Monotone path shells

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This construction retains exponentially many choices at each scale while proving balance and coverage. It does not give a uniform representation bound. The exact remaining condition concerns simultaneous interactions with all earlier blocks, rather than any one pair of blocks.

A flexible finite cover

For an integer $q\ge7$, choose $h_q(0)=0$ with increments in $\{1,2\}$ through index $4q$. Put $D=\{0,1,2,3\}$ and

$$A_q=\{q^2+qx+h_q(x):0\le x\le4q\}+D, \quad B_q=\{q^2+qy-h_q(y):0\le y\le4q\}+D. \tag{1}$$

Both cardinalities are $4(4q+1)$. Their corresponding ranks differ by at most $16q$, and

$$r_{A_q,B_q}\le272, \qquad [3q^2-q,9q^2+q]\cap\mathbb Z\subseteq A_q+B_q. \tag{2}$$

Indeed, consecutive bare points have gaps at least $q-2>3$, so the four-point clusters are disjoint and their ordering agrees on both sides. The rank discrepancy is $2h_q(x)\le16q$. A bare sum is $2q^2+qs+h_q(x)-h_q(s-x)$. Its correction lies in $[-8q,8q]$, so a fixed target permits at most seventeen integer values of $s$. For each $s$, the correction increases strictly with $x$, in steps between two and four. Hence there is at most one representation for that $s$; the sixteen shift pairs give the bound in (2).

For $q\le s\le7q$, the admissible $x$-interval has length at least $q$. The correction has symmetric endpoints $-\Delta_s$ and $\Delta_s$, with $\Delta_s\ge q$. Adding $D+D=[0,6]$ fills every gap between its successive values. Thus the entire interval $[2q^2+qs-q,2q^2+qs+q]$ is covered. These intervals overlap for consecutive $s$, proving (2).

A schedule with controlled pairwise interactions

Use the increasing sequence

$$\mathcal Q=\{2^k,3\cdot2^{k-1}:k\ge4\}.$$

Its successive ratios are $3/2$ and $4/3$, and for $p<Q$ in this sequence, $p/\gcd(p,Q)\le3$.

For arbitrary increment words in (1), every mixed block pair has an absolute representation bound. To prove this for a fixed shift pair in $A_Q+B_p$, compare two representations. Their large-block indices differ by at most $4p^2/Q$, since a small bare block has span at most $4p^2$. Consequently the correction $h_Q(x)-h_p(y)$ ranges over an interval of length at most $16p$. The base quantity $Qx+py$ is a multiple of $g=\gcd(p,Q)$, so at most $16p/g+1\le49$ such base values are possible. Along each line $Qx+py=\text{constant}$, the correction is strictly monotone. There is at most one solution on each line.

For $A_p+B_Q$, the same comparison uses large-block gaps at least $Q-2$. The large index range is at most $(4p^2+8p)/(Q-2)\le8p^2/Q$, giving correction range at most $24p$. Thus at most $24p/g+1\le73$ base values occur, again with one solution on each line. Restoring the sixteen shift pairs gives a uniform bound of $1168$ for both orientations.

Balance of the full unions

To control deduplication within a component, further require each increment word to contain neither $111$ nor $22$. Every window of length $t$ then satisfies

$$\tfrac43t-1\le h_q(i+t)-h_q(i)\le\tfrac32t+\tfrac12. \tag{3}$$

There are still exponentially many choices: arbitrary concatenations of the two length-five words $12112$ and $11212$, followed by an alternating final segment if necessary, are admissible.

That particular entropy source is excluded: it fixes $h_q(5t)=7t$, and the resulting progression cores already force representation peaks of every fixed height on a set of positive lower density. Sparse deletions and uniformly bounded perturbations do not remove that obstruction. The broader no-$111$, no-$22$ family is not excluded by that proof.

An equal-index obstruction rules out taking the paths from one common infinite initial word, however nonperiodic, and more generally from finitely many uniformly approximated prefix germs. This argument is independent of the progression anchors above. Opposite scale-dependent translations can remove that exact resonance, but no full packing bound is known for them either. The arbitrary-offset affine obstruction excludes $h_q(t)=t$ even with arbitrary opposite translations $o(q^2)$; genuinely nonlinear translated paths are not covered.

Put $A=\bigcup_{q\in\mathcal Q}A_q$ and $B=\bigcup_{q\in\mathcal Q}B_q$. Same-side intersections of distinct blocks have bounded cardinality. Here are the details needed to justify that claim. All blocks lie in $[q^2,6q^2]$, so only $Q/p\le\sqrt6$ need be considered. Fixing the two cluster shifts, an equality between $A_Q$ and $A_p$ has base quantity $Qx-py$ in an interval of length $8(Q+p)+6$. This interval contains only an absolute number of multiples of $g$, since $p/g\le3$ and $Q/p\le\sqrt6$.

On one base line, the solutions have steps $(x,y)\mapsto(x+p/g,y+Q/g)$. For two line parameters separated by $u\ge1$, (3) gives

$$\Delta\bigl(h_Q(x)-h_p(y)\bigr) \le-\frac5{18}\frac pg\,u+\frac32,$$

using $Q/p\ge4/3$. In particular, a fixed correction can occur at at most six line parameters. The corresponding argument for the two $B$-blocks reverses the sign. Summing over the finitely many base lines and cluster shifts proves the claimed intersection bound.

