, Problem 20.52, D. MacHale, 20th issue, 2022).