EDIT- result. westzynthius(1931) showed that we can create a $p\_x$ denizen longer than $p\_x \times log(log(log(p\_x)))$... Meaning for large enough prime numbers, the maximum denizen is much larger than $2p\_x, 3p\_x$ etc.

**Definition - Denizen**

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> A sequence $\lbrace a\_k \rbrace$ is a **denizen** if all of it's members are prime numbers, i.e $a\_0, a\_1, ... a\_n \in \mathbb{P} $; and it satisfies the following condition; 
>  **if** "$a\_{x\_1} =y\_1 $", "$a\_{x\_2} =y\_2$", "$x\_1 \pm m\_1y\_1 \neq x\_2 \pm m\_2 y\_2 $ when $y\_2<y\_1 $" and "$m\_3$ isn't divisible by $y\_1$"; **then** "$a\_{x\_1 \pm m\_1y\_1}=y\_1$" and $"a\_{x\_1 \pm m\_3} \neq y\_1$" (where $m \in\mathbb{N}$ where $y \in\mathbb{P}$ and where $x \in\mathbb{Z}$). 
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Let a denizen consisting of prime numbers up to and including $p\_\alpha$ be denoted $\lbrace a\_k \rbrace ^{p\_\alpha} $.
For example; a denizens that can be denoted as $\lbrace a\_k \rbrace^7 $ is {2,7,2,3,2,5,2}.

**Question**

What is the maximum length $\lbrace a\_k \rbrace ^{p\_\alpha} $ can take?

**Attempt**

In order to find the maximum length a denizen $\lbrace a\_k \rbrace ^{p\_\alpha} $ can take I considered denizens of two different types.

I first considered a denizen $\lbrace a\_k \rbrace ^{p\_\alpha} $ of a type which corresponds to the sequence of natural numbers $ \lbrace 2,3,4,...,p\_{\alpha +1} -1 \rbrace $. The corresponding denizen is $\lbrace 2,3,2,...,2 \rbrace $. This type of denizen had been created such that $a\_i=d\_i|i$, where $d\_i$ is some divisor of $i$, implies $a\_i \in $$\lbrace a\_k \rbrace$. The consequence of this property is that this type of denizen can be considered a sequence of the lowest prime divisors of the natural numbers from $2$ to $p\_{\alpha +1} -1$ respectively. 
For example, of this type; $\lbrace a\_k \rbrace ^{11} = \lbrace 2,3,2,5,2,7,2,3,2,11,2 \rbrace $ and corresponds to $\lbrace 2,3,4,5,6,7,8,9,10,11,12 \rbrace$.

A denizen $\lbrace a\_k \rbrace ^{p\_\alpha} $ of this type which has a length of $ p\_{\alpha +1} -2 $. 

However I found a larger type of denizen corresponding to the sequence of integers $\lbrace -(p\_{\alpha -1} -1), ..., -4,-3,-2,-1,0,1,2,3,4, ..., p\_{\alpha -1} -1 \rbrace $. The corresponding denizen is $\lbrace 2, ..., 2,3,2,p\_\alpha,2,p\_{\alpha -1},2,3,2, ..., 2 \rbrace $. This type of denizen can also be considered a sequence of lowest prime factors of the integer sequence above, however it also requires the replacement of $-1$ and $1$ with the two primes $p\_{\alpha}$ and $p\_{\alpha -1}$ respectively and involves replacing $0$ with $2$. For example, of this type $\lbrace a\_k \rbrace ^{11} = \lbrace 2,5,2,3,2,7,2,11,2,3,2,5,2 \rbrace $ and corresponds to $\lbrace -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6 \rbrace$. 

This type of denizen has a length of $ 2p\_{\alpha -1} -1 $ and also has an apparent symmetry as defined below. this type of denizen has a length greater or equal to the length of the previous type because of the identity $2p\_{x-2} \geq p\_{x}-1$ [proved here](https://math.stackexchange.com/questions/1384588/improved-betrands-postulate).

So my next question is; is the second type of denizen the largest lengthed $\lbrace a\_k \rbrace ^{p\_\alpha} $ denizen possible? How could you prove it was?

I have tried to prove this, using the concept of symmetry.
I defined the **symmetric depth** as the largest prime number $p\_N$ such that $a\_{x+ p\_N}=a\_{x- p\_N} = p\_N$ and $a\_{x+ p\_{i}}=a\_{x- p\_{i}} =p\_i$ for all prime numbers $p\_i$ less than $p\_N$, centred on some $a\_x\in \lbrace a\_k \rbrace$. I let $\lbrace a\_k \rbrace ^{p\_\alpha} \_ {p\_N} $ be a denizen consisting prime numbers upto and including $p\_\alpha$ with symmetric depth $p\_N$, and hoped to prove that the length of a denizen $\lbrace a\_k \rbrace ^{p\_\alpha} \_ {p\_N} $ is always greater than or equal to the length of a denizen $\lbrace a\_k \rbrace ^{p\_\alpha} \_ {p\_{N-1}} $. However I found a counter example to this, due to the fact that the gaps between large prime numbers tend to be greater than the gaps between smaller prime numbers.

However I have yet to produce a denizen $\lbrace a\_k \rbrace ^{p\_\alpha} $ greater in length than $\lbrace a\_k \rbrace ^{p\_\alpha} \_ {p\_{\alpha -2}} $.

So any ideas to further this?

**Further Questions**

**Definition - Apparent width**

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> In a denizen $\lbrace a\_k \rbrace$, **if** $p\_A$ is the largest prime number such that $a\_{i+ mp\_A}$ and $a\_{i- mp\_A}$ are prime number greater than or equal to $p\_A$ **then** $a\_i$ has an apparent width of $p\_A$. Furthermore; **if** the apparent width of $a\_i$ is larger then all other apparent widths of members in $\lbrace a\_k \rbrace$ **then** the apparent width of $\lbrace a\_k \rbrace$ is $p\_A$. 
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Can there be a denizen $\lbrace a\_k \rbrace ^{p\_x}$ that has an apparent width strictly less than some $p\_\beta$, but that has a length that is greater than another denizen $\lbrace a\_k \rbrace^{p\_x}$ with apparent width of $p\_\beta$?

Furthermore, are all denizen $\lbrace a\_k \rbrace ^{p\_x}$ with an apparent width of $p\_\beta$ less than or equal to in length to the smallest denizen $\lbrace a\_k \rbrace ^{p\_x}\_{p\_\beta}$?