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Statement

arXiv v5, p. 2 (journal p. 414): "Proposition 1.1. For any n≥1n\ge1 there exists a∈Sym([n])a\in\mathrm{Sym}([n]) such that ∣S(a)∣≥14n2|S(a)|\ge\frac14n^2."

Here [n]={1,2,…,n}[n]=\{1,2,\ldots,n\}, Sym([n])\mathrm{Sym}([n]) is the set of permutations a=(ai)i=1na=(a_i)_{i=1}^n of [n][n], and S(a)={∑i=uv−1ai:1≤u<v≤n+1}S(a)=\{\sum_{i=u}^{v-1}a_i:1\le u<v\le n+1\} is the set of sums of consecutive terms (p. 1), so ∣S(a)∣|S(a)| is the site's S(π)S(\pi) for π=a\pi=a: the sums ∑u≤i≤vπ(i)\sum_{u\le i\le v}\pi(i) over 1≤u≤v≤n1\le u\le v\le n, 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 ai=(i+1)/2a_i=(i+1)/2 for odd ii and ai=n+1−i/2a_i=n+1-i/2 for even ii, the permutation 1,n,2,n−1,3,n−2,…1,n,2,n-1,3,n-2,\ldots, so that ai+ai+1=n+1a_i+a_{i+1}=n+1 for each odd i<ni<n. Let S~\tilde S be the set of consecutive sums of odd length (v−uv-u odd). Each s∈S~s\in\tilde S has a unique representation: writing s=(n+1)l+ks=(n+1)l+k with l=(v−u−1)/2l=(v-u-1)/2 and k=av−1k=a_{v-1} for odd uu, k=auk=a_u for even uu, the pair (l,k)(l,k) is determined by ss (since 1≤k≤n1\le k\le n, l=⌊s/(n+1)⌋l=\lfloor s/(n+1)\rfloor), kk determines the position ww with aw=ka_w=k, and u≡w(mod2)u\equiv w\pmod2 forces u=wu=w or v−1=wv-1=w, hence (u,v)(u,v). So ∣S(a)∣≥∣S~∣=⌈(n+1)/2⌉⌊(n+1)/2⌋≥n2/4|S(a)|\ge|\tilde S|=\lceil(n+1)/2\rceil\lfloor(n+1)/2\rfloor\ge n^2/4. The paper adds that the constant 1/41/4 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 S(π)=o(n2)S(\pi)=o(n^2) for all π∈Sn\pi\in S_n; the proposition gives, for every nn, a permutation with S(π)≥n2/4S(\pi)\ge n^2/4, so no bound S(π)≤ϵn2S(\pi)\le\epsilon n^2 with ϵ<1/4\epsilon<1/4 holds for all large nn and all π\pi. The paper (p. 3) notes that Hegyvári's 1986 construction already gives a permutation with at least (1/18+o(1))n2(1/18+o(1))n^2 distinct consecutive sums, "an analogue of Proposition 1.1 with a slightly worse constant".