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The term for this is primorial numbering. What you see is that each digit in the "primorial numbering system" corresponds to a residue in each cyclic ring with prime modulus. So in fact if you are looking for a prime $p_n$ then we know that each residue for the $p_i$ when $i < n$ implies the order $p_i < p_n$.

First, $x_n \equiv p_n \equiv 0 \pmod{p_n}$ and $x_i \not \equiv 0 \pmod{p_i}$.

Then observe that $p_n \equiv 1 \pmod{2}$ for all the odd primes and that $p_n \in (\mathbb{Z}/p_i \# \mathbb{Z})^* $ in general which is equivalent to $x_i \not \equiv 0 \pmod{p_i}$. Unfortunately for you, when the greatest common factor between your $(x_i, p_i) = 1$ there are infinitely many primes $\{ p \equiv x_i \pmod{p_i} \}$ via Dirichlet's Theorem.

As a result, whenever you form your primorial numbering with the Chinese Remainder Theorem, $\bigcap_{i<n} \{x_i \pmod{p_i} \}$ you won't be able to uniquely determine a value for $p_n \in \mathbb{P}$ below $k(n) = n$, because that intersection forms a residue class and $\{ p_n \equiv x_i \pmod{p_i} \}$ implies that $x_i \in (\mathbb{Z} / p_i \mathbb{Z})^*$ because $p_n$ and $p_i$ are relatively coprime. The intersection formed by $\bigcap_{i<n} \{x_i \pmod{p_i} \}$ when $i \in [0,n)$ is a prime residue class and, again, as a result of Dirichlet's Theorem, it contains infinitely many prime numbers. $x_i \equiv 0 \pmod{p_i}$ if and only if $x_i = p_i = p_n$. Otherwise there exist at least two residue classes for which $p_n \equiv 0 \pmod{p_i}$ and as a result either $p_n$ is not a prime number, or $p_i \ne p_n$, while $p_i \mid p_n$. This is a contradiction.

Therefore $k(n) = n$.

The term for this is primorial numbering. What you see is that each digit in the "primorial numbering system" corresponds to a residue in each cyclic ring with prime modulus. So in fact if you are looking for a prime $p_n$ then we know that $i < n$ implies the order $p_i < p_n$, because the $i$-th prime numbers $p_i$ are an increasing sequence.

First, $x_n \equiv p_n \equiv 0 \pmod{p_n}$ and $x_i \not \equiv 0 \pmod{p_i}$.

Then observe that $p_n \equiv 1 \pmod{2}$ for all the odd primes and that $p_n \in (\mathbb{Z}/p_i \# \mathbb{Z})^* $ in general which is equivalent to $x_i \not \equiv 0 \pmod{p_i}$. Unfortunately for you, when the greatest common factor between your $(x_i, p_i) = 1$ there are infinitely many primes $\{ p \equiv x_i \pmod{p_i} \}$ via Dirichlet's Theorem.

As a result, whenever you form your primorial numbering with the Chinese Remainder Theorem, $\bigcap_{i<n} \{x_i \pmod{p_i} \}$ you won't be able to uniquely determine a value for $p_n \in \mathbb{P}$ below $k(n) = n$, because that intersection forms a residue class and $\{ p_n \equiv x_i \pmod{p_i} \}$ implies that $x_i \in (\mathbb{Z} / p_i \mathbb{Z})^*$ because $p_n$ and $p_i$ are relatively coprime. The intersection formed by $\bigcap_{i<n} \{x_i \pmod{p_i} \}$ when $i \in [0,n)$ is a prime residue class and, again, as a result of Dirichlet's Theorem, it contains infinitely many prime numbers. $x_i \equiv 0 \pmod{p_i}$ if and only if $x_i = p_i = p_n$. Otherwise there exist at least two residue classes for which $p_n \equiv 0 \pmod{p_i}$ and as a result either $p_n$ is not a prime number, or $p_i \ne p_n$, while $p_i \mid p_n$. This is a contradiction.

Therefore $k(n) = n$.

The term for this is primorial numbering. What you see is that each digit in the "primorial numbering system" corresponds to a residue in each cyclic ring with prime modulus. So in fact if you are looking for a prime $p_n$ then we know that each residue for the $p_i$ when $i < n$ implies the order $p_i < p_n$.

First, $x_n \equiv p_n \equiv 0 \pmod{p_n}$ and $x_n \not \equiv 0 \pmod{p_i}$.

Then observe that $p_n \equiv 1 \pmod{2}$ for all the odd primes and that $p_n \in (\mathbb{Z}/p_i \# \mathbb{Z})^* $ in general. Unfortunately for you, when the greatest common factor between your $(x_i, p_i) = 1$ there are infinitely many primes $\{ p \equiv x_i \pmod{p_i} \}$ via Dirichlet's Theorem.

As a result, whenever you form your primorial numbering with the Chinese Remainder Theorem, $\bigcap_{i<n} \{x_i \pmod{p_i} \}$ you won't be able to uniquely determine a value for $p_n \in \mathbb{P}$ below $k(n) = n$, because that intersection forms a residue class and $\{ p_n \equiv x_i \pmod{p_i} \}$ implies that $x_i \in (\mathbb{Z} / p_i \mathbb{Z})^*$ so that the intersection formed when $i \in [0,n)$ is a prime residue class and, again, as a result of Dirichlet's Theorem, it contains infinitely many prime numbers, forming a contradiction. $x_i \equiv 0 \pmod{p_i}$ iff $x_i = p_i = p_n$. Otherwise $p_n$ is not a prime number.

