Problem packetResearch packetR756
The six listed orders are the exact exception set
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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 perfect squares
Authored record and scope
- Authored title
- The six listed orders are the exact exception set
- Record type
- claim
- Stored status
- established
- Evidence grade
- sourced
- Recorded scope data
- { "kind": "universal", "statement": "every positive integer n" }
2Authored explanation
Robert Israel gave a complete construction and obstruction argument in 2012. Suppose a working permutation is known on \([r]\). Choose a square \(q\) and set \(r=q-n-1\), with \(0\leq r<n\). Keep the permutation on \([r]\), and set \(\sigma(i)=q-i\) on \([r+1,n]\). The latter interval maps to itself and every new sum is \(q\).
Strong induction uses seven base witnesses at \(n=3,5,8,9,10,13,17\). For \(n\geq12\), the square immediately above \(n\), or the following square in the six exceptional remainder cases, reduces the order to an earlier admissible one.
Hall's theorem proves failure at the six exceptions. Deficient left sets are \(\{1\}\) for \(n=1,2\), \(\{4\}\) for \(n=4\), \(\{1,6\}\) for \(n=6,7\), and \(\{4,11\}\) for \(n=11\). Their neighbor sets have sizes \(0,0,0,1,1,1\), respectively. The candidate's proposed classification is therefore a solved theorem.
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3Evidence
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Verification source: mathoverflow.net ↗, Robert Israel, MathOverflow answer 117234, 26 December 2012; construction, base permutations, and Hall obstructions
4How it connects
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"title": "The six listed orders are the exact exception set",
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}6Provenance
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