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We prove L28. A dissection means a finite union of closed cubes with pairwise disjoint interiors and positive side lengths. All cubes in the construction are translates of positive scalar multiples of the unit cube.

Replacing any one tile by a $j$ by $\cdots$ by $j$ grid increases the tile count by $j^d-1$. Repeated replacements realize $1+t$ for every $t$ in the additive semigroup generated by these increments.

Put $a=2^d-1$ and $B=M^d-1$. On $\mathbb Z/a\mathbb Z$, draw directed edges adding each allowed increment. The residues reachable from zero form a submonoid of a finite group, hence a subgroup: the inverse of a reachable element is a nonnegative multiple of that element. The gcd hypothesis makes this subgroup the whole group. A shortest path to any residue has at most $a-1$ edges, since a repeated vertex could be removed. Its sum $w_r$ of integer increments therefore satisfies

$$0\le w_r\le(a-1)B,\qquad w_r\equiv r\pmod a.$$

For every integer $t\ge(a-1)B$, take $r=t\bmod a$. Then $t=w_r+qa$ with an integer $q\ge0$. The additional $a$ increments are also allowed, proving the asserted all-count threshold.

Now suppose $d\ge3$ is odd and put $M=\lfloor\sqrt{3d}\rfloor+1$. If a prime $p$ divides every increment, then $p>M$: otherwise the increment with $j=p$ is $-1$ modulo $p$. The subgroup

$$H=\{x\in\mathbb F_p^*:x^d=1\}$$

contains $1,\ldots,M$ and has at most $d$ elements, by the root bound for a nonzero polynomial over a field. Every element of $H$ is a square, because $x=(x^{(d+1)/2})^2$. In particular $p$ is odd.

If $p\le M^2$, let $r$ be the least positive quadratic nonresidue. It exists since the nonzero squares are exactly half the nonzero field elements. Thus $2\le r<p$. Put $m=\lceil p/r\rceil$ and $u=mr-p$. Primality gives $1\le u<r$, so $u$ is a square. The congruence $mr=u$ then makes $m$ a nonresidue, and hence $m\ge r$. Since $p>r(m-1)$, we obtain $p>r(r-1)$ and therefore $r<\sqrt p+1\le M+1$. The integer $r$ is at most $M$, contradicting $1,\ldots,M\subset H$.

If $p>M^2$, the reduced fractions $b/c$ with $1\le b,c\le M$ inject into $H$ modulo $p$. Indeed, equality of two residues makes $bc'-b'c$ a multiple of $p$ of absolute value less than $M^2$, so the fractions agree as rational numbers. Their number $C(M)$ exceeds $M^2/3$: by the union bound, the number of noncoprime ordered pairs is at most

$$\sum_{j=2}^M\lfloor M/j\rfloor^2 \le M^2\sum_{j=2}^{\infty}j^{-2} <\frac23 M^2.$$

For the last strict inequality, the tail from $j=6$ is less than $\int_5^\infty x^{-2}\,dx$, and

$$\frac14+\frac19+\frac1{16}+\frac1{25}+\frac15 =\frac{2389}{3600}<\frac23.$$

Thus $|H|\ge C(M)>M^2/3>d$, again a contradiction. The gcd is one.

Finally $M=\sqrt{3d}+O(1)$, so the threshold gives

$$\log c(d)\le\tfrac12 d\log d+d\log(2\sqrt3)+O(\sqrt d)$$

through odd $d$. Subtracting $d\log d$ tends to minus infinity. This proves the claimed ratio limit only along those dimensions.