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On the Generalized Honeymoon Oberwolfach Problem

Published 5 Mar 2026 in math.CO | (2603.05736v1)

Abstract: The generalized Honeymoon Oberwolfach Problem (HOP) asks whether it is possible to seat $2n$ participants consisting of nn newlywed couples at a conference with ss tables of size $2$ and tt ''round'' tables of sizes 2m1,2m2,…,2mt2m_1, 2m_2, \ldots, 2m_t, where n=s+∑i=1<sup>t</sup>min = s + \sum_{i=1}<sup>{t}</sup> m_i with all mi≥2m_i \geq 2, over several nights so that each participant sits next to their spouse every time and next to each other participant exactly once. We denote this problem by HOP(2<sup>⟨</sup>s⟩,2m1,…,2mt)\mathrm{HOP}(2<sup>{\langle</sup> s \rangle}, 2m_1, \ldots, 2m_t). This paper is the first of two papers investigating the generalized HOP. While the second paper will deal with the generalized HOP with a single round table (i.e. table of size at least $4$), the present work develops solutions for the generalized HOP with multiple round tables. In particular, we present solutions to certain cases with two round tables, showing that a solution to HOP(2<sup>⟨</sup>s⟩,2m1,2m2)\mathrm{HOP}(2<sup>{\langle</sup> s \rangle}, 2m_1, 2m_2) exists when n≡1(mod(2m1+2m2))n \equiv 1 \pmod{(2m_1 + 2m_2)} or n≡m1+m2(mod(2m1+2m2))n \equiv m_1 + m_2 \pmod{(2m_1 + 2m_2)}. We also develop solutions for cases with small round tables, showing that HOP(2<sup>⟨</sup>s⟩,2m1,…,2mt)\mathrm{HOP}(2<sup>{\langle</sup> s \rangle}, 2m_1, \dots, 2m_t) has a solution whenever m=m1+⋯+mt≤10m = m_1 + \dots + m_t \leq 10, n=s+mn = s + m is odd, and n(n−1)≡0(mod2m)n(n - 1) \equiv 0 \pmod{2m}.

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