Necessary asymptotic shape of a one-rare-distance family

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New long-scale restriction. The angular matching argument shows that every fixed $k$ satisfies $(\Delta-r_k)/\Delta=O_k(n^{-2})$ in such a sequence. This sharpens the long-scale description but does not control the intermediate palette or contradict the short lattice structure below.

Subsequent strengthening. The translated-triangle argument now shows that each lattice orientation has only $O(n^{5/7})$ good centers, so there are $\Omega(n^{2/7})$ orientations and no macroscopic minimum-distance component. It also gives $\Omega(n^{9/14})$ exceptional interior vertices and minimum pairs touching the outer two layers. These conclusions do not contradict the $o(n)$ errors below. A common lattice of mesh $\delta$ is excluded, and any common finer triangular lattice must have superpolynomial refinement. Arbitrarily fine lattices remain unexcluded. The quantitative short-incidence bound also supplies complete patches of radius $c n^{1/112}(\log(2n))^{-1/14}$ at linearly many centers.

This is a conditional consequence of the short-edge incidence inverse, not a contradiction or an answer to either unrestricted question.

Suppose $|A|=n\to\infty$ along a sequence with $R(A)=1$. Let $L_i$ be its convex layers, $h=|L_1|$, $s=|L_2|$, $\delta$ its minimum distance, and $r_2$ its second-largest distance. Then

$$h+s=(2/3+o(1))n,\qquad m_A(r_2)=m_A(\delta)+o(n)=(1+o(1))n, \tag{1}$$ $$(1/3-o(1))n\le h\le(1/2+o(1))n, \tag{2}$$

and every subsequent fixed layer has $o(n)$ points. Furthermore, all but $o(n)$ points of $I=A\setminus(L_1\cup L_2)$ have exactly six minimum-distance neighbors in $I$.

Proof of the layer proportions

Since the diameter is the unique rare distance, both $\delta$ and $r_2$ are rich for sufficiently large $n$. The distance count is at most $n/2$. The inverse theorem and circle theorem imply that the number of minimum-distance pairs touching any strictly convex subset is $o(n)$, uniformly over that subset. In particular, for every fixed $j$,

$$e_\delta(L_1\cup\cdots\cup L_j,A)=o(n). \tag{3}$$

Here the expression counts unordered pairs with at least one endpoint in the displayed union. Degenerate layers of size at most two cause no problem, since minimum-distance degree is at most six.

Put $I_j=A\setminus(L_1\cup\cdots\cup L_j)$. The same degree bound gives

$$n<m_A(\delta)\le 3|I_j|+o(n).$$

Thus $|L_1\cup\cdots\cup L_j|\le(2/3+o(1))n$ for each fixed $j$. On the other hand, the checked second-distance bound gives

$$n<m_A(r_2)\le \min\{\tfrac32(h+s),\ \tfrac43h+2s,\ 2h+s\}. \tag{4}$$

The first term proves (1); the other two, after substituting $s=2n/3-h+o(n)$, prove (2). Combining the $j=2$ equality with the upper bound for each larger fixed $j$ proves the assertion about later layers. Bound (4) is Theorem 1.3 and its proof in Section 2.2 of the supplied Clemen--Dumitrescu--Liu paper, whose full passage was checked. The $3/2$ bound also follows by applying Vesztergombi's theorem to $L_1\cup L_2$.

Exact local triangular-lattice structure

Set $I=I_2$, so $|I|=n/3+o(n)$. By (3),

$$e_\delta(I)=m_A(\delta)-o(n)\ge n-o(n)=3|I|-o(n).$$

Every vertex of this graph has degree at most six. Hence

$$\sum_{p\in I}(6-\deg_{I,\delta}(p))=o(n).$$

Only $o(n)$ vertices have degree less than six. At a degree-six vertex, the six neighbors lie on its radius-$\delta$ circle, with consecutive angular gaps at least $\pi/3$, because $\delta$ is the global minimum. The gaps sum to $2\pi$, so they are all exactly $\pi/3$. The neighbors therefore form a regular hexagon.

