gpt pro 5.4 attempt on Math-5550

Round 2 continuation for Erdős Problems #282, #283, #287

Erdős Problem #282

1) ROUND-2 OBJECTIVE

I pursue (C) obstruction/correction for the broad "for which pairs $(x,A)$ does the greedy process terminate?" part of the problem. Round 1 left the Stein case $(A=$ all odd numbers$,\ x$ of odd denominator$)$ unresolved, but it did not yet separate that case from the behaviour of sparse allowed sets. The most promising Round-2 target is therefore to produce an explicit infinite family of rational pairs $(x,A)$ for which the greedy process provably does not terminate, and to sharpen the arithmetic obstruction for finite odd expansions.

2) Round-1 FOUNDATION USED

I use the following Round-1 items without reproving them.

1. The corrected distinct-denominator greedy algorithm: at each step choose the least unused $n\in A$ with $1/n\le x$.
2. The basic odd-denominator obstruction: if a finite sum of distinct odd unit fractions equals $a/b$ in lowest terms, then $b$ is odd.
3. The classical Fibonacci--Sylvester termination proof for $A$ equal to the full set of positive integers.
4. The Round-1 computational fact that the odd greedy algorithm terminated for all reduced $a/b\in(0,1)$ with odd $b\le 51$.

3) NEW INSIGHT / TOOL (ROUND-2)

There are two genuinely new ingredients in this round.

1. A Sylvester-tail nontermination family: for every integer $m\ge 3$ there is an explicit infinite set $A_m$ and a rational $x_m=1/(m-1)$ such that the restricted greedy algorithm on $(x_m,A_m)$ follows a closed-form infinite trajectory and never terminates.
2. A mod $8$ terminal invariant for odd expansions: if \[ \frac{a}{b}=\frac1{n_1}+\cdots+\frac1{n_t} \] with all $n_i$ odd and $a/b$ in lowest terms, then \[ n_1+\cdots+n_t\equiv a b^{-1}\pmod 8, \] where $b^{-1}$ is the inverse of $b$ modulo $8$.

The first result shows that rational nontermination is abundant for sparse allowed sets, even inside odd sets and even inside a single residue class. The second result strengthens the Round-1 parity obstruction.

4) ATTACK PLAN (ROUND-2)

Round 1 identified the main gap as the absence of either a global monotone quantity or an explicit nonterminating trajectory. I therefore prove the following two precise claims.

1. For each $m\ge 3$, the pair \[ x_m=\frac1{m-1},\qquad A_m=\{a_1,a_2,a_3,\dots\} \] with \[ a_1=m,\qquad a_{r+1}=(m-1)a_1a_2\cdots a_r+1 \] produces an infinite greedy expansion.
2. Any finite odd Egyptian expansion satisfies a stronger congruence modulo $8$.

These claims do not settle the Stein case, but they overcome two specific Round-1 obstacles: they exhibit explicit nontermination for rational input, and they sharpen the arithmetic constraints on any hypothetical terminal odd expansion.

5) WORK (ROUND-2)

5.1. An explicit infinite nonterminating family.

Fix an integer $m\ge 3$. Define \[ a_1=m,\qquad a_{r+1}=(m-1)a_1a_2\cdots a_r+1\quad (r\ge 1), \] and let \[ A_m:=\{a_1,a_2,a_3,\dots\},\qquad x_m:=\frac1{m-1}. \] Let \[ P_r:=a_1a_2\cdots a_r. \] I claim that the greedy algorithm for $(x_m,A_m)$ chooses $a_1,a_2,\dots$ in order and has remainder \[ R_r:=x_m-\sum_{j=1}^r \frac1{a_j}=\frac1{(m-1)P_r}\qquad (r\ge 0), \] where by convention $P_0=1$.

For $r=0$ this is exactly $R_0=x_m=1/(m-1)$. Assume the formula holds for some $r\ge 0$. Then \[ \frac1{R_r}=(m-1)P_r, \] so by definition of the sequence, \[ a_{r+1}=\frac1{R_r}+1. \] Hence \[ R_r-\frac1{a_{r+1}} =\frac1{(m-1)P_r}-\frac1{(m-1)P_r+1} =\frac1{(m-1)P_r\left((m-1)P_r+1\right)} =\frac1{(m-1)P_{r+1}}. \] Thus the remainder formula holds for all $r$ by induction.

It remains to check the greedy choice. Since $a_{r+1}=1/R_r+1$, we have $a_{r+1}>1/R_r$, so $1/a_{r+1}<R_r$. Also $a_1<a_2<a_3<\cdots$, because \[ a_{r+1}=(m-1)P_r+1>(m-1)a_r+1>a_r. \] Therefore every previously used denominator is $<a_{r+1}$, and every unused element of $A_m$ is at least $a_{r+1}$. So among unused elements of $A_m$, the least one satisfying $1/n\le R_r$ is exactly $a_{r+1}$. This is precisely the corrected greedy rule.

