Set $A$ with $|2A| \geq 100|A|$ but $|3A| < 1000|A|$ Let $kA$ denote the sumset $\{ a_1 + \cdots + a_k \mid  a_i \in A \}$.
I want to show that $|2A| \geq 100|A|$ does not imply $|3A| \geq 1000|A|$. [I know this to be true experimentally, but am looking for a more elegant (or more generalisable) way to demonstrate this which does not rely on writing a brute force python script.]

I know that it suffices to find an example with $|2B| \geq 10|B|$ and $|3B| < 31|B|$, say. Then, $A := B^2 \subseteq \mathbb{Z}^2$ satisfies $|2A| \geq 10^2|A|$ but $|3A| < 31^2|A|$. Then, for sufficiently large $N$, the projection map $(x_1, x_2) \mapsto x_1 + N^1 \cdot x_2$ applied to $A \subseteq \mathbb{Z}^2$ gives an example in $\mathbb{Z}$.
(In general we could apply the same trick with $\mathbb{Z}^n$, not just $\mathbb{Z}^2$. For example, it would also suffice to find $B$ with $|2B| \geq 5|B|$ and $|3B| < 10|B|$, since $5^3 \geq 100$ and $10^3 = 1000$.)
However, I can't think of any good ways to come up with such examples systematically (as opposed to generating them programmatically).
 A: Let $X = \{ 0,1,\ldots,6 \}$ and $\phi : X^7 \to \mathbb{N}$ be the map:
$$X^7 \ni (x_0,\ldots,x_6) \quad\mapsto\quad \sum_{k=0}^6 x_k 7^k \in \mathbb{N}$$
It is easy to see the map $\phi$ is injective.
Let $B = \{ 0, 1, 3 \} \subset X$ and $A$ be the image of $B^7$ under $\phi$, i.e
$$A = \phi(B^7) = \left\{ \sum_{k=0}^6 b_k 7^k : b_0,\ldots,b_6 \in B \right\}$$
Since $\phi$ is injective, $|A| = |B^7| = 3^7$.
Notice $C \stackrel{def}{=} 2B = \{0,1,2,3,4,6\}$ is also a subset of $X$ and
$$2A = \left\{ \sum_{k=0}^6 c_k 7^k : c_0,\ldots, c_6 \in C\right\} = \phi(C^7)$$
This implies $|2A| = |C^7| = |C|^7 = 6^7$ and hence
$$|2A| = \frac{6^7}{3^7} |A| = 128 |A| > 100|A|$$
One the other hand, the largest element in $A$ is $3 \sum_{k=0}^6 7^k = 3\frac{7^7 - 1}{7-1} = \frac12(7^k - 1)$. So the largest element in $3A$ is $\frac32(7^7 - 1)$. This leads to
$$3A \subset \mathbb{N} \cap \left[ 0, \frac32(7^7 - 1) \right]
\quad\implies\quad |3A| \le \frac{\frac32 (7^7 - 1) + 1}{3^7} |A| < 565 |A|$$
By brute force, $|3A| = 1152909 \sim 527.2 |A|$. Around $93.3\%$ of the integer interval $\mathbb{N} \cap \left[ 0, \frac32(7^7 - 1) \right]$ is covered by $3A$.
