Editing Openai/692c9ea1-b3cc-800e-94da-4224ac78a36a (section)

==== 4. Where the real BEGL difficulty is ====

The original BEGL paper works with sequences of pure powers

Pow⁡(A;s):=sorted {ak:a∈A, k≥s}\operatorname{Pow}(A;s) := \text{sorted }\{a^k : a\in A,\ k\ge s\}Pow(A;s):=sorted {ak:a∈A, k≥s}
for some s≥1s\ge1s≥1 and a set AAA of integers >1>1>1. They conjectured (roughly) that for finite AAA, Pow⁡(A;s)\operatorname{Pow}(A;s)Pow(A;s) is complete iffRivMat<ref>{{cite web|title=RivMat|url=https://www.rivmat.unipr.it/fulltext/2004-3s/pdf/16.pdf|publisher=rivmat.unipr.it|access-date=2025-12-01}}</ref>
# ∑a∈A1a−1≥1\displaystyle \sum_{a\in A} \frac{1}{a-1}\ge1a∈A∑a−11≥1, and
# gcd⁡{a:a∈A}=1\gcd\{a:a\in A\}=1gcd{a:a∈A}=1.

Melfi later gave counterexamples for infinite AAA, showing that this “if and only if” cannot hold in full generality, though the finite-AAA case remains open.RivMat<ref>{{cite web|title=RivMat|url=https://www.rivmat.unipr.it/fulltext/2004-3s/pdf/16.pdf|publisher=rivmat.unipr.it|access-date=2025-12-01}}</ref>

Crucially, BEGL’s hard version of your problem:
* removes the term 1=a01=a^01=a0 from the allowed powers (i.e. uses s≥1s\ge1s≥1), and
* adds a gcd condition instead.

Once you throw away all the 1’s, the neat Brown argument above breaks: the sequence no longer starts at 1, and the initial partial sums are too small. One could try to “shift” or rescale the sequence and reapply Brown, or to extract a clever subsequence that does satisfy Brown’s inequality, but that becomes just as hard as the original conjecture.

So, to answer your original meta‑question:
* Yes, Brown’s criterion absolutely attacks (and essentially solves) the version of the problem you wrote down, where 1’s are allowed. You take the union of all powers of the did_idi, check Brown’s inequality via the sum ∑1/(di−1)≥1\sum 1/(d_i-1)\ge1∑1/(di−1)≥1, and then group by base to get the desired “digits 0/1” representation.
* For the sharper BEGL version (no 1’s, gcd condition, completeness of Pow⁡(A;s)\operatorname{Pow}(A;s)Pow(A;s) for s≥1s\ge1s≥1), Brown’s criterion in this simple form no longer suffices, and the problem is still open for finite sets AAA. Brown’s framework is still conceptually useful—you can phrase the conjecture as “does the Brown inequality eventually hold for this power sequence?”—but proving that inequality is exactly the deep part.

If you’d like, next step could be to write out a fully rigorous Brown‑criterion proof for the “with 1’s” version in your favourite level of detail, or to look at small concrete examples (like {3,4,7}\{3,4,7\}{3,4,7}) and see the inequality at work by hand.