First exceptional length and triangular-lattice propagation

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Subsequent strengthening. The translated-triangle theorem shows that the first exceptional length must eventually exist along any hypothetical unbounded $R=1$ sequence. If every normalized squared distance were a triangular-lattice norm, distances to a good equilateral triangle would put all points in a common mesh-one lattice, contrary to the new orientation bound. The finite propagation lemma below retains its stated conditional hypothesis.

The first lemma is an exact finite statement. Its subsequent asymptotic consequences assume the hypothetical $R=1$ sequence studied in one_rare_asymptotics. Neither assertion of Problem 132 is resolved.

Closed-ball propagation

Normalize the minimum distance to one, and put

$$\mathcal N=\{\sqrt{a^2+ab+b^2}:a,b\in\mathbb Z, (a,b)\ne(0,0)\}.$$

Suppose $d=\min(D(A)\setminus\mathcal N)$ exists. A point $p$ is a patch center if $A$ contains the complete triangular-lattice graph ball of radius two about $p$, in some mesh-one triangular lattice $\Lambda_p$. Then

$$A\cap\overline B(p,d)\subseteq\Lambda_p. \tag{1}$$

Translate $p$ to zero. For $x\in A$ with $1<t=|x|\le d$, choose adjacent unit lattice vectors $u,v$ bracketing its direction. They satisfy

$$x\cdot u,x\cdot v\ge t/2>1/2, \qquad x\cdot(u+v)\ge3t/2>3/2.$$

Since $|u|=|v|=1$ and $|u+v|^2=3$, the distances from $x$ to $u,v,u+v$ are all strictly less than $t$. They therefore have integer squares $s_u,s_v,s_{u+v}$; a possible zero distance causes no problem. The identity

$$|x|^2=s_u+s_v-s_{u+v}+1$$

makes $|x|^2$ an integer. It follows also that $U=2x\cdot u$ and $V=2x\cdot v$ are integers. Now

$$3|x|^2=U^2-UV+V^2\equiv(U+V)^2\pmod3.$$

Consequently $U+V\equiv0\pmod3$, and

$$x=\frac{2U-V}{3}u+\frac{2V-U}{3}v\in\Lambda_p.$$

The cases $|x|=0,1$ follow from the patch, because minimum separation one and its six unit neighbors exclude any other unit neighbor. There are no distances in $(0,1)$. This proves (1), without a bounded-rank hypothesis or a margin below $d$.

Two consequences will be useful if a global palette argument is found. First, if $|x-y|=d$, then

$$\overline B(x,d)\cap\overline B(y,d)$$

contains no patch center. Otherwise (1) would put both endpoints in one triangular lattice, contradicting $d\notin\mathcal N$.

Second, two patch centers $p,q$ with $|p-q|<d$ have the same lattice. Choose a unit neighbor $z$ of $q$ pointing within $\pi/6$ toward $p$. Minimum separation gives $|p-q|\ge1$, so

$$|z-p|^2\le |q-p|^2+1-\sqrt3|q-p|<|q-p|^2<d^2.$$

Both $q,z$ belong to $\Lambda_p$. Their unit difference determines the orientation of a mesh-one triangular lattice, so $\Lambda_q=\Lambda_p$. The point $z$ need not itself be a patch center.

The first exceptional length is not near either end of the spectrum

Suppose $n\to\infty$ with $R(A)=1$, and let $d$ exist as above. The established exact bounded-window spectrum already makes its smallest-distance rank tend to infinity. Its largest-distance rank also tends to infinity.

Fix $k$, write $r_k$ for the $k$-th largest distance, and let $B_k$ be the union of the first $k$ convex layers. The checked layer-depth bound in layers_and_obstructions says that every edge of length at least $r_k$ has both endpoints in $B_k$. There is a unit equilateral triangle entirely outside $B_k$: almost $n/3$ interior points are degree-six lattice vertices, while the fixed layers after the second contain only $o(n)$ points, whose unit-distance neighborhood also has only $o(n)$ points. For $k=1,2$ the same triangle can be chosen outside $B_2$.

All distances incident with this triangle are strictly smaller than $r_k$. If $d\ge r_k$, their squared values are integers. Integer squared distances to a unit equilateral triangle force a point into its triangular lattice: translate one vertex to zero and apply the same $U,V$ calculation above, now using the integer squared distance to zero directly. Thus every point of $A$ would belong to one lattice, contradicting occurrence of $d$. Hence $d<r_k$ eventually for every fixed $k$.

This conclusion is conditional on existence of a nonlattice distance; it does not exclude a set lying entirely in a triangular lattice.

A growing deficit in the global palette

Another consequence of the checked shell formula is

$$\frac n2-q(A)\longrightarrow\infty \qquad\text{if }R(A)=1, n\to\infty. \tag{2}$$

Fix $M$. The $M$ distinct lattice lengths $2^j\sqrt7$, $0\le j<M$, each have at least twelve lattice-vector representations. At each of these fixed lengths the shell formula gives at least $(2-o(1))n$ unordered pairs. These are not the diameter for large $n$. Every other nondiameter class has at least $n+1$ pairs, and the diameter has at least one. Consequently

$$\binom n2\ge1+(q(A)-1)(n+1)+M(n-o(n)),$$

and therefore $q(A)\le n/2-M+o(1)$. Since $M$ was arbitrary, (2) follows. This is stronger than the elementary bound $q(A)\le n/2$, but remains compatible with $q(A)$ of order $n$.

What is not proved

The first exceptional graph avoids the large deep core but can still have its edges on the approximately $2n/3$ outer points. Its empty lenses alone do not bound its multiplicity by $n$. Even for a strictly convex set, having an interior point outside all such lenses does not force its length-$r$ graph to have at most as many edges as vertices. The minimum-graph construction already provides counterexamples to that proposed geometric shortcut.

For example, a universal estimate $m_A(r_2)\le2q(A)+C$ with an absolute constant $C$ would contradict (2) along an $R=1$ sequence. No such estimate is proved here. The tempting value $C=2$ is false: the odd regular two-ring construction on ten points has $q=6$ and second-distance multiplicity $15$. No inference from this suggested estimate, the empty lenses, or the local lattices to either target assertion is made.

The author checked the closed-ball arithmetic, the nearby-center argument, the two-sided rank conclusion, and the pair-budget deduction; the argument is author-recorded. The underlying shell and layer results retain their precise hypotheses in their linked notes.