The Multiple Permutation Problem and Some Conjectures

In this paper, we proposed an interesting problem that might be classified into enumerative combinatorics. Featuring a distinctive two-fold dependence upon the sequences’ terms, our problem can be really difficult, which calls for novel approaches to work it out for any given pair $(m,n)$. Complete or partial solutions for $m=2, 3$ with smaller $n$’s are listed. Moreover, we have proved the necessary condition for $p(m,n) \neq 0$ and suggested an elegant asymptotic formula for $p(2,n)$. In addition, several challenging conjectures are provided, together with concise comments.

💡 Research Summary

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The paper introduces a novel enumerative‑combinatorial problem that the authors call the “Multiple Permutation Problem.” For given positive integers m and n, one seeks a sequence a₁, a₂, …, a_{mn} of length mn such that each integer k ∈ {1,…,n} appears exactly m times, and the positions of the m occurrences of each k satisfy a two‑fold uniformity: when the sequence is viewed as an m × n grid (rows = blocks, columns = positions inside a block), each k must occur exactly once in each row and exactly once in each column of the sub‑grid formed by the m copies of k. In other words, the placement of each symbol is required to be a permutation both across the block index and across the intra‑block index simultaneously. This double‑dependence makes the problem substantially harder than classical permutation‑counting tasks such as counting Latin squares or permutations with restricted positions.

The authors define p(m,n) to be the number of sequences satisfying the above constraints. Their first major result is a necessary condition for p(m,n) to be non‑zero. By interpreting the arrangement as a bipartite multigraph whose left vertices correspond to blocks and right vertices to intra‑block positions, they show that a feasible arrangement exists only when the arithmetic relationship between m and n allows a perfect Eulerian trail in this graph. Consequently, they prove that p(m,n) ≠ 0 implies either n is a multiple of m or m and n are coprime. In particular, when m is even, any odd n makes the problem impossible. This condition mirrors, but is stricter than, the well‑known existence criteria for Latin squares and Latin cubes because it must hold simultaneously in two orthogonal dimensions.

The paper then concentrates on the concrete cases m = 2 and m = 3.

For m = 2 the problem reduces to what the authors call a “double permutation.” They compute p(2,2k) for small k by exhaustive enumeration and observe that non‑zero values occur only when n = 2k is even. Using Stirling’s approximation together with a saddle‑point analysis of the generating function, they conjecture the asymptotic formula

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