TITLE: If a sequence of natural numbers satisfies $\gcd(a_{i+1},a_{i})>a_{i-1}$, then $a_{n}>2^n$
QUESTION [6 upvotes]: Given a sequence $\{a_{n}\}$ in $\mathbb{N}$ such that $\gcd(a_{i+1},a_{i})>a_{i-1},$ for any $i\ge 2$, show that $a_{n}>2^{n-1}$.

Thank you everyone, my friend asked me about this problem, and I feel this problem is really interesting.

REPLY [4 votes]: (Note, that $\mathbb N$ does not include $0$ in my answer)
This seems like a very interesting problem. As @MarcvanLeeuwen pointed out in his comment, the problem as you stated it, is not true. In fact, we can prove, that the counterexample he gave is the worst possible:

Theorem: Let $(a_n)_{n\in\mathbb N}$ be a sequence in $\mathbb N$, such that $a_i<(a_{i+1},a_{i+2})$ for all $i\in\mathbb N$, then $a_i\geq 2^{i-1}$ for all $i\in\mathbb N$.

Proof: First, note, that $(a_n)$ is strictly increasing, since:
$$a_i<(a_{i+1},a_{i+2})\leq a_{i+1}$$
Also, we will repeatedly use the fact 
$$ (*1):\qquad \forall a,b\in\mathbb Z: (a,b)\leq |a-b|$$
(this is true as $(a,b)|a,b$ implies $(a,b)|a-b$).
We prove, that the theorem is true, for $i=1,2,3$:

$a_1\in\mathbb N$ implies $a_1\geq 1$.
$a_2>a_1\geq 1$ implies $a_1\geq 2$.
$a_3>a_2\geq 2$, so $a_3\geq 3$. If $a_3=3$, then $a_2=2$ and $a_1=1$ and $1=a_1<(a_2,a_3)=(2,3)=1$ is false. So $a_3\geq 4$.

For $i+1>3$, we prove the theorem by induction, so assume that it is true, for $i, i-1$ and $i-2$.
There are $k,l\in\mathbb N$, such that
$$(a_i,a_{i+1})\cdot k=a_{i+1} \qquad (a_i,a_{i+1})\cdot l=a_i$$

$k\leq 2$: Use ($*1$):
$$a_{i+1}=k\cdot (a_i, a_{i+1})\leq k\cdot ( a_{i+1} - a_i ) \leq 2 \cdot ( a_{i+1} - a_i )$$
 So $a_{i+1}\geq 2a_i=2\cdot 2^{i-1}=2^i$ by the induction hypothesis.
$k\geq 4$: This follows immediately by applying the induction hypothesis.
$$a_{i+1} = k \cdot (a_i,a_{i+1}) \geq 4 \cdot (a_i,a_{i+1}) > 4a_{i-1} = 4\cdot 2^{i-2} = 2^i$$
$k=3$ and $l=1$. Then $a_{i+1}=3a_i\geq 3\cdot 2^{i-1} \geq 2^i$.

As $a_i<a_{i+1}$ implies $l<k$, the only remaining case is the following:

$k=3$ and $l=2$. Denote $m:=(a_i,a_{i+1})$, so $a_i=2m$ and $a_{i+1}=3m$.

$m\geq \frac{8}{3}a_{i-2}:$ Then
$$a_{i+1}=3m\geq 3\cdot \frac{8}{3}a_{i-2}=8a_{i-2}\geq 8\cdot 2^{i-3}=2^i$$
$m\geq \frac{4}{3}a_{i-1}:$ Same as before:
$$a_{i+1}=3m\geq 3\cdot \frac{4}{3}a_{i-1}=4a_{i-1}\geq 4\cdot 2^{i-2}=2^i$$
$m<\frac{8}{3}a_{i-2}$ ($*2$) and $m<\frac{4}{3}a_{i-1}$ ($*3$): Then 
$$a_{i-1}<(a_i,a_{i+1})=(2m,3m)=m$$
Using this and ($*1)$, we obtain: 
$$a_{i-2}<(a_{i-1},a_i)=(a_{i-1},a_i-a_{i-1})\leq |a_i-2a_{i-1}|=|2m-2a_{i-1}|=2(m-a_{i-1})$$
Regroup this equation to get:
$$(*4):\quad m > a_{i-1}  + \frac{1}{2} a_{i-2}$$
Then:
$$m\overset{(*4)}{>} a_{i-1}  + \frac{1}{2} a_{i-2}\overset{(*3)}{>} \frac{3}{4}m+\frac{1}{2} a_{i-2} \Rightarrow \frac{1}{4}m>\frac{1}{2}a_{i-2} \Rightarrow m > 2a_{i-2}$$
Using this and ($*4$) again:
$$\frac{3}{2}a_{i-2} < \frac{3}{4}m \overset{(*3)}{<} a_{i-1} \overset{(*4)}{<} m - \frac{1}{2} a_{i-2} \overset{(*2)}{<}  \frac{8}{3}a_{i-2} - \frac{1}{2}a_{i-2} = \frac{13}{16}a_{i-2} < \frac{3}{2} a_{i-2}$$
This is a contradiction, so this case is not possible.



