OBJECTIVE AND REVIEW OF ALL VERSIONS
Erdős #769; independent attempt, 2026-09-23. Let $c(n)$ be the least threshold above which every integer occurs as the number of smaller homothetic cubes in a decomposition of the unit $n$-cube. Seek good bounds, especially whether $c(n)\gg n^n$. I read both
769.tex and
769_v2.tex completely. Their argument correctly uses the semigroup generated by $j^n-1$, but its residue threshold and initial tiling can both be sharpened.
WORK: THE EXACT THRESHOLD FOR A FIXED REFINEMENT FAMILY
Let $h$ be the least integer such that $\gcd(2^n-1,\ldots,h^n-1)=1$, put $a=2^n-1$, and let $S$ be the additive semigroup generated by those increments. For each residue $r$ modulo $a$, let $s_r$ be the smallest element of $S$ in that residue. Such elements exist because the generators span the finite cyclic group; nonnegative combinations suffice since the inverse of any residue is a nonnegative multiple of it. The largest integer outside $S$ is exactly \[
F=\max_{0\le r<a}s_r-a.
\] Indeed in residue $r$, every $s_r+ja$, $j\ge0$, is represented and no smaller nonnegative integer in that residue is represented. Thus the last omission in that residue is $s_r-a$, with negative values harmless. For every positive $u\in S$, the count $1+u$ is realizable: start with the whole cube and apply the grid refinements corresponding to a representation of $u$. Its first positive increment already subdivides the original cube into smaller cubes. Consequently \[
c(n)\le F+2.
\tag{1}
\] Set $b=h^n-1$. A shortest path to any residue using the generator residues visits no residue twice, so it has at most $a-1$ edges. Its weight is at most $(a-1)b$, and minimal-weight $s_r$ is no larger. Therefore \[
c(n)\le(a-1)b-a+2
=(2^n-2)(h^n-1)-2^n+3.
\tag{2}
\] This subtracts $2^{n+1}-3$ from the supplied bound. If $h=3$, coprimeness makes $0,b,\ldots,(a-1)b$ exactly the least representatives, giving $F=ab-a-b$ and the sharper displayed form $c(n)\le(2^n-2)(3^n-2)+1$.
ADVERSARIAL VERIFICATION AND REMAINING GAP
All arguments concern a sufficient family of grid refinements. An omitted semigroup count need not be an impossible geometric tiling count, so $F+2$ is not asserted equal to $c(n)$. For example the two-generator bound at $n=2$ is $15$, while the official entry gives $c(2)=6$. The improvement to the supplied estimate is finite and does not change its exponential scale or settle the conjectured lower bound. The residue formula is the standard numerical-semigroup argument, proved here rather than claimed new.
SOURCES CHECKED
Both complete local versions; official indexed statement and background
https://www.erdosproblems.com/history/769. All inequalities newly derived above are elementary and do not depend on the cited geometric bounds.
FINAL
UNRESOLVED. The refinement-family threshold is made exact and its elementary upper bound sharpened; the growth of $c(n)$ remains open.
COMPLETION ESTIMATE
Subjective progress toward the original asymptotic problem.
COMPLETION: 10%