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Statement
arXiv v5, p. 2 (journal p. 414): "Proposition 1.1. For any there exists such that ."
Here , is the set of permutations of , and is the set of sums of consecutive terms (p. 1), so is the site's for : the sums over , single terms included.
Source. Jakub Konieczny, On consecutive sums in permutations, arXiv:1504.07156v5 (27 August 2021), p. 2; J. Combinatorics 12 (2021), no. 3, 413--477, pp. 414--415 (the statement at the foot of p. 414, the proof on p. 415). The wording is identical in both editions. Library home: konieczny_2015_consecutive_sums_permutations.
Read depth. Claims checked: the statement and the definitions it uses were read clause by clause in the text layer of the arXiv v5 and on the journal pages. The half-page proof was read for structure and not independently checked; nothing here is independently reviewed.
Proof pointer
Page 2 (journal p. 415), half a page. Take for odd and for even , the permutation , so that for each odd . Let be the set of consecutive sums of odd length ( odd). Each has a unique representation: writing with and for odd , for even , the pair is determined by (since , ), determines the position with , and forces or , hence . So . The paper adds that the constant can be improved by a randomized variant of the construction.
Dependencies
None; the argument is self-contained.
Bears on
- Problem 34: the status-defining counterexample. The site's question asks whether for all ; the proposition gives, for every , a permutation with , so no bound with holds for all large and all . The paper (p. 3) notes that Hegyvári's 1986 construction already gives a permutation with at least distinct consecutive sums, "an analogue of Proposition 1.1 with a slightly worse constant".