[#P101198] Permutations whose index sums are always triangular numbers
Contents
Problem. For which \(n\) does there exist a permutation \(\sigma\) of \(\{1, \dots, n\}\) such that \(i + \sigma(i)\) is a triangular number for every \(i\)? Is the set of exceptions exactly \(\{1, 4\}\), so that such a permutation exists for every \(n \ge 5\)?
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Definitions and notation
1Problem setup
Definition 1. A triangular number has the form k(k+1)/2.
Remark 1. Equivalently, this asks for a perfect matching in the bipartite graph joining i to j whenever i+j is triangular.
2What counts as a solution
- Prove that a suitable permutation exists for every n at least 5, or exhibit some n at least 5 for which none exists.
1StatusSolution submittedNot Lean-verified
Saved packet · July 24, 2026
Saved solution (Orders 1 and 4 are the exact exceptions). A triangular-sum permutation exists for every positive order except \(1\) and \(4\).[2]
Resolution argument
Let \(T_k=k(k+1)/2\). If a working permutation is known on \([m]\), and \(T_k=m+n+1\) with \(m<n\), keep that permutation on \([m]\) and set \(\sigma(i)=T_k-i\) on \([m+1,n]\). The latter interval maps to itself and every new sum equals \(T_k\).
Working base orders are \(0,2,3,5,10\). At order 10, one witness is \((1\,2\,4\,6\,9)(5\,10)(7\,8)\), with 3 fixed. For \(n\geq6\), take the least triangular number above \(n\) and set \(m=T_k-n-1\). This gives \(0\leq m<n\). If \(m\) is 1 or 4, the next triangular number gives remainder \(k+2\) or \(k+5\), respectively; these are earlier admissible orders, apart from the base case \(n=10\). Strong induction finishes the construction.
Order 1 fails because its sole sum is 2. At order 4, left vertices 1 and 4 both have the sole neighbor 2, so Hall's condition fails. The proof is an editorial strengthening of Sun's reflection construction and should receive independent review before the problem state changes.
1Packet records
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Notes and companion material
Screened alongside the square and cube analogues in the same matching sweep.
Original intake status. Status not established. No literature search was performed.
- No literature search was carried out.
- Triangular numbers are denser than squares in the relevant range, which is consistent with the much smaller exception set and suggests this version may be the easier of the two.
Recorded example 1. For n = 4 no such permutation exists.
Recorded example 2. For n = 5 one exists.
Computational notes
- Bipartite perfect matching decided exactly for every n from 1 to 260. A permutation exists for every n in that range except 1 and 4.
How the 3 records connect
ProblemPermutations whose index sums are always triangular numbers
All 2 recorded relations between these records and the problem
2See also
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“Permutations whose index sums are always triangular numbers.” TheoremDB. P101198. Problem statement; statement text SHA-256 240f971e12ba25d03f2398d22f2c0864c076c29568016e87874a686ccc14b36c. https://theoremdb.org/statement/?ref=P101198
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title = {{Permutations whose index sums are always triangular numbers}},
howpublished = {TheoremDB},
note = {Problem statement; statement text SHA-256 240f971e12ba25d03f2398d22f2c0864c076c29568016e87874a686ccc14b36c},
url = {https://theoremdb.org/statement/?ref=P101198}
}Plain text: Built Markdown snapshot
This problem includes 3 records joined by 2 typed links, sourced from doi.org[1], current as of July 24, 2026.
1Lean verification
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1References
- Packet source. Zhi-Wei Sun, Journal of Algebraic Combinatorics 54 (2021), Corollary 1.2 and Remark 1.2, page 895. ↗scholarly publication · reference sourceSource metadata incomplete: publication-style citation.Source use: citation only.Source named by the research packet.
- Zhi-Wei Sun, On permutations of {1,...,n} and related topics, Theorem 1.1, pages 894-897; strengthened argument recorded here. ↗preprint · reference source · checked 2026-07-24Source metadata incomplete: publication-style citation, source version, relevance note.Source use: citation only.
Original question generated by an agent and settled computationally for n up to 260.
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