Therefore $k(n) = n$.

The term for this is primorial numbering. What you see is that each digit in the "primorial numbering system" corresponds to a residue in each cyclic ring with prime modulus. So in fact if you are looking for a prime $p_n$ then we know that each residue for the $p_i$ when $i < n$ implies the order $p_i < p_n$.

First, $x_n \equiv p_n \equiv 0 \pmod{p_n}$ and $x_i \not \equiv 0 \pmod{p_i}$.

Then observe that $p_n \equiv 1 \pmod{2}$ for all the odd primes and that $p_n \in (\mathbb{Z}/p_i \# \mathbb{Z})^* $ in general which is equivalent to $x_i \not \equiv 0 \pmod{p_i}$. Unfortunately for you, when the greatest common factor between your $(x_i, p_i) = 1$ there are infinitely many primes $\{ p \equiv x_i \pmod{p_i} \}$ via Dirichlet's Theorem.

As a result, whenever you form your primorial numbering with the Chinese Remainder Theorem, $\bigcap_{i<n} \{x_i \pmod{p_i} \}$ you won't be able to uniquely determine a value for $p_n \in \mathbb{P}$ below $k(n) = n$, because that intersection forms a residue class and $\{ p_n \equiv x_i \pmod{p_i} \}$ implies that $x_i \in (\mathbb{Z} / p_i \mathbb{Z})^*$ because $p_n$ and $p_i$ are relatively coprime. The intersection formed by $\bigcap_{i<n} \{x_i \pmod{p_i} \}$ when $i \in [0,n)$ is a prime residue class and, again, as a result of Dirichlet's Theorem, it contains infinitely many prime numbers. $x_i \equiv 0 \pmod{p_i}$ if and only if $x_i = p_i = p_n$. Otherwise there exist at least two residue classes for which $p_n \equiv 0 \pmod{p_i}$ and as a result either $p_n$ is not a prime number, or $p_i \ne p_n$, while $p_i \mid p_n$. This is a contradiction.

Therefore $k(n) = n$.

The term for this is primorial numbering. What you see is that each digit in the "primorial numbering system" corresponds to a residue in each cyclic ring with prime modulus. So in fact if you are looking for a prime $p_n$ then we know that each residue for the $p_i$ when $i < n$ implies the order $p_i < p_n$.

First, $p_n$ is $p_n \equiv 0 \pmod{p_n}$ and $p_n \not \equiv 0 \pmod{p_i}$.

Then observe that $p_n \equiv 1 \pmod{2}$ for all the odd primes and that $p_n \in (\mathbb{Z}/p_i \# \mathbb{Z})^* $ in general. Unfortunately for you, when the greatest common factor between your $(x_i, p_i) = 1$ there are infinitely many primes $\{ p_n \equiv x_i \pmod{p_i} \}$ via Dirichlet's Theorem.

As a result, whenever you form your primorial numbering with the Chinese Remainder Theorem, $\bigcap_{i<n} \{x_i \pmod{p_i} \}$ you won't be able to uniquely determine a value for $p_n \in \mathbb{P}$ below $k(n) = n$, because that intersection forms a residue class and $\{ p_n \equiv x_i \pmod{p_i} \}$ implies that $x_i \in (\mathbb{Z} / p_i \mathbb{Z})^*$ so that the intersection formed when $i \in [0,n)$ is a prime residue class and, again, as a result of Dirichlet's Theorem, it contains infinitely many prime numbers, forming a contradiction.

Therefore $k(n) = n$.

The term for this is primorial numbering. What you see is that each digit in the "primorial numbering system" corresponds to a residue in each cyclic ring with prime modulus. So in fact if you are looking for a prime $p_n$ then we know that each residue for the $p_i$ when $i < n$ implies the order $p_i < p_n$.

First, $x_n \equiv p_n \equiv 0 \pmod{p_n}$ and $x_n \not \equiv 0 \pmod{p_i}$.

Then observe that $p_n \equiv 1 \pmod{2}$ for all the odd primes and that $p_n \in (\mathbb{Z}/p_i \# \mathbb{Z})^* $ in general. Unfortunately for you, when the greatest common factor between your $(x_i, p_i) = 1$ there are infinitely many primes $\{ p \equiv x_i \pmod{p_i} \}$ via Dirichlet's Theorem.

As a result, whenever you form your primorial numbering with the Chinese Remainder Theorem, $\bigcap_{i<n} \{x_i \pmod{p_i} \}$ you won't be able to uniquely determine a value for $p_n \in \mathbb{P}$ below $k(n) = n$, because that intersection forms a residue class and $\{ p_n \equiv x_i \pmod{p_i} \}$ implies that $x_i \in (\mathbb{Z} / p_i \mathbb{Z})^*$ so that the intersection formed when $i \in [0,n)$ is a prime residue class and, again, as a result of Dirichlet's Theorem, it contains infinitely many prime numbers, forming a contradiction. $x_i \equiv 0 \pmod{p_i}$ iff $x_i = p_i = p_n$. Otherwise $p_n$ is not a prime number.

Therefore $k(n) = n$.