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Is there a better way of doing this? If so, how?

Yes, there is. Your current algorithm is \$O \left( n + \left( n \times \dfrac{n}{2} \right) + \left( \dfrac{n}{2} \right) \right) \$, or \$ O(n^{2}) \$. (The first \$n\$ is the iterating of the numbers, the \$ \left( n \times \dfrac{n}{2} \right) + \left( \dfrac{n}{2} \right) \$ is the time complexity of a triangular number.)

If you use the Sieve of Eratosthenes, which works like this...

_{(image courtesy of linked Wikipedia article)}

you get \$ O \left(n \left( \log n \right) \left( \log \log n \right) \right) \$, which is faster. See this Stack Overflow question on how the time complexity was determined.

Is there a better way of doing this? If so, how?

Yes, there is. Your current algorithm is \$O \left( n + \left( n \times \dfrac{n}{2} \right) + \left( \dfrac{n}{2} \right) \right) \$, or \$ O(n^{2}) \$. (The first \$n\$ is the iterating of the numbers, the \$ \left( n \times \dfrac{n}{2} \right) + \left( \dfrac{n}{2} \right) \$ is the time complexity of a triangular number.)

If you use the Sieve of Eratosthenes, which works like this...

_{(image courtesy of linked Wikipedia article)}

you get \$ O \left(n \left( \log n \right) \left( \log \log n \right) \right) \$, which is faster. See this Stack Overflow question on how the time complexity was determined.

Is there a better way of doing this? If so, how?

Yes, there is. Your current algorithm is \$ \BigO{n + \left( n \times \dfrac{n}{2}} + \left( \dfrac{n}{2} \right) \right) \$, or \$ \BigO{n^{2}}\$. (The first \$n\$ is the iterating of the numbers, the \$ \left( n \times \dfrac{n}{2} \right) + \left( \dfrac{n}{2} \right) \$ is the time complexity of a triangular number.)

If you use the Sieve of Eratosthenes, which works like this...

_{(image courtesy of linked Wikipedia article)}

you get \$ \BigO{n \left( \log n \right) \left( \log \log n \right)} \$, which is faster. See this Stack Overflow question on how the time complexity was determined.

\$ \newcommand{\BigO}[1]{\operatorname{O}\left(#1\right)} \$

Is there a better way of doing this? If so, how?

Yes, there is. Your current algorithm is \$O \left( n + \left( n \times \dfrac{n}{2} \right) + \left( \dfrac{n}{2} \right) \right) \$, or \$ O(n^{2}) \$. (The first \$n\$ is the iterating of the numbers, the \$ \left( n \times \dfrac{n}{2} \right) + \left( \dfrac{n}{2} \right) \$ is the time complexity of a triangular number.)

If you use the Sieve of Eratosthenes, which works like this...

_{(image courtesy of linked Wikipedia article)}

you get \$ O \left(n \left( \log n \right) \left( \log \log n \right) \right) \$, which is faster. See this Stack Overflow question on how the time complexity was determined.

Is there a better way of doing this? If so, how?

Yes, there is. Your current algorithm is \$O \left( n + \left( n \times \dfrac{n}{2} \right) + \left( \dfrac{n}{2} \right) \right) \$, or \$ O(n^{2}) \$. (The first \$n\$ is the iterating of the numbers, the \$ \left( n \times \dfrac{n}{2} \right) + \left( \dfrac{n}{2} \right) \$ is the time complexity of a triangular number.)

If you use the Sieve of Eratosthenes, which works like this...

_{(image courtesy of linked Wikipedia article)}

you get \$ O \left(n \left( \log n \right) \left( \log \log n \right) \right) \$, which is faster. See this Stack Overflow question on how the time complexity was determined.

Is there a better way of doing this? If so, how?

Yes, there is. Your current algorithm is \$ \BigO{n + \left( n \times \dfrac{n}{2}} + \left( \dfrac{n}{2} \right) \right) \$, or \$ \BigO{n^{2}}\$. (The first \$n\$ is the iterating of the numbers, the \$ \left( n \times \dfrac{n}{2} \right) + \left( \dfrac{n}{2} \right) \$ is the time complexity of a triangular number.)

If you use the Sieve of Eratosthenes, which works like this...

_{(image courtesy of linked Wikipedia article)}

you get \$ \BigO{n \left( \log n \right) \left( \log \log n \right)} \$, which is faster. See this Stack Overflow question on how the time complexity was determined.

\$ \newcommand{\BigO}[1]{\operatorname{O}\left(#1\right)} \$