Each block overlaps numerically with only an absolute number of other blocks. Hence replacing the sum of block counting functions by the union counting function costs only $O(\log x)$. For one block, the rank discrepancy bound $16q$, together with bare gaps at least $q-2$, gives $A_q(x)-B_q(x)=O(1)$ uniformly in the cutoff. Consequently

$$A(x)-B(x)=O(\log x),\qquad A(x),B(x)=\Theta(\sqrt x). \tag{4}$$

The upper growth bound follows by summing $O(q)$ over the geometric scales. For the lower bound one complete block of scale comparable to $\sqrt x$ suffices.

Both unions have gaps $O(\sqrt x)$ near $x$. Choose the largest $q\in\mathcal Q$ with $q^2\le x$. Then $x<(3q/2)^2$, whereas each bare block extends beyond $5q^2-8q\ge(9/2)q^2$. Its consecutive gaps are at most $q+2$. Combining this gap bound with (4) yields matched-rank displacement $O(\sqrt x\log x)=o(x)$, and hence $a_n/b_n\to1$.

Coverage and the unresolved uniform bound

The bands in (2) overlap on this scale schedule, proving that $A+B$ is cofinite. A representation of $n$ whose maximal block scale is $Q$ has $Q^2\le n\le12Q^2$. There are only an absolute number of such maximal scales. Summing the fixed block-pair bounds over the $O(\log n)$ smaller scales proves only

$$1\le r_{A,B}(n)=O(\log n)\quad(n\text{ sufficiently large}).$$

This estimate is not uniform and does not answer Problem 1145.

The exact-resonance construction, proved as L126 at tier 0, shows that this logarithmic order is attained by an admissible infinite family. For any prescribed path at a dyadic $Q$, admissible smaller paths can give $\gg\log Q$ distinct representations of $3Q^2+h_Q(2Q)$, with exact carries. This disproves a uniform bound derived solely from the local path restrictions. It does not exclude a globally coordinated choice satisfying (5).

Locally sparse paired pruning preserves coverage after enlarging the clusters, but no bounded-capacity choice of such masks has been proved. Independent constrained paths, including independent uniform central bridges, have unbounded representations even after arbitrary $o(\sqrt x)$ deletions. Consequently any successful alteration in that setting must delete a nonvanishing proportion or use a globally coupled choice of paths.

Let $P^A_{<Q}=\bigcup_{q<Q}A_q$, and similarly for $B$. The exact additional condition that would complete this construction is

$$\sup_{Q,n}\left( r_{A_Q,P^B_{<Q}}(n)+r_{P^A_{<Q},B_Q}(n)\right)<\infty. \tag{5}$$

The same maximal-scale argument shows why (5), together with (2), would suffice. Earlier blocks with $6q^2\le Q$ collectively cost at most eight representations in either orientation: all their points lie in $[0,Q]$, and an interval of length $Q$ meets at most two four-point clusters of the large block. The unresolved range is $\sqrt{Q/6}<q<Q$, containing an unbounded number of scales.

An exact carry formulation may help retain the issue. For dyadic $Q=2^k$, $q=2^j$, $j\ge k/2$, set $m=Q/q$. In $A_Q+B_q$, write $y=mt+s$, $0\le s<m$, and $n-Q^2-q^2=QV+R$, $0\le R<Q$. Then a representation is equivalent to

$$x=V-t-c,\qquad R=qs+h_Q(V-t-c)-h_q(mt+s)+d+e-cQ, \tag{6}$$

with the original index ranges, $d,e\in D$, and an absolutely bounded integer $c$. What is missing is a choice of the paths that makes these tests, over all smaller scales, overlap only a bounded number of times. No such coding, existence theorem, or invariant has been established. Separate pairwise bounds and random-looking tests do not prove (5).

Limitation of an exact-grid scale-gap proposal

One proposed way to separate smaller scales was to force every correction branch onto an exact arithmetic grid, with stride $3m/2$ for scale ratio $m$. For fixed $V,s,c$ in (6), put $G_t=h_Q(V-t-c)-h_q(mt+s)$. Requiring $G_t-G_{t+1}=3m/2$ forces

$$\frac{3m}{2}-\bigl(h_q(m(t+1)+s)-h_q(mt+s)\bigr) =h_Q(V-t-c)-h_Q(V-t-c-1)\in\{1,2\}.$$

Imposing these deficits on all aligned blocks at all relevant dyadic scales is inconsistent.

Precisely, let $m$ be even. If four consecutive aligned blocks satisfy

$$\frac{3m}{2}-\bigl(h(i+(j+1)m)-h(i+jm)\bigr) \in\{1,2\}\qquad(0\le j<4),$$

their sum gives

$$6m-\bigl(h(i+4m)-h(i)\bigr)\ge4.$$

The union therefore cannot also have its deficit in $\{1,2\}$. This rules out only imposing those exact-grid deficits simultaneously at all scales, not general nonlinear paths or branch-dependent routing. The scale-gap tag formulation retains a possible bounded-overlap objective, but has no proved integer realization.