Problem packetResearch packetR805
Orders 1 and 4 are the exact exceptions
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Recorded status: established
Recorded scope: every positive integer n
Complete recorded scope and conditions
{
"kind": "universal",
"statement": "every positive integer n"
}Originating problem: Permutations whose index sums are always triangular numbers
Authored record and scope
- Authored title
- Orders 1 and 4 are the exact exceptions
- Record type
- claim
- Stored status
- established
- Evidence grade
- sourced
- Recorded scope data
- { "kind": "universal", "statement": "every positive integer n" }
2Authored explanation
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.
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3Evidence
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Verification source: arxiv.org ↗, Zhi-Wei Sun, On permutations of {1,...,n} and related topics, Theorem 1.1, pages 894-897; strengthened argument recorded here
4How it connects
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"body": "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\\).\n\nWorking 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.\n\nOrder 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.",
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}6Provenance
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