Since $R_r=1/((m-1)P_r)>0$ for every $r$, the algorithm never reaches $0$. Therefore \[ \frac1{m-1}=\sum_{r=1}^{\infty}\frac1{a_r} \] is an infinite greedy expansion and the greedy process does not terminate.

5.2. Two useful corollaries of the family.

1. If $m$ is odd, then every $a_r$ is odd. Indeed $a_1=m$ is odd, and if $a_1,\dots,a_r$ are odd then $(m-1)P_r$ is even, so $a_{r+1}=(m-1)P_r+1$ is odd. Thus for every odd $m\ge 3$ we obtain an explicit nonterminating rational pair with $A_m$ contained in the odd positive integers.
2. If $m\equiv 1\pmod d$, then every $a_r\equiv 1\pmod d$. This is immediate from \[ a_{r+1}=(m-1)P_r+1\equiv 1\pmod d. \] Hence for every modulus $d\ge 1$ there are nonterminating rational pairs $(x,A)$ with $A$ contained in a single residue class modulo $d$.

These corollaries do not settle the open odd/full-set question, because the first family has $x_m=1/(m-1)$ and therefore even denominator when $A_m$ is odd, and the second family lies outside Graham's admissible congruence condition in the natural example $A\subseteq\{n\equiv 1\pmod d\}$ with $x=1/d$.

5.3. A stronger terminal obstruction for odd expansions.

Suppose \[ \frac{a}{b}=\frac1{n_1}+\cdots+\frac1{n_t} \] with all $n_i$ odd and $a/b$ in lowest terms. Because $b$ is odd, the residue class $b^{-1}\pmod 8$ is defined.

For any odd integer $u$ one has $u^2\equiv 1\pmod 8$, hence $u^{-1}\equiv u\pmod 8$. Therefore, in the ring of rationals with odd denominator modulo $8$, \[ \frac1{n_i}\equiv n_i\pmod 8. \] Summing over $i$ gives \[ \frac{a}{b}=\sum_{i=1}^t \frac1{n_i}\equiv \sum_{i=1}^t n_i\pmod 8. \] Multiplying by $b$ yields the equivalent integer congruence \[ n_1+\cdots+n_t\equiv a b^{-1}\pmod 8. \] This strictly strengthens the Round-1 parity statement: reducing modulo $2$ only recovers the fact that $b$ must be odd, whereas modulo $8$ one gets a definite residue class for the total denominator sum.

6) ADVERSARIAL VERIFICATION

I checked the new work against the main failure modes.

1. Greedy-rule ambiguity. The proof uses the Round-1 corrected greedy rule with distinct denominators. This matters: the sequence $A_m$ is strictly increasing, so the least unused allowed denominator is unambiguous at every step.
2. Could an earlier unused element of $A_m$ be large enough to interfere? No. There are no earlier unused elements: the greedy trajectory uses $a_1,a_2,\dots$ in order, and $A_m$ is strictly increasing.
3. Could the family accidentally solve the Stein case? No. When $A_m$ is odd, $x_m=1/(m-1)$ has even denominator. So this family lies deliberately outside Stein's hypothesis.
4. Could the residue-class corollary refute the admissible congruence-class question? No. For the natural choice $A\subseteq\{n\equiv 1\pmod d\}$ and $x=1/d$, Graham's admissibility condition fails, so these examples do not contradict the open admissible case.
5. Is the mod $8$ congruence really stronger than Round 1? Yes: it gives a specific residue class for the terminal denominator sum, not merely oddness of the reduced denominator.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The Stein problem for the full odd set remains open, but Round 2 makes two substantive advances beyond Round 1.

1. It gives an explicit infinite family of rational nonterminating greedy trajectories, including examples with $A$ odd and examples with $A$ contained in one residue class modulo an arbitrary $d$.
2. It upgrades the odd-denominator terminal obstruction from a parity statement to a sharp congruence modulo $8$.

So the broad "for which pairs $(x,A)$?" question is now sharply separated from the special full-set odd case: sparse allowed sets can force rational nontermination, while the Stein case remains the real unresolved core.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 45%

9) REFERENCES

Checked or used in this round:

1. T. F. Bloom, Erdős Problem #282, webpage accessed 2026-03-07.
2. W. van Doorn, Partitions with prescribed sum of reciprocals: computational results, arXiv:2502.01409 (checked for context; the congruence used above was reproved directly).