TITLE: Is this greedy sequence optimal?
QUESTION [8 upvotes]: Consider this strictly increasing sequence $(x_n)$ of 21 natural numbers:
$$1, 3, 4, 11, 12, 27, 28, 59, 60, 123, 124, 251, 252, 507, 508, 1019, 1020, 2043, 2044, 4091, 4092$$
i.e., $$x_n=\begin{cases}1&n=1\\2^{k+2}-5&n=2k\\2^{k+2}-4&n=2k+1>1\end{cases}$$
It has the property 
$$\tag{$\star$} \text{There is no index $k$ with $x_{k-1}x_{k+1}<2x_k^2<4x_{k-1}x_{k+1}$.}$$
Questions: 

Is there a sequence with property $(\star)$ and $x_{21}<4092$? 
What is the minimal possibly value of $x_{21}$?
More general, for $n\gg 1$, what is the minimal possible value of $x_n$? 
Is there an infinite sequence with property $(\star)$ that achieves the minimal value  for all $n\gg 1$?

Note that numbers from $1$ to $2046$ can (approximately) be interpreted as showing that $x_{21}\ge 2047$. In my answer to that question, I improved this first to $x_{21}\ge 2559$ and then to $x_{21}\ge 3071$. There's probably still room for improvement along the methods I used,  but the case distinctions started to become convoluted.
The sequence above was constructed greedily, i.e., starting with $x_1=1$ and $x_2=3$ we recursively find the smallest $x_{k+1}$ that is $>x_k$ and is either $\ge \frac{2x_k^2}{x_{k-1}}$ or $\le  \frac{x_k^2}{2x_{k-1}}$.
Note that being too greedy is bad: Starting with $x_1=1$, $x_2=2$, we arrive at the much worse $1,2,8,9,21,22,47,\ldots, x_{21}=6647$.

REPLY [2 votes]: Lemma 1. Let $a<b<c$ be three naturals for which $ac<2b^2<4ac$ is false.
Then one of the following holds:
$$\tag1c\ge 2a+5$$
$$\tag2c=b+2=2a+4$$
$$\tag3c=b+1\in\{2a+2,2a+3,2a+4\}.$$
Proof.
If $2b^2\le ac$, then $c\ge \frac{2b^2}a\ge\frac{2(a+1)^2}a=2a+4+\frac1{a^2}$ and $(1)$ follows.
Assume therefore that $b^2\ge 2ac$. Then $b^2\ge 2a(b+1)$, so $b-1+\frac1{b+1}=\frac{b^2}{b+1}\ge \frac{2ac}{b+1}\ge 2a$ and hence
$c\ge 2a+2$.

If $c=2a+2$, then $2ac=4a^2+4a>(2a)^2$ and hence $b>2a$.
If $c\ge2a+3$, then $2ac=4a^2+6a>4a^2+4a+1=(2a+1)^2$ and hence $b>2a+1$.

