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This would follow from the Dickson's conjecture or Schinzel's hypothesis H.

Let $M:=2^k \prod_\limits{\text{prime }p\leq 2^{k-1}+1} p$ and let us fix any primes $p_1<p_2<\dots<p_{k-1}$, each of which $\equiv -1\pmod{M}$. We will be looking for $p_k$ of the form $f(x):=2^kx-1$ for some integer $x$. In addition, consider $2^{k-1}$ polynomials indexed by subsets $A\subseteq [k-1]$: $$g_A(x) := \frac1{2^k}\bigg(f(x) \prod_{i\in A} p_i + \prod_{j\in [k-1]\setminus A} p_j\bigg).$$ It can be seen that each such polynomial $g_A(x)$ has integer and coprime coefficients, and its free term is divisible by each prime below $2^{k-1}+1$. This makes the set of polynomials $F:=\{f(x)\}\cup\{g_A(x)\ :\ A\subseteq [k-1]\}$ to miss any "fixed divisor" as defined in Schinzel's hypothesis H, which then would imply that there are infinitely many values of $x$ such that all polynomials from $F$ simultaneously take prime values. Hence, we can take a value of $x=x_0$ such that $p_k:=f(x_0) > p_{k-1}$ to have a required collection of primes.

This would follow from the Dickson's conjecture or Schinzel's hypothesis H.

Let $M:=2^k \prod_\limits{\text{prime }p\leq 2^{k-1}+1} p$ and let us fix any primes $p_1<p_2<\dots<p_{k-1}$, each of which $\equiv 1\pmod{M}$. We will be looking for $p_k$ of the form $f(x):=2^kx-1$ for some integer $x$. In addition, consider $2^{k-1}$ polynomials indexed by subsets $A\subseteq [k-1]$: $$g_A(x) := \frac1{2^k}\bigg(f(x) \prod_{i\in A} p_i + \prod_{j\in [k-1]\setminus A} p_j\bigg).$$ It can be seen that each such polynomial $g_A(x)$ has integer and coprime coefficients, and its free term is divisible by each prime below $2^{k-1}+1$. This makes the set of polynomials $F:=\{f(x)\}\cup\{g_A(x)\ :\ A\subseteq [k-1]\}$ to miss any "fixed divisor" as defined in Schinzel's hypothesis H, which then would imply that there are infinitely many values of $x$ such that all polynomials from $F$ simultaneously take prime values. Hence, we can take a value of $x=x_0$ such that $p_k:=f(x_0) > p_{k-1}$ to have a required collection of primes.

This would follow from the Dickson's conjecture or Schinzel's hypothesis H.

Let $M:=2^k \prod_\limits{\text{prime }p\leq 2^{k-1}+1} p$ and let us fix any primes $p_1<p_2<\dots<p_{k-1}$, each of which $\equiv -1\pmod{M}$. We will be looking for $p_k$ of the form $f(x):=2^kx-1$ for some integer $x$. In addition, consider $2^{k-1}$ polynomials indexed by subsets $A\subseteq [k-1]$: $$g_A(x) := \frac1{2^k}\bigg(f(x) \prod_{i\in A} p_i + \prod_{j\in [k-1]\setminus A} p_j\bigg).$$ It can be seen that each such polynomial $g_A(x)$ has integer and coprime coefficients, and its free term is divisible by each prime below $2^{k-1}+1$. This makes the set of polynomials $F:=\{f(x)\}\cup\{g_A(x)\ :\ A\subseteq [k-1]\}$ to miss any "fixed divisor" as defined in Schinzel's hypothesis H, which then would imply that there are infinitely many values of $x$ such that all polynomials from $F$ simultaneously take prime values. Hence, we can take a value of $x=x_0$ such that $p_k:=f(x_0) > p_{k-1}$ to have a required collection of primes.

This would follow from the Dickson's conjecture or Schinzel's hypothesis H.

Let $M:=2^k \prod_\limits{\text{prime }p\leq 2^{k-1}+1} p$ and let us fix any primes $p_1<p_2<\dots<p_{k-1}$, each of which $\equiv -1\pmod{M}$. We will be looking for $p_k$ of the form $f(x):=2^kx-1$ for some integer $x$. In addition, consider $2^{k-1}$ polynomials indexed by subsets $A\subseteq [k-1]$: $$g_A(x) := \frac1{2^k}\bigg(f(x) \prod_{i\in A} p_i + \prod_{j\in [k-1]\setminus A} p_j\bigg).$$ It can be seen that each such polynomial $g_A(x)$ has integer and coprime coefficients, and its free term is divisible by each prime below $2^{k-1}+1$. This makes the set of polynomials $F:=\{f(x)\}\cup\{g_A(x)\ :\ A\subseteq [k-1]\}$ to miss any "fixed divisor" as defined in Schinzel's hypothesis H, which then would imply that there are infinitely many values of $x$ such that all polynomials from $F$ simultaneously take prime values. Hence, we can take a value of $x=x_0$ such that $p_k:=f(x_0) > p_{k-1}$ to have a required collection of primes.

This would follow from the Dickson's conjecture or Schinzel's hypothesis H.

Let $M:=2^k \prod_\limits{\text{prime }p\leq 2^{k-1}+1} p$. Fix any primes $p_1<p_2<\dots<p_{k-1}$, each of which $\equiv -1\pmod{M}$. We will be looking for $p_k$ of the form $f(x)=2^kx-1$ for some integer $x$. In addition, consider $2^{k-1}$ polynomials indexed by subsets $A\subseteq [k-1]$: $$g_A(x) := \frac1{2^k}\bigg(f(x) \prod_{i\in A} p_i + \prod_{j\in [k-1]\setminus A} p_j\bigg).$$ It can be seen that each such polynomial $g_A(x)$ have integer and coprime coefficients, and its free term is divisible by each prime below $2^{k-1}+1$. This makes the set of polynomials $F:=\{f(x)\}\cup\{g_A(x)\ :\ A\subseteq [k-1]\}$ to miss any "fixed divisor" as defined in Schinzel's hypothesis H, which then would imply that there is a value of $x$ such that all polynomials from $F$ simultaneously take prime values as required.

This would follow from the Dickson's conjecture or Schinzel's hypothesis H.

Let $M:=2^k \prod_\limits{\text{prime }p\leq 2^{k-1}+1} p$ and let us fix any primes $p_1<p_2<\dots<p_{k-1}$, each of which $\equiv -1\pmod{M}$. We will be looking for $p_k$ of the form $f(x):=2^kx-1$ for some integer $x$. In addition, consider $2^{k-1}$ polynomials indexed by subsets $A\subseteq [k-1]$: $$g_A(x) := \frac1{2^k}\bigg(f(x) \prod_{i\in A} p_i + \prod_{j\in [k-1]\setminus A} p_j\bigg).$$ It can be seen that each such polynomial $g_A(x)$ has integer and coprime coefficients, and its free term is divisible by each prime below $2^{k-1}+1$. This makes the set of polynomials $F:=\{f(x)\}\cup\{g_A(x)\ :\ A\subseteq [k-1]\}$ to miss any "fixed divisor" as defined in Schinzel's hypothesis H, which then would imply that there are infinitely many values of $x$ such that all polynomials from $F$ simultaneously take prime values. Hence, we can take a value of $x=x_0$ such that $p_k:=f(x_0) > p_{k-1}$ to have a required collection of primes.