Balogh–Linz–Patkós' odd-parity conjecture for intersecting k-Sperner families

Let [n]={1,,n}[n]=\{1,\ldots,n\}, let 2[n]2^{[n]} denote the Boolean lattice, and let a family F2[n]\mathcal{F}\subseteq 2^{[n]} be tt-intersecting if ABt|A\cap B|\ge t for all A,BFA,B\in\mathcal{F}, and kk-Sperner if it contains no chain of length k+1k+1. When n+tn+t is odd, set

q=n+t12.q=\frac{n+t-1}{2}.

For a fixed tt-set T[n]T\subseteq[n], define the qq-uniform tt-star and its upper shadow by

ST={S([n]q):TS},\mathcal{S}_T=\left\{S\in\binom{[n]}{q}:T\subseteq S\right\}, s(F)={A([n]s): there exists FF with FA}\nabla_s(\mathcal{F})=\left\{A\in\binom{[n]}{s}:\text{ there exists }F\in\mathcal{F}\text{ with }F\subseteq A\right\}

for F([n]r)\mathcal{F}\subseteq\binom{[n]}{r} and srs\ge r, and let

BT(t,k)=STi=1k1([n]q+i)(([n]q+k)q+k(ST)).\mathcal{B}_T(t,k)=\mathcal{S}_T\cup\bigcup_{i=1}^{k-1}\binom{[n]}{q+i}\cup\left(\binom{[n]}{q+k}\setminus\nabla_{q+k}(\mathcal{S}_T)\right).

Its size is

BT(t,k)=(ntqt)+i=1k(nq+i)(ntq+kt).|\mathcal{B}_T(t,k)|=\binom{n-t}{q-t}+\sum_{i=1}^{k}\binom{n}{q+i}-\binom{n-t}{q+k-t}.

Balogh–Linz–Patkós' conjecture. There exists a positive integer n0=n0(k,t)n_0=n_0(k,t) such that if n+tn+t is odd, n>n0n>n_0, and F2[n]\mathcal{F}\subseteq 2^{[n]} is a tt-intersecting kk-Sperner family, then

F(ntqt)+i=1k(nq+i)(ntq+kt).|\mathcal{F}|\le\binom{n-t}{q-t}+\sum_{i=1}^{k}\binom{n}{q+i}-\binom{n-t}{q+k-t}.

This is the odd-parity counterpart of the even-parity extremal problem, with the fixed-star construction providing the predicted sharp bound. The source presents it as a conjecture; no resolution is supplied there.

Sources & referencesView supporting material

Primary source

Jia-Bao Yang and Leilei Zhang, “Counterexamples to the Balogh-Linz-Patkós Conjecture”, arXiv:2607.03026 (2026).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.