gpt pro 5.4 attempt on Math-739

Erdős Problem 271 --- Round 2

1) ROUND-2 OBJECTIVE

Path (C): the general problem is still open, so the most promising strict advance is to correct the Round 1 computational slip, then extend the fully explicit theory from $A(1)$ to the next nontrivial seed $A(2)$ by a complete proof. This adds a genuinely new explicit family beyond Round 1.

2) Round-1 FOUNDATION USED

From Round 1 I use only the following established facts.

• $A(1)$ is exactly the set $S$ of nonnegative integers whose base-$3$ expansion uses only the digits $0$ and $1$.
• The key representation lemma for $S$: for every integer $u\ge 0$ there exist $a,b\in S$ with \[ u=2b-a, \] and if $u\notin S$ then one may choose $b<u$.
• The growth of $A(1)$ is $a_k=\Theta(k^{\log_2 3})$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new object is \[ T:=3S\cup(3S+2). \] Equivalently, $T$ is the set of nonnegative integers whose ternary expansion has digits $0$ or $1$ in every position except the units digit, which is allowed to be $0$ or $2$.

The Round-2 claim is that \[ A(2)=T. \] This yields an explicit binary-to-ternary formula and an exact two-line recursion for the sequence generated by $\{0,2\}$.

4) ATTACK PLAN (ROUND-2)

There were two Round 1 gaps relevant here.

• The small illustrative sample for $A(4)$ in Round 1 was computational only, and needed verification.
• No seed beyond $n=1$ had been treated explicitly.

The plan is therefore:

• first correct the $A(4)$ sample by direct greedy checking;
• then prove that $T=3S\cup(3S+2)$ is 3-free;
• finally show that every integer $m>2$ outside $T$ is rejected by the greedy algorithm because it completes a 3-term progression with earlier elements of $T$.

5) WORK (ROUND-2)

Round-2 correction of the Round-1 sample.

The Round 1 sample for $A(4)$ was incorrect. The correct initial segment is \[ A(4):\ 0,4,5,7,11,12,16,23,26,31,33,37,38,44,49,56,\dots \] A quick check already shows why $6$ could not have appeared: after choosing $0,4,5$, the candidate $6$ completes the arithmetic progression $4,5,6$ and must be rejected.

This correction affects only the heuristic illustration from Round 1; none of the actual theorems from Round 1 depended on the erroneous list.

Theorem 5.1.

Let $S=A(1)$, and define \[ T:=3S\cup(3S+2). \] Then the Stanley sequence generated by $\{0,2\}$ is exactly $T$.

Proof. We first show that $T$ is 3-free.

Suppose $x<y<z$ lie in $T$ and satisfy $x+z=2y$. Reduce modulo $3$.

• If $x\equiv z\equiv 0\pmod 3$, then also $y\equiv 0\pmod 3$, so \[ x=3x',\quad y=3y',\quad z=3z' \] with $x',y',z'\in S$, and then $x'+z'=2y'$, contradicting the 3-freeness of $S$.

• If $x\equiv z\equiv 2\pmod 3$, then also $y\equiv 2\pmod 3$, so \[ x=3x'+2,\quad y=3y'+2,\quad z=3z'+2 \] with $x',y',z'\in S$, and again $x'+z'=2y'$, contradiction.

• If $x\equiv 0\pmod 3$ and $z\equiv 2\pmod 3$ (or vice versa), then \[ x+z\equiv 2\pmod 3, \] so $2y\equiv 2\pmod 3$, hence $y\equiv 1\pmod 3$, impossible because every element of $T$ is congruent to $0$ or $2$ modulo $3$.

Thus $T$ contains no nontrivial 3-term arithmetic progression.

Now let $m>2$ be an integer with $m\notin T$. We show that $m$ is rejected by the greedy algorithm for seed $\{0,2\}$. There are three residue classes modulo $3$.

Case 1: $m\equiv 0\pmod 3$. Write $m=3u$. Since $m\notin T$, we have $u\notin S$. By the Round 1 representation lemma for $S$, there exist $a,b\in S$ with \[ u=2b-a, \qquad b<u. \] Then \[ m=3u=2(3b)-3a. \] Both $3a$ and $3b$ lie in $T$, and $3b<m$ because $b<u$. Hence $m$ completes a 3-term progression with earlier elements of $T$.

