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Definition - Denizen

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 $x \in\mathbb{Z}$).

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.

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?

Definition - Denizen

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 $x \in\mathbb{Z}$).

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.

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?

Definition - Denizen

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 : y_2<y_1 $ and $m_z$ isn't divisible by $y_1$; then $ a_{x_1 \pm m_1y_1}=y_1$ and $a_{x_1 \pm m_z} \neq y_1$ (where $m_i \in\mathbb{N}$).

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.

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?

Definition - Denizen

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 $x \in\mathbb{Z}$).

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.

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?

Definition - Denizen

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 condition, that 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 : y_2<y_1 $ and $m_z$ isn't divisible by $y_1$ ; then $ a_{x_1 \pm m_1y_1}=y_1$ and $a_{x_1 \pm m_z} \neq y_1$ (where$m_i \in\mathbb{N} $

Now suppose we wanted to find the maximum lengthed denizen consisting of prime numbers up to and including $p_\alpha$, and let a denizen consisting of prime numbers up to and including $p_\alpha$ be denoted $\lbrace a_k \rbrace ^{p_\alpha} $.

Question

What is the maximum length $\lbrace a_k \rbrace ^{p_\alpha} $ can take?

Attempt

I first considered a denizen $\lbrace a_k \rbrace ^{p_\alpha} $ of the form $\lbrace 2,3,2,...,2 \rbrace $ corresponding to the sequence of natural numbers $ \lbrace 2,3,4,...,p_{\alpha +1} -1 \rbrace $ which has a length of $ p_{\alpha +1} -2 $. The correspondence is due to the fact that $a_i \in \lbrace a_k \rbrace $ obeys the following $a_i=d|i$

However I found a larger denizen $\lbrace a_k \rbrace ^{p_\alpha} $ of the form $\lbrace 2,..., 2,3,2, p_\alpha , 2, p_{\alpha -1}, 2,3, 2, ..., 2\rbrace $ 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 $.

Note. This denizen involves the replacement of  $-1$ and $1$ of its corresponding sequence with the two largest primes of $\lbrace a_k \rbrace ^{p_\alpha} $ creating a symmetry upto and not including $p_{\alpha -1} $. This denizen has a length of $ 2p_{\alpha -1} -1 $ and this denizen is found to be greater or equal in length to the previous one mentioned because of the property $2p_{x-2} \geq p_{x}-1$ proved here.

So is this the largest lengthed $\lbrace a_k \rbrace ^{p_\alpha} $ denizen? How could you prove it was?

I have tried to prove this, using the concept of symmetry; defining symmetry as the largest prime number $p_N$ such that $p_N = a_{x+ p_N}=a_{x- p_N}$ for some $a_x\in \lbrace a_k \rbrace$ and for all prime numbers $p_i$ less than $p_N$, the following condition holds; $a_{x+ p_{i}}=a_{x- p_{i}}$. I let $\lbrace a_k \rbrace ^{p_\alpha} _ {p_N} $ be a denizen consisting prime numbers upto and including $p_\alpha$ with symmetry upto and including $p_N$, and hoped to prove that the length of a denizen with the following conditions $\lbrace a_k \rbrace ^{p_\alpha} _ {p_N} $ is always greater than or equal to the length of a denizen with the following conditions $\lbrace a_k \rbrace ^{p_\alpha} _ {p_{N-1}} $. However I found a counter example as the gaps between larger primes tend to be greater than the gaps between smaller primes.

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}} $.

Any ideas to further this?

Definition - Denizen

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 : y_2<y_1 $ and $m_z$ isn't divisible by $y_1$; then $ a_{x_1 \pm m_1y_1}=y_1$ and $a_{x_1 \pm m_z} \neq y_1$ (where $m_i \in\mathbb{N}$).

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.

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?