GPT 6 Astra Ultra attempt on Math-727

Erdős Problem #243: independent continuation

Date: 23 September 2026. Source versions read in full: 243.tex in gpt_pro_5.4.

1) OBJECTIVE AND SOURCE AUDIT

Suppose increasing integers $a_n$ satisfy $a_{n+1}/a_n^2\to1$ and $\sum1/a_n\in\mathbb Q$. Must the sequence eventually obey $a_{n+1}=a_n^2-a_n+1$? The supplied file proves a tail asymptotic and invokes an external bounded-lcm theorem. I give a direct arithmetic criterion in terms of reduced tail numerators, without using that theorem.

2) WORK: RECIPROCAL TAIL RIGIDITY

Put $T_n=\sum_{j>n}1/a_j=b_n/c_n$ in lowest positive terms, and $r_n=1/T_n=c_n/b_n$. Write $a=a_{n+1}$, $b=a_{n+2}$ (the temporary $b$ here is a sequence term, not the tail numerator $b_n$). Quadratic growth gives \[ T_n=1/a+1/b+O(b^{-2}),\qquad r_n=a-\frac{a^2}{b}+O(a^{-1}). \] The tail bound follows since eventually each new term is at least half the square of the previous one; the terms after $b$ contribute $O(b^{-2})$. Expanding the reciprocal, with $b\sim a^2$, proves the second formula. In particular $r_n-(a-1)\to0$.

If the reduced numerator $b_n$ is bounded for all sufficiently large $n$, the integer $c_n-b_n(a_{n+1}-1)=b_n(r_n-a_{n+1}+1)$ tends to zero and therefore eventually vanishes. Hence $r_n=a_{n+1}-1$ exactly. The identity \[ T_{n+1}=T_n-1/a_{n+1} \] now gives $r_{n+1}=a_{n+1}(a_{n+1}-1)$. Comparing it with $r_{n+1}=a_{n+2}-1$ proves the desired Sylvester recurrence.

The same proof permits growing numerators under the explicit condition \[ b_n\left(\left|\frac{a_{n+2}}{a_{n+1}^2}-1\right| +\frac1{a_{n+1}}\right)\longrightarrow0. \tag{1} \] Indeed the reciprocal expansion bounds $|r_n-(a_{n+1}-1)|$ by a constant times the expression inside parentheses. Multiplication by $b_n$ again forces the integer difference to be zero eventually.

3) ADVERSARIAL VERIFICATION

The assumption that the reciprocal sum is rational makes every $T_n$ rational, but does not bound its reduced numerator. Approaching an integer does not imply eventual equality for rationals with growing denominators. Condition $1$ explicitly supplies that missing discrete separation. Bounded numerator only along a subsequence would not justify the two consecutive exact identities used in the recurrence argument.

7) FINAL AND REFERENCE

UNRESOLVED. A self-contained sufficient condition is proved, along with a quantitative description of what prevents the rounding argument. No argument establishing $1$ from the original hypotheses is available here. The full target was checked in the indexed primary history: https://www.erdosproblems.com/history/243.

8) COMPLETION ESTIMATE

COMPLETION: 15% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.