Attack log — thin cyclic additive bases (Green 100 #33)

2026-08-17 — literature (start with Green #33)

Fetched, in order:

• Green, 100 Open Problems (Dec 2025 update), Problem 33: infinitely many $q$ with $A\subset\mathbb Z/q\mathbb Z$, $|A|=(\sqrt2+o(1))q^{1/2}$, and $A+A=\mathbb Z/q\mathbb Z$? Comment: “Potentially yes, but I’m not sure I know how to construct such sets.” Same question with $q=p$ prime. Then $S=\{(x,x^2):x\in A\}\cup\{(1,0)\}$ determines a line in every direction in $\mathbb F_p^2$. Counting forbids anything smaller than $(\sqrt2+o(1))p^{1/2}$. Asked by Granville (Croot–Lev, Problem 5.2) and Caprace–de la Harpe, Question 6.1 (arxiv numbering 5.e / the $n_p$ estimate).

• Croot–Lev, Open problems in additive combinatorics, CRM 43 (2007), Problem 5.2: smallest $E\subset\mathbb F^r$ determining every direction. Counting $|E|\ge\sqrt2\,q^{(r-1)/2}$. Konyagin: a difference basis $D\subset\mathbb F^{r-1}$ of size $C q^{(r-1)/2}$ gives $D\oplus\{0,1\}$ of size $2C q^{(r-1)/2}$. Constant factor only.

• Caprace–de la Harpe, arXiv:1807.04992 / Confluentes Math. 12 (2020): $n_p=\alpha(p,1)$ is the least size of $F\subset\mathbb F_p^2$ whose difference set meets every line through the origin. $n_p>\sqrt{2p}$, and Fitch–Jamison give $n_p\le 2\lceil\sqrt p\rceil+1$. Exact value open. Green is the one who pointed them at Croot–Lev.

• Terminology: a sum cover of $G$ is $A$ with $A+A=G$ (Haanpää, J. Integer Seq. 7 (2004), #04.2.6). Equivalently $\varphi(G,2)$ in Bajnok, Additive Combinatorics: A Menu of Research Problems (arXiv:1705.07444 / CRC 2018), and $\mathrm{SS}(n,2)$ in Bevan–Erskine–Lewis, arXiv:1506.04962. Bajnok’s identity: $\varphi(G,2)=\varphi(G,[0,2])+1$.

Counting (absolute, every $n$):

No perfect sum cover for $|A|\ge3$ (Haanpää Thm 2.2; Bajnok Thm B.8 for the $[0,2]$-form).

Published upper bounds I must beat or match with a new family:

Singer difference sets give $A-A=G$ at size $(1+o(1))\sqrt n$, not $A+A=G$. A symmetric Singer set of size $k>5$ is impossible ($k(k+1)/2<k(k-1)+1$). Converting a difference cover by $A\mapsto A\cup(-A)$ costs a factor 2 and lands at the elementary constant, not below $\sqrt{8/3}$.

Haanpää $n'_s(k)$ (largest cyclic order with a $k$-element sum cover): 3,5,9,13,19,21,30,35,43,51,63,67 for $k=2,\ldots,13$. Ratios $k/\sqrt n$ sit in $1.33$–$1.59$. Isolated small-$q$ tables are residue unless they imply an infinite family (house rule).

Green still open as of the Dec 2025 update. No later paper found that beats BEL’s $\sqrt{8/3}$ for infinitely many $n$, or Jia–Shen’s $\sqrt3+\varepsilon$ for all large $n$.

Tonight’s dent, in order:

1. An infinite parametric family with $|A|\le(c+o(1))\sqrt q$ and $c<\sqrt{8/3}$.
2. The Green constant $c=\sqrt2$.
3. A certified finite $q$ beating a published asymptotic density, not just Haanpää’s table.

Do not claim a dent from a handful of small-$q$ optima.

2026-08-17 — handle

BEL’s $\sqrt{8/3}$ is a 4-family product: $\mathbb Z_{r_1}\times\mathbb Z_{r_2}\times\mathbb Z_6$ with a 4-element strict sum cover $\{0,1,2,4\}$ of $\mathbb Z/6\mathbb Z$, order $6q^2$, degree $4q+O(1)$. Observation 11 of that paper caps this class at $n\le(3/8)d^2$. Beating $\sqrt{8/3}$ needs a different template.

Candidates:

• three APs $I\cup dI\cup eI$ whose three rectangles tile more than $3.375\,\ell^2$ of $\mathbb Z/n\mathbb Z$;
• Singer / Bose Sidon set plus a short repair set of size $\ge(\sqrt2-1)\sqrt n$;
• a 2-parameter algebraic family (quadratic window, discrete log of an interval, planar-function graph projected to a cyclic coordinate).

