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Is there a real number $A$ such that $$\left \lfloor n^{A} \right \rfloor$$ is a prime number ($\forall n \in \mathbb{N})$for all natural numbers $n$)? It is obvious that $A>1+\epsilon$ from the prime number theorem.
Is there a real number $A$ such that $$\left \lfloor n^{A} \right \rfloor$$ is a prime number ($\forall n \in \mathbb{N})$? It is obvious that $A>1+\epsilon$ from the prime number theorem.
Is there a real number $A$ such that $$\left \lfloor n^{A} \right \rfloor$$ is a prime number (for all natural numbers $n$)? It is obvious that $A>1+\epsilon$ from the prime number theorem.
Is there a real number $A$ such that $$\left \lfloor n^{A} \right \rfloor$$ is a prime number ($\forall n \in \mathbb{N})$? It is obvious that $A>1+\epsilon$ from the prime number theorem.