These findings are logically equivalent to the claims. $\square$
For naturals $a,b$ with $a<b$ let $S(a,b)$ denote the smallest natural $c>b$ for which $ac<2b^2<4ac$ is false.
Then we extract from the above proof that
$$ \tag4 S(a,b)=\begin{cases}b+1&\text{if }b\ge 2a+1\\
\left\lceil\frac{2b^2}{a}\right\rceil&\text{otherwise}\end{cases}$$
In particular,
$$\tag5S(a,a+1)=2a+5\qquad\text{if }a\ge 2.$$
Indeed, as $a+1$ is not $\ge 2a+1$, $S(a,a+1)=\left\lceil\frac{2(a+1)^2}{a}\right\rceil=\left\lceil2a+4+\frac2a\right\rceil=2a+5$.
Let $\mathscr X$ be the set of all increasing sequences of natural numbers with property $(\star)$, i.e., such that $x_nx_{n+2}<2x_{n+1}^2<4x_nx_{n+2}$ is false for all $n$.
For naturals $a<b$, we can define the greedy sequence $(G_{a,b;n})_{n\in\Bbb N}\in\mathscr X$ with initial values $a,b$ by letting $G_{a,b;1}=a$,$G_{a,b;2}=b$, and recursively $G_{a,b;n}=S(G_{a,b;n-2},G_{a,b;n-1})$ for $n\ge 3$.
In particular, let $(\alpha_n)=(G_{1,2;n})\in\mathscr X$, $(\beta_n)=(G_{1,3;n})\in\mathscr X$, and define the sequence $(\mu_n)$ by letting $\mu_n=\min\{\alpha_n,\beta_n\}$.
Playing with $(4)$ we find that these sequences start
$$\begin{align}
\alpha\colon\quad&1, 2, 8,\hphantom{0} 9, 21, 22, 47, 48, 99,100, 203,\ldots\\
\beta\colon\quad&1, 3, 4, 11, 12, 27, 28, 59, 60, 123, 124,\ldots\\
\mu\colon\quad & 1,2,4,\hphantom{0} 9,12,22,28,48,60,100,124,\ldots
\end{align} $$
In a more usable form we have
Lemma 2.
In closed form,
$$\begin{align}\alpha_n=&\begin{cases}
13\cdot 2^{k-1}-5&n=2k+1,k\ge1\\
13\cdot 2^{k-2}-4&n=2k,k\ge2\\
1&n=1\\
2&n=2
\end{cases} \\
 \beta_n=&\begin{cases}
2^{k+2}-5&n=2k,k\ge1\\
2^{k+2}-4&n=2k+1,k\ge1\\
1&n=1
\end{cases}\\
\mu_n=&\begin{cases}
13\cdot 2^{k-2}-4&n=2k,k\ge2\\
2^{k+2}-4&n=2k+1,k\ge1\\
1&n=1\\
2&n=2
\end{cases}\end{align}
$$
Proof. For $\alpha,\beta$ use induction with $(4)$ and $(5)$. For $\mu$ use $2^3<13<2^4$. $\square$
Corollary. For $n\ge 3$, we have $\mu_{n+2}=2\mu_n+4$ and $\mu_{n+1}>\mu_n+2$.
Theorem. Let $(x_n)\in\mathscr X$. Then
$$\tag{$P_n$} x_n\ge\mu_n$$
and
$$\tag{$Q_n$} \text{if }x_n=\mu_n\text{ and }n>1\text{, then }x_{n-1}=x_n-1.  $$
Proof.
Clearly $x_1\ge 1$ ($\to P_1$) and $x_2\ge 2$ ($\to P_2$), and if $x_2=2$ then necessarily $x_1=1$ ($\to Q_2$).
From $S(1,2)=8$, we see: We cannot have $x_3=3$ ($\to P_3$), and if $x_3=4$ then $x_2=3$ ($\to Q_3$).
To show $P_4$ and $Q_4$, assume $x_4\le 8$, or $x_4=9$ and $x_3<x_4-1$. Then $x_3\le 7$. Then lemma 1 implies $x_2\le 3$. 
We conclude $x_2=3$ from $S(1,2)=8>x_3$, 
$x_1=1$ from $S(2,3)=9>x_3$,
$x_3\ge 5$ from $S(3,4)=11>x_4$.
But then $x_1x_3\le 7<18=2x_2<20=4x_1x_3$.
We conclude that both $P_4$ and $Q_4$.
Then $x_3=4$ or $x_3\ge 7$
If $4\le x_3<7$ then $S(3,x_3)=\left\lceil\frac{2x_3^2}{3}\right\rceil\ge \left\lceil\frac{2\cdot 4^2}{3}\right\rceil=11>9$. Hence $x_3=7$.
But then 
 and $x_4\ge S(3,7)$Then from $S(2,3)=9>x_3$, $x_1=1$. Now $x_1x_3\le 7<2x_2^2=18
Let $n\ge 5$ and assume that we already know $P_\nu, Q_\nu$ for $1\le \nu<n$.
We apply lemma 1 to $(a,b,c)=(x_{n-2},x_{n-1},x_n)$.

Assume $(1)$ holds. Using $P_{n-2}$ and the corollary we have $x_n\ge 2x_{n-2}+5\ge 2\mu_{n-2}+5=\mu_n+1$ so that $P_n$ and (vacuously) $Q_n$.
Assume $(2)$ holds. 
If $x_{n-2}=\mu_2$ then by $Q_{n-2}$ and $(5)$, $x_{n-1}\ge 2x_{n-3}+5=2\mu_{n-2}+3$ and $x_n=x_{n-1}+2\ge 2\mu_{n-2}+5>\mu_n$ so that $P_n$ and (vacuously) $Q_n$.
Assume $(3)$ holds. Then $Q_n$ follows immediately.
If $x_{n-2}>\mu_{n-2}$, then $x_n\ge 2x_{n-2}+2\ge 2\mu_{n-2}+4=\mu_4$ and so $P_n$.
If $x_{n-2}=\mu_{n-2}$, then by $Q_{n-2}$ and $(5)$, $x_n=x_{n-1}+2\ge 2x_{n-3}+6=2\mu_{n-2}+4=\mu_n$ and again $P_n$.

$\square$
Answers to the original questions
We can interprete the first part of the theorem as
$$
\min\{\,x_n\mid(x_k)_{k\in\Bbb N}\in\mathscr X\,\}=\mu_n.$$
This answers the third question for all $n\ge 1$.
For example, we compute $x_{21}\ge \mu_{21}=2092$, which answers the second question as well as (negatively) the first.
Finally, the corollary says that $(\mu_n,\mu_{n+1},\mu_{n+2})$ cannot occur as consecutive terms in a sequence $\in\mathscr X$ as that would contradict lemma 1. Hence every sequence in $\mathscr X$ differs from $(\mu_n)$ in infinitely many places, which answers the fourth question negatively.
Nevertheless, the greedy sequence in the OP (sequence $\beta$) is "optimal" insofar as it coincides with $\mu_n$ for all odd $n$ (and $\alpha$ for even $n$).