2026-08-17 — BEL replayed

Implemented Bevan–Erskine–Lewis Theorem 10(a) in the product group $T=\mathbb Z_q\times\mathbb Z_{q-2}\times\mathbb Z_6$, $q\equiv1\pmod6$, $q\ge7$, then CRT-embedded into $\mathbb Z_n$ with $n=6q(q-2)$. Connection set from Theorem 9 with $B=\{0,1,2,4\}$, $c_o=q+2$, $c_u=q$. $A=X\cup\{0\}$ has size $4q$. Product-group coverage and the independent cyclic verifier both pass for

Ratios fall toward $\sqrt{8/3}\approx1.633$. Greedy deletion drops zero points on the $q=13$ and $q=19$ certificates. The construction is rigid at these orders. This is the published infinite family, not a dent.

Haanpää 2004 Table 3 (cyclic rows, $n\le67$) independently re-verified. Those ratios sit in $1.15$–$1.59$. Isolated small $n$, residue.

2026-08-17 — named failures

1. Equal-length 3-AP $I\cup dI\cup eI$. Linear coefficient scan: $n=A\ell^2+B\ell+C$ for $A\in\{3,4,5\}$, $B\in[-4,4]$, $C\in[-3,3]$, differences linear in $\ell$. 56,700 formulas, required a cover on $\ell=6,7,8$ simultaneously, 0 hits. Wider $(d,e)$ scan for each $\ell=5,\ldots,12$ in the window $3.375\ell^2\le n\le\binom{3\ell+1}{2}$ also 0 hits. Three equal APs do not reach below BEL in this window.

2. Singer difference set as a Sidon seed. Corrected $\mathrm{Tr}_{q^3/q}$ construction: $|D|=q+1$, $D-D=G$, and $D$ is a sum packing, so $|D+D|=(q+1)(q+2)/2$ and the missed set has size $q(q-1)/2$. Interval or AP thickenings of length $\le(\sqrt{8/3}\sqrt v-|D|)$ never completed. Single-point gain on the missed set is only $\sim|D|/2$ (histogram: $q=13$, max gain 8 on $|D|=14$). Greedy therefore wants $\sim q$ extra points and lands at ratio $\sim1.7$–$1.85$, not below BEL, and the added residues have no visible pattern.

3. Bose Sidon set in $\mathbb Z/(q^2-1)$. Same picture: $|A+A|=q(q+1)/2$, interval/AP repair failed inside the BEL budget.

4. Elementary two-AP prune. Starting from $\{0,\ldots,a-1\}\cup a\{0,\ldots,a-1\}$ with $a=\lceil\sqrt n\rceil$, greedy deletion drops 0 or 1 point on $n\le210$. The construction is essentially unique-representation and cannot be thinned by 18% toward $\sqrt{8/3}$.

5. Quadratic window / geometric progression in $\mathbb F_p$. Squares of $[0,m)$ plus a short interval cover only 2 of 15 tested primes, both at ratio $>2$. Geometric progressions of counting size cover a constant fraction, not the whole field.

6. Random-style local search at sizes between counting and BEL for $n=31,57,72$: found covers at $n=31,m=9$ (ratio 1.616) and $n=57,m=12$ (ratio 1.589). Both below Haanpää's table range only for $n=57$ at a size Haanpää already permits ($n'_s(12)=63$). Isolated, not a family.

2026-08-17 — why BEL's 3/8 is the wall of this template

A directed diameter-2 generating set in $\mathbb Z_{r_1}\times\mathbb Z_{r_2}\times\mathbb Z_w$ needs a strict sum cover $B\subset\mathbb Z_w$ of size $k+2=4$. The largest such $w$ is 6 (Haanpää: $n'_{ss}(4)=6$, and $\{0,1,2,4\}$ realises it). Then $n=6r_1r_2$, $|X|=4r_1-1$, and $r_2\sim r_1$ gives $n/|X|^2\to3/8$. Larger $B$ with larger $w$ is worse: $n'_{ss}(5)=9$ gives at most $9/25=0.36<3/8$. Observation 11 of BEL is sharp for this class. Beating $\sqrt{8/3}$ needs a different coupling (finite-field mixing of the two large coordinates, or a non-AP seed whose repair tiles the missed sums). Neither was found.

2026-08-17 — residue

No infinite family with $c<\sqrt{8/3}$. No certified $q$ that beats a published asymptotic density. Green #33 remains open. The product is the BEL replay, the Haanpää replay, and the failed-search log.