| $\frac{3}{2^{2/3}} \approx 1.8898815748$ | [NS2017] | They prove $\lvert\mathcal{F}\rvert \le 3(n+1) \sum\_{k=0}^{\lfloor n/3\rfloor} \binom{n}{k} \le \left(\frac{3}{2^{2/3}}\right)^{n(1+o(1))}$ for sunflower-free $\mathcal{F}\subseteq 2^{[n]}$. |tracked since 2026-07-24 — no movement seen yet
Lower bound, last-listed row
| $\ge 1.554$ (unpublished) | [NS2017] | The arXiv preprint version of [NS2017] records $\mu^{\mathrm S}\_3\ge 1.554$, citing an unpublished manuscript of the first author. |tracked since 2026-07-24 — no movement seen yet
Cite this rowBounds Ledger — Erdős–Szemerédi $3$-sunflower-free capacity (49a)
upper bound: $\frac{3}{2^{2/3}} \approx 1.8898815748$ (tracked since 2026-07-24 — no movement seen yet)
lower bound: $\ge 1.554$ (unpublished) (tracked since 2026-07-24 — no movement seen yet)
Mirrored from teorth/optimizationproblems@2c1968c. These are the last-listed rows of that constant's bounds table — a listing position, not a statement that this bound is the strongest or most recent. A snapshot at that sha, not a live read.
https://u00dxk2.github.io/bounds-ledger/c/49a.html