Case 2: $m\equiv 2\pmod 3$. Write $m=3u+2$. Since $m\notin T$, again $u\notin S$. Choose $a,b\in S$ with $u=2b-a$ and $b<u$. Then \[ m=3u+2=2(3b+2)-(3a+2). \] Now $3a+2,3b+2\in T$, and $3b+2<m$ because $b<u$. So $m$ is rejected.

Case 3: $m\equiv 1\pmod 3$. Write $m=3u+1$. Since $m>2$, we have $u\ge 1$. Apply the Round 1 representation lemma to $u-1$: choose $a,b\in S$ with \[ u-1=2b-a, \qquad b\le u-1. \] Then \[ m=3u+1=3(u-1)+4=2(3b+2)-3a. \] Again $3a\in T$ and $3b+2\in T$, while \[ 3b+2\le 3(u-1)+2=3u-1<m. \] So $m$ is rejected.

Therefore every integer $m>2$ outside $T$ is rejected, while every integer in $T$ is accepted because $T$ is 3-free. Since the seed is $\{0,2\}$, the greedy sequence is exactly $T$.

Corollary 5.2 (explicit formula for $A(2)$).

Let \[ A(2)=\{b_0<b_1<b_2<\cdots\}. \] If \[ k=\sum_{i\ge 0}\varepsilon_i 2^i \qquad (\varepsilon_i\in\{0,1\}) \] is the binary expansion of $k$, then \[ b_k=2\varepsilon_0+\sum_{i\ge 1}\varepsilon_i 3^i. \] Equivalently, if $A(1)=\{a_0<a_1<a_2<\cdots\}$, then \[ b_{2m}=3a_m, \qquad b_{2m+1}=3a_m+2. \]

Proof. The set $T=3S\cup(3S+2)$ is the disjoint union of two increasing translates of $S$, and for every $s\in S$ we have \[ 3s<3s+2<3s' \qquad\text{whenever } s'\ge s+1. \] Hence the increasing enumeration of $T$ interleaves as \[ 3a_0,\ 3a_0+2,\ 3a_1,\ 3a_1+2,\ 3a_2,\ 3a_2+2,\dots, \] which is exactly the stated recursion. Rewriting this in terms of the binary digits of $k$ gives the explicit formula.

Corollary 5.3 (growth of $A(2)$).

The sequence $A(2)$ satisfies \[ b_k=\Theta(k^{\log_2 3}). \]

Proof. From Corollary 5.2, \[ b_{2m}=3a_m, \qquad b_{2m+1}=3a_m+2. \] Since Round 1 proved $a_m=\Theta(m^{\log_2 3})$, the same asymptotic follows immediately for $b_k$.

6) ADVERSARIAL VERIFICATION

• The correction of the $A(4)$ sample is genuine: the earlier list was wrong, but only in a heuristic example, not in any proved statement.
• In the proof of Theorem 5.1, the only nontrivial imported fact is the Round 1 representation lemma for $S=A(1)$. I used it exactly in the two forms already proved there: for $u\notin S$, one may choose $u=2b-a$ with $b<u$; and for arbitrary $u\ge 0$, one may choose $u=2b-a$ with $b\le u$.
• The residue-$1$ case is the delicate one. The formula \[ 3u+1=2(3b+2)-3a \] requires a representation of $u-1$, not of $u$. This was checked explicitly; the inequality $3b+2<m$ follows from $b\le u-1$.
• The proof is self-contained once Round 1 is accepted.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED)

Round 2 corrects the faulty $A(4)$ sample and gives a full explicit determination of $A(2)$, including the exact recursion \[ b_{2m}=3a_m, \qquad b_{2m+1}=3a_m+2, \] and the growth law $A(2)=\Theta(k^{\log_2 3})$. The general problem for arbitrary $n$ remains open.

8) COMPLETION ESTIMATE

COMPLETION: 65%

9) REFERENCES

• A. M. Odlyzko and R. P. Stanley, Some Curious Sequences Constructed with the Greedy Algorithm, January 1978. Checked in this round for the classical regular/irregular context and the corrected initial examples.
• T. F. Bloom, Erdős Problem #271, checked 2026-03-07.