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Source. Proposition 4.1 and its proof, preprint p. 10, with the sentence before it (p. 10). Read on the rendered page. The paper is cited by its record on the source card.

Statement

With the notation of Corollary 4.2 (positive integers $a_n$ and $b_n$ with $\sum b_n/a_n$ convergent, and $A_n=\operatorname{lcm}(a_1,\ldots,a_n)$; the proposition says only "Let the notation be as in Corollary 4.2", and its proof uses $a_n\to\infty$, which that convergence gives): if $b_n=1$ and

$$a_{n+1}=a_n\Bigl(\frac{A_n}{A_{n-1}}-1\Bigr)+\gcd(A_n,a_{n+1})\qquad\text{for all }n$$

and $\gcd(A_n,a_{n+1})>1$ for infinitely many $n$, then

$$\limsup_{n\to\infty}\frac{a_n^2}{a_{n+1}}>1.$$

The sentence before the proposition (p. 10) draws the consequence that under the conditions of Corollary 4.2 with $b_n=1$ for all $n$ and $\limsup a_n^2/a_{n+1}\le1$, the gcd equals $1$ from some $n_1$ on, so that $a_{n+1}=a_n^2-a_n+1$ for all larger $n$. The proposition is printed with the equality for all $n$, whereas Corollary 4.2 yields it for $n\ge n_0$; the paper applies it in the latter setting without comment. (Once $\gcd(A_n,a_{n+1})=1$ for all large $n$, also $\gcd(A_{n-1},a_n)=1$, so $A_n/A_{n-1}=a_n$; that is how the recurrence takes the stated form.)

Proof pointer (p. 10)

The proof first shows $A_n>A_{n-1}$, since otherwise the sequence would be bounded, so that $\gcd(A_n,a_{n+1})\le a_{n+1}/2$, and then bounds $a_n^2/a_{n+1}$ from below separately when the gcd equals $a_{n+1}/2$ and when it lies strictly between $1$ and $a_{n+1}/2$.

Relation to problem 243

The hypothesis $a_n/a_{n-1}^2\to1$ of problem 243 gives $\limsup a_n^2/a_{n+1}=1$, so it meets the limsup condition of the consequence above; what it does not supply is the lower bound of Corollary 4.2, without which the equality case need not arise. The proposition alone does not settle the problem.

Bears on. #243 (context: with Corollary 4.2 it yields the problem's recurrence under that corollary's lower bound, which the problem's hypothesis does not imply).