Walkthrough — Thin cyclic additive bases

• Problem: problems/thin-cyclic-bases
• Quest: Green 100 #33, infinitely many $q$ with $|A|=(\sqrt2+o(1))\sqrt q$ and $A+A=\mathbb Z/q\mathbb Z$
• Model: SuperGrok CLI grok-4.6 --reasoning-effort xhigh
• Date: 2026-08-17
• Argument status: published infinite family independently replayed; no thinner family found
• Problem status: open

0. What was actually missing

The counting degree of freedom is real: unordered pairs plus doubles give at most $\binom{m+1}{2}$ sums, so any sum cover has $m\ge(\sqrt2+o(1))\sqrt q$. Difference covers already meet the companion bound $m\sim\sqrt q$ along Singer and Bose orders (Banakh–Gavrylkiv). The missing object is a sum packing that is also a cover — a Sidon-like set of size $\sqrt{2q}$ whose pairwise sums hit every residue, or a structured thickening of a known Sidon set that adds only $(\sqrt2-1)\sqrt q$ extra points.

Green’s parabola remark makes the same gap geometric: on $y=x^2$, slopes are pairwise sums. A subset of the parabola of size $\sqrt{2p}$ determines every finite direction if and only if those sums exhaust $\mathbb F_p$.

1. False starts (named obstacles)

• Convert a Singer difference set by $A\cup(-A)$. A reversible Singer set of size $k>5$ is impossible, because $k(k+1)/2<k(k-1)+1$. The symmetric hull costs a factor 2 and returns the elementary constant, not $\sqrt2$.
• Thicken Singer or Bose by an interval or a short AP. Both seeds are sum packings, so they miss about half the group. A translate of a Singer set always meets the existing sumset ($D+D-D=G$), and a single extra point covers only about $|D|/2$ missed residues. Interval/AP repairs inside the $\sqrt{8/3}$ budget never completed.
• Three equal arithmetic progressions $I\cup dI\cup eI$. Three $\ell\times\ell$ rectangles have just enough cells to cover $n\sim3.4\ell^2$ if they tile. A 56,700-formula linear scan and a wider $(d,e)$ scan for $\ell\le12$ produced no cover in the BEL-beating window.
• Delete points from the elementary two-AP cover. That cover is essentially a collision-free rectangle. Greedy deletion drops 0 or 1 point for $n\le210$. You cannot walk from constant 2 down to $1.63$ by local thinning.
• Quadratic windows and geometric progressions in $\mathbb F_p$. Near-covers at ratio $\ge2$, not thin covers.

2. The useful failure

Singer is the right shape of seed — a perfect sum packing of size $\sqrt q$ — and the first-order counting for a repair set $B$ of size $\alpha q$ says $\alpha\ge\sqrt2-1\approx0.414$ is enough if $D+B$ and $B+B$ tile the missed half. The gain histogram kills the fantasy. Each new point realises only $\sim|D|/2$ missed sums, so a hitting-set estimate wants $\alpha\sim1$ and the completed ratio drifts back to 2. Random $B$ is worse: $D+B$ hits each missed residue with probability $1-e^{-\alpha}$, and a union bound forces an extra $\sqrt{\log q}$ .

The same arithmetic explains why BEL stops at $3/8$. A diameter-2 generating set in $\mathbb Z_{r_1}\times\mathbb Z_{r_2}\times\mathbb Z_w$ needs a 4-element strict sum cover of $\mathbb Z_w$. The largest such $w$ is 6. That pair $(|B|,w)=(4,6)$ maximises $w/|B|^2$ among Haanpää’s table. Larger $B$ loses.

3. The click

There was no construction click. The organisational click was to stop treating “small $n$ below $\sqrt{8/3}$” as progress and to name the published infinite-family constant correctly: Bevan–Erskine–Lewis, Corollary 18, $\liminf \mathrm{SS}(n,2)/\sqrt n\le\sqrt{8/3}$, not Jia–Shen’s $\sqrt3+\varepsilon$ (which is the all-$n$ bound). A new bound tonight is a family with a strictly smaller liminf, or a proof that the liminf is $\sqrt2$.

4. The argument, in the order it was found

Green #33, then Croot–Lev 5.2, then Caprace–de la Harpe’s $n_p$, then Bajnok’s $\varphi(\mathbb Z_n,[0,2])$, then Jia–Shen $\sqrt3$ for every large $n$, then BEL’s directed circulant template. The last of these is the live record for infinitely many $n$.

The template, for $q\ge7$ and $q\equiv1\pmod6$:

• $T=\mathbb Z_q\times\mathbb Z_{q-2}\times\mathbb Z_6$, cyclic of order $n=6q(q-2)$;
• $B=\{0,1,2,4\}$ is a strict sum cover of $\mathbb Z/6\mathbb Z$;
• generators: the first axis, the second axis shifted by $1$, the diagonal $(t,t,2)$ of length $q+2$, and the slope-2 diagonal $(t,2t,4)$ of length $q$;
• $A=X\cup\{0\}$ has size $4q$.

Lemma 6 of BEL guarantees directed diameter 2, hence $A+A=T$. CRT writes $A$ as a subset of $\mathbb Z/n\mathbb Z$. Independently, compute/verify.py recomputes the cyclic sumset from the listed residues and checks $|A+A|=n$.

5. Computer search

Replayable:

python3 compute/make_bel_certs.py
python3 compute/haanpaa_replay.py
python3 compute/verify.py

BEL certificates through $q=61$ ($n=21594$, ratio $1.660$) all verify. Haanpää’s twelve cyclic table rows all verify. Search logs: linear_family_hits.json (0 hits), three_ap_wide.json (0 hits), sidon_repair.json (no interval/AP completion), greedy_singer.json (unstructured extras, ratio $\ge1.62$), two_ap_prune.json (almost nothing drops), algebraic_eval.json (no thin prime-field cover).

6. What is proved vs still open

Proved here: the BEL family really does give cyclic sum covers of size $(4+o(1))q$ in order $6q(q-2)$, hence $\mathrm{SS}(n,2)\le(\sqrt{8/3}+o(1))\sqrt n$ along $n=6q(q-2)$, $q\equiv1\pmod6$. This is the published statement, independently checked, not a new constant.

Still open: Green’s $\sqrt2+o(1)$ for infinitely many $q$; any liminf strictly below $\sqrt{8/3}$; Jia–Shen’s challenge of a universal constant below $\sqrt3$. Isolated Haanpää-scale tables do not move any of those.