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Proposed set and outcome
Put R={0,1,5}. Start with the seven-adic outer skeleton A7, using residues r modulo 2q on [4q^2/49,4q^2], for q=7^k and r in R. Add the following sets at each q:
common repair: a = r+7q (mod 14q), 5q^2<a<=49q^2;
even repair: a = r+jq (mod 14q), 4q^2<a<=8q^2,
j in {4,6,8,10}.Call the resulting union C. This attempted rule was motivated by the common reflected root 7q-M, and by selecting four of the six nonzero even lift classes for the small-root gap. It fails coverage, with the exact witness 26522 below. Independently, its own prefix count has upper limit at least 195/(28*sqrt(2))>9/2. Neither its boundary count nor the four-class choice supplies a repaired complement.
All calculations are author-recorded, without independent acceptance.
A middle lift omitted by the proposed routing
Take q=49, m=29, r=5, M=4q-m=167, and
t=M^2+r-2*(14q)=26522.The old root 2q+m=127 gives residual 10393. This lies in (4q^2,8q^2] but has class r+2q modulo 14q, which the even repair omits. The next smaller same-parity roots 2q-m=69 and m=29 give residuals 21761 and 25681, both exceeding 8q^2=19208. The common reflected root is 7q-M=176>sqrt(t), so its residual is nonpositive.
This is a middle-gap failure: m/q lies between 1/2 and 2/3. The two old roots that were both available in the smaller gap no longer both have residual <=8q^2. A subclass argument that checked only the small gap cannot be applied here.
All other scales and centers are excluded
We have t=6 (mod 7), so among R only r=5 is possible: t-0=6 and t-1=5 are nonresidues modulo 7, while t-5=1 is a nonzero square. This applies to every scale >=7 in C.
At scale 1, all points of the proposed set contributed by that scale are at most 49. But floor(sqrt(t))=162 and t-162^2=278>49.
At scale 7, the outer annulus ends at 196, so even the nearest square leaves too large a residual. The common repair is at most 2401 and requires n^2=8 (mod 98). Its roots are 20 and 78 modulo 98. The largest eligible n below sqrt(t) is 118, leaving residual 12598>2401. For the even repair, a residual in (196,392] permits only n=162. Its residual 278 is 82 modulo 98, whereas the allowed center-5 classes are 33,47,61,75.
At scale 49, the outer class requires n=29 or 69 (mod 98). The largest eligible root below sqrt(t) is 127, leaving 10393>9604. For the even repair, parity restricts the eligible roots modulo 49 to 29,69,127. The first two residuals exceed 19208; the last has the omitted class r+2q. The common class has bounded root 176 and fails as above.
At scale 343, the outer bounded root is 167>sqrt(t). By the fixed-center monotonicity argument, every later bounded root for the same center is at least 167. All repairs from scale 343 onward start above 4*343^2=470596>t. Outer annuli from scale 2401 onward also start at least at 470596. This exhausts all scales and proves that t is not a square plus an element of C.
The count also fails between the boundaries
For large q, the common repair contributes (66/7)q+O(1) points and the even repair (24/7)q+O(1). Their classes are distinct. The repair supports at different scales are disjoint, and none overlaps the outer skeleton except for bounded initial exceptions.
Through 4q^2 the outer skeleton contributes (48/7)q+O(log q), and the earlier repairs contribute (15/7)q+O(log q). Thus
C(4q^2)=9q+O(log q).On (4q^2,5q^2], the next outer annulus and the four even repair classes give density 15/(14q). On (5q^2,8q^2], the common repair raises the density to 18/(14q). Consequently
C(8q^2) = [9 + 15/14 + 3*(18/14)]q+O(log q)
= (195/14)q+O(log q),and hence
limsup C(X)/sqrt(X) >= 195/(28*sqrt(2)) > 9/2.The inequality is strict because 195^2=38025 is greater than (126*sqrt(2))^2=31752. This count failure does not depend on whether the proposed routing could be repaired. It concerns C as a set.
What changes the next attempt
The two defects require different repairs. Coverage needs an option for middle lift positions where only one old residual is short enough. The counting target requires less mass near the first added bands, or a proved sharing rule with smaller scales. Enlarging the present intervals would not correct its already excessive prefix count.