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

==== 3. Constructing the “power sequence” and applying Brown ====

Form the multiset of all powers of the did_idi:

P={die:1≤i≤k, e≥0}.\mathcal{P} = \{d_i^e : 1\le i\le k,\ e\ge0\}.P={die:1≤i≤k, e≥0}.
Sort it in nondecreasing order (with multiplicity if the same integer occurs as a power of two different bases), and call the resulting sequence

a1≤a2≤a3≤…a_1 \le a_2 \le a_3 \le \dotsa1≤a2≤a3≤…
We clearly have a1=1a_1 = 1a1=1 (some di0=1d_i^0=1di0=1), so condition (1) of Brown’s criterion holds.

Fix nnn. For each base did_idi, let ei,ne_{i,n}ei,n be the smallest exponent such that diei,nd_i^{e_{i,n}}diei,n has not yet appeared among a1,…,ana_1,\dots,a_na1,…,an. Then:
* The powers of did_idi that have appeared are exactly di0,…,diei,n−1d_i^0,\dots,d_i^{e_{i,n}-1}di0,…,diei,n−1,
* so the contribution of base did_idi to the partial sum Sn=a1+⋯+anS_n = a_1+\dots+a_nSn=a1+⋯+an is at least 1+di+⋯+diei,n−1  =  diei,n−1di−1.1 + d_i + \dots + d_i^{e_{i,n}-1} \;=\; \frac{d_i^{e_{i,n}}-1}{d_i-1}.1+di+⋯+diei,n−1=di−1diei,n−1.

Summing over all bases,

Sn  ≥  ∑i=1kdiei,n−1di−1.(★)S_n
\;\ge\;
\sum_{i=1}^k \frac{d_i^{e_{i,n}}-1}{d_i-1}.
\tag{★}Sn≥i=1∑kdi−1diei,n−1.(★)
On the other hand, the next term an+1a_{n+1}an+1 must be the smallest unused power among all bases, so

an+1=min⁡1≤i≤kdiei,n.a_{n+1} = \min_{1\le i\le k} d_i^{e_{i,n}}.an+1=1≤i≤kmindiei,n.
Call this minimum MMM. Then for every iii we have diei,n≥Md_i^{e_{i,n}}\ge Mdiei,n≥M, and so

diei,n−1di−1  ≥  M−1di−1.\frac{d_i^{e_{i,n}}-1}{d_i-1} \;\ge\; \frac{M-1}{d_i-1}.di−1diei,n−1≥di−1M−1.
Plugging this into (★),

Sn  ≥  (M−1)∑i=1k1di−1  ≥  (M−1)if∑i=1k1di−1≥1.S_n
\;\ge\;
(M-1)\sum_{i=1}^k \frac{1}{d_i-1}
\;\ge\;
(M-1)
\quad\text{if}\quad
\sum_{i=1}^k \frac{1}{d_i-1}\ge1 .Sn≥(M−1)i=1∑kdi−11≥(M−1)ifi=1∑kdi−11≥1.
But M=an+1M=a_{n+1}M=an+1, so we have exactly Brown’s inequality

Sn  ≥  an+1−1for all n.S_n \;\ge\; a_{n+1}-1
\quad\text{for all }n.Sn≥an+1−1for all n.
Thus Brown’s criterion is satisfied (and we also have gcd=1=1=1 because 1 is in the sequence), so the power sequence (an)(a_n)(an) is complete: every positive integer is a sum of distinct powers died_i^edie.MathWorld<ref>{{cite web|title=MathWorld|url=https://mathworld.wolfram.com/BrownsCriterion.html|publisher=mathworld.wolfram.com|access-date=2025-12-01}}</ref>

As explained in §2, grouping powers by base turns this into exactly the representation you want, with each summand having only 0/1 digits in its own base.

So:

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This argument is essentially the one written down in the Erdős Problem #124 discussion and formalised in Lean (“Aristotle’s” proof).Erdős Problems<ref>{{cite web|title=Erdős Problems|url=https://www.erdosproblems.com/forum/thread/124|publisher=Erdős Problems|access-date=2025-12-01}}</ref>