Call these vertices good. Each connected component of the graph induced by the good vertices lies in one translated and rotated triangular lattice of mesh $\delta$: along an edge joining two good vertices, the two regular hexagonal direction sets agree, and this propagates along paths. The lattices of different components need not agree.

There are only $o(n)$ such components. A finite component has a vertex missing at least one of its six lattice neighbors inside the component (maximize a generic linear functional). That neighbor exists in $I$, because the vertex is good, and must be bad. Each bad vertex has at most six neighbors. Charging a component to one such incidence bounds the number of components by six times the number of bad vertices. Consequently, for every fixed $M$, only $o(n)$ good vertices belong to components of size at most $M$.

Exact spectrum at every fixed short scale

Let $Q(a,b)=a^2+ab+b^2$ be the triangular-lattice norm. For every fixed $L>0$, eventually along the hypothetical sequence,

$$D(A)\cap(0,L\delta] =\delta\{\sqrt{Q(a,b)}:a,b\in\mathbb Z,\ 0<Q(a,b)\le L^2\}. \tag{5}$$

Thus this is exact equality of finite sets of lengths, not convergence of their normalized numerical values. Moreover, for each fixed represented $r>0$, put $t(r)=|\{(a,b)\in\mathbb Z^2:Q(a,b)=r^2\}|$. Then

$$m_A(\delta r)=\left(\frac{t(r)}6+o(1)\right)n. \tag{6}$$

Complete patches at most interior vertices

Let $B\subset I$ be the bad vertices, $b=|B|=o(n)$. In the internal minimum-distance graph on $I$, remove the vertices at graph distance at most $k$ from $B$. For fixed $k$, there are at most $(1+6+\cdots+6^k)b=o(n)$ removed vertices. Denote the remaining set by $H_k$.

At each $v\in H_k$, propagation of the six lattice directions along good vertices gives a complete triangular-lattice graph ball of radius $k$, contained in $I$. Choose $k\ge\lceil2L+2\rceil$. A nearest point $z$ of that lattice to any $x\in\overline B(v,L\delta)$ is within $\delta/\sqrt3$ of $x$: the lattice's equilateral triangles have circumradius $\delta/\sqrt3$. Moreover, $|z-v|\le(L+1/\sqrt3)\delta$. In axial coordinates, the path length $|a|+|b|$ is at most $2\sqrt{Q(a,b)}$, so $z$ lies in the complete graph ball and belongs to $I$.

If $x\in A$, minimum separation forces $x=z$, since otherwise $0<|x-z|\le\delta/\sqrt3<\delta$. Consequently $A\cap\overline B(v,L\delta)$ is exactly the corresponding complete lattice patch, and all its points lie in $I$.

Excluding a first nonlattice length

Suppose a distance $d\le L\delta$ not on the right of (5) occurs, and choose the least such distance. Its global smallest-distance rank is at most $N_L+1$, where $N_L=|\{Q(a,b):0<Q(a,b)\le L^2\}|$ is fixed. No $d$-edge can touch $H_k$. The short-edge inverse says that the $d$-edges touching $L_1\cup L_2$ number $o(n)$, uniformly over these bounded ranks. All other edges are among the $o(n)$ exceptional interior vertices, whose $d$-degrees are at most $6(N_L+1)$. Thus $m_A(d)=o(n)$.

Packing the disjoint radius-$\delta/2$ disks gives $n\le(2\Delta/\delta+1)^2$, so $\Delta/\delta\to\infty$. For large $n$ the distance $d$ is not the diameter and must be rich under $R(A)=1$, a contradiction. Every lattice length on the right of (5) occurs in the complete patches, since $H_k$ is nonempty. This proves (5).

For a fixed represented $r$, each deep vertex has exactly $t(r)$ neighbors at distance $\delta r$, all in $I$. Exceptional interior vertices have bounded degree, and the outer-layer incidences are $o(n)$ by (5) and the fixed-short-rank inverse. Therefore $2e_{\delta r}(I)=t(r)|I|+o(n)$; use $|I|=n/3+o(n)$ to obtain (6). In particular, the classes at $\delta,\sqrt3\delta,2\delta$ each have $n+o(n)$ pairs, and the class at $\sqrt7\delta$ has $2n+o(n)$.