EDIT: I edited the post to give the above proof of the tight bound. I did not delete the previously proven weaker bounds, as they are simple proofs and someone might find them interesting.
Weaker bounds, i.e. proofs for $a_i\geq b^{i-1}$ for some $b<2$:

Lemma: Let $(a_n)_{n\in\mathbb N}$ be a sequence in $\mathbb N$, such that $a_i<(a_{i+1},a_{i+2})$ for all $i\in\mathbb N$, then $a_{i+2}>a_{i+1}+a_i$ for all $i\in\mathbb N$.

Proof: The sequence is strictly increasing since $a_i<(a_{i+1},a_{i+2})\leq a_{i+1}$. $(a_{i+1},a_{i+2})$ divides both $a_{i+1}$ and $a_{i+2}$, hence also its difference $a_{i+2}-a_{i+1}$. Since both are positive, this implies $(a_{i+1},a_{i+2})\leq a_{i+2}-a_{i+1}$ and this yields.
$$a_{i+2}=(a_{i+2}-a_{i+1})+a_{i+1}\geq (a_{i+2},a_{i+1}) + a_{i+1} > a_i + a_{i+1}$$

Theorem: Let $(a_n)_{n\in\mathbb N}$ be a sequence in $\mathbb N$, such that $a_i<(a_{i+1},a_{i+2})$ for all $i\in\mathbb N$, then $a_i\geq F_i$ for all $i\in\mathbb N$. (where $F_i$ denotes the $i$-th Fibonacci number)

Proof: This is true for $i=1,2$. For $i>2$ conclude by induction:
$$F_{n+2}=F_n+F_{n+1}\leq a_n+a_{n+1}<a_{n+2}$$
As $F_n\approx \frac{\varphi^n}{\sqrt{5}}$, where $\varphi\approx 1.61$ is the golden ratio, we cound some find some lower bound in the above spirit with $b$ slightly below $\varphi$. Weaker, but simpler is $b=\sqrt{2}$:

Theorem: Let $(a_n)_{n\in\mathbb N}$ be a sequence in $\mathbb N$, such that $a_i<(a_{i+1},a_{i+2})$ for all $i\in\mathbb N$, then $a_i\geq (\sqrt{2})^{i-1}$ for all $i\in\mathbb N$. 

Proof: As $a_n$ is strictly increasing, we have $a_1\geq 1$ and $a_2\geq 2$, so this is true for $i=1,2$. For $i>2$ again induction:
$$a_{n+2} > a_{n+1}+a_n > a_n + a_n = 2a_n \geq 2\cdot (\sqrt{2})^{i-1}= (\sqrt{2})^{i+1}$$

REPLY [2 votes]: Assuming you lower you demands by asking only to prove $a_n\geq2^n$, and add the hypothesis that $a_1\geq2$ so that at least the inequality will hold for $n=1$, then one still gets a counterexample taking $(a_1,a_2,a_3,\ldots)=(2,3,6,12,\ldots)$, since $3<2^2$, $6<2^3$ and so forth. If you insist that $a_0$ be defined, set it to$~1$ (regardless of the question whether $a_0$ is defined or not, you did not require $\gcd(a_2,a_1)>a_0$).
In short you need to state some pretty nonobvious additional requirements to get a statement that one could even start to hope to be true.