125. $(a_{i+2},a_{i+1})>a_i$

[nobu]

Sia $a_0,a_1,a_2,...$ una successione di numeri interi positivi tali che $(a_{i+2},a_{i+1})>a_i$ per ogni $i\geq0$.
Dimostrare che $a_n\geq 2^n$ per ogni $n\geq 0$.

[jordan]

Sono bboypa su Mathlinks, e ricordavo di aver risolto questo problema tempo addietro, difatti:

Let $ a_0$, $ a_1$, $ a_2$, $ \ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $ \gcd (a_i, a_{i + 1}) > a_{i - 1}$. Prove that $ a_n\ge 2^n$ for all $ n\ge 0$.

Our aim $ a_n \ge 2^n$(*) is true for $ n \in \{0,1\}$, in fact $ a_1=1$ would imply $ 1= (a_2,a_1)>a_0 \in \mathbb{N}_0$, absurd. It is also true that $ \{a_n\}_{n \in \mathbb{N}}$ is strictly incrasing since $ \min\{a_{n+1},a_{n+2}\}$ $ \ge (a_{n+1},a_{n+2})$ $ >a_n$. Now if (*) is true for $ n \in \{0,1,\ldots,n-1\}$ then it is true also for $ n$: in fact we have $ a_n-a_{n-1}$ $ \ge (a_n-a_{n-1},a_{n-1})$ $ =(a_n,a_{n-1}) >a_{n-2} \implies$ $ a_n > a_{n-1}+a_{n-2}$. Now if $ \frac{a_{n-1}}{3} \ge (a_{n-1},a_n)> a_{n-2}$ then we are done since $ a_n>a_{n-1}+a_{n-2} \ge 4a_{n-2} \ge 2^n$. Otherwise $ \frac{a_{n-1}}{2}=(a_n,a_{n-1})$. Now if $ a_n \ge 2a_{n-1}$ we are done, otherwise it means that $ 2 \mid a_{n-1}$, $ 3 \nmid a_{n-1}$ and $ a_n=\frac{3}{2}a_{n-1}$. Now if $ \frac{a_{n-1}}{4} \ge a_{n-2} \implies a_n>a_{n-1} \ge 2^n$, in the last last case (since $ 3 \nmid a_{n-1}$) we must have $ (a_{n-2},a_{n-1})=\frac{a_{n-1}}{2}$, but $ \frac{a_{n-1}}{2}=(a_n,a_{n-1})>a_{n-2}=\frac{a_{n-1}}{2}$, contradiction.

Original thread: qui (c'è anche una soluzione che usa un'altra strada se non ricordo male)
Proposed by Morteza Saghafian, Iran, IMO Shortlist 2008