Entire minimum-distance components lie in lattices

Equation (5) with $L=2$ strengthens the earlier good-component statement. Eventually each connected component of the entire minimum-distance graph on $A$ lies in one translated, rotated triangular lattice of mesh $\delta$, including its bad vertices.

Indeed, on a two-edge path $x-y-z$ with $x\ne z$, (5) gives $|x-z|^2/\delta^2\in\{1,3,4\}$. The two directed edges consequently differ in direction by $0$, $\pm\pi/3$, or $\pm2\pi/3$. Directions and lattice coordinates propagate along paths from a first edge. Isolated points can be placed in any lattice of this mesh. Separate components need not have the same lattice.

Total diameter of good components

For a good component $S$, take its own axial lattice coordinates $a,b$. Connectivity makes the occupied $a$- and $b$-values integer intervals. The boundary incidences in directions $\pm e_1$ number at least $2(\operatorname{span}b+1)$; those in directions $\pm e_2$ number at least $2(\operatorname{span}a+1)$. Hence

$$|\partial S|\ge2(\operatorname{span}a+\operatorname{span}b+2) \ge 2\operatorname{diam}(S)/\delta+4.$$

Every external lattice neighbor of a good vertex exists in $I$ and is bad, since otherwise it belongs to the same good component. Each bad vertex has at most six minimum-distance neighbors, so $\sum_S|\partial S|\le6b$. It follows that

$$\sum_S\operatorname{diam}(S)\le3b\delta=o(n\delta). \tag{7}$$

The linear-aspect-ratio theorem further gives $\Delta/(n\delta)\to\infty$. Thus these component diameters also have sum $o(\Delta)$. This does not bound the separation of their centers or force their lattices to agree.

The long-core surplus argument additionally forces $m_A(\Delta)\ge(1/6-o(1))n$. In particular, the remaining asymptotic obstruction does not have only a bounded number of diameter pairs.

What is still missing

This does not prove that the triangular-lattice components share a common lattice, that one component contains a positive fraction of all points, or that their distance palette is incompatible with the two outer layers. The $o(n)$ minimum-distance incidences on those layers can still exceed the boundary deficit of a large lattice patch. Replacing $o(n)$ by zero, or silently deducing a common lattice, would invalidate the argument.

In (6), a six-vector shell has multiplicity $n+o(n)$, not necessarily at most $n$. The error can still make every such shell rich. The exact short spectrum and the diameter estimate (7) do not control overlaps of the long pinned-distance spectra between the distributed lattice components and the outer layers. That is the remaining palette compatibility question, not a consequence of local lattice structure.

Nor does the first distance joining two minimum-graph components have to create a new palette value. For a concrete example, take the two triangular-lattice patches $T_m=\{a+b\omega:a,b\in\mathbb Z_{\ge0},\ a+b\le m\}$ and $T_m+(m+2)$, where $\omega=e^{i\pi/3}$ and $m\ge2$. Their minimum-distance components are separate, their shortest cross-distance is two, and distance two already occurs inside each patch. Bounding only the cross-pairs therefore does not bound the full multiplicity of that class. This example tests that inference, not the $R=1$ hypotheses.

Nor does near-extremality of $r_2$ on $L_1\cup L_2$ alone force a circular fraction. The even-order deformation construction attains exact $3N/2$ second-distance multiplicity with no five points concyclic. The additional global bound $q(A)\le n/2$ has not been combined with extremality to obtain the missing rigidity.

The reduction only applies to an unbounded sequence with $R=1$. It neither excludes a single finite counterexample nor treats general bounded $R>1$. The author checked the pair factors, degree-six rigidity, component count, and full source bound against these precise hypotheses; the argument is author-recorded. The exact bounded-window spectrum, nearest-lattice exclusion, fixed-shell factors, eventual whole-component propagation, and component-diameter bound were also checked by the author and are author-recorded.