We all know the enormous and tedious formula for primes by Willan. I fail to understand why was it necessary to set the limit as high as $2^n$. The equation runs into Zero's halfway. Is this the only way? Or is there anything else that can substitute $2^n$? Any help &/ clarification would be beneficial. Thanks in advance.
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1Is the explanation in this YouTube video sufficient to your question? – Duong Ngo Dec 16 '24 at 14:27
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It doesn't explain the significance of 2^n – Tejas parashar Dec 16 '24 at 14:41
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And is there a way to replace it if I want to minimise thread time if I want to use it for python – Tejas parashar Dec 16 '24 at 14:42
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The video does explain why $2^{n}$ is there, it is due to Bertrand's postulate. – Duong Ngo Dec 16 '24 at 14:45
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However he also mentioned that in reality, $n^{th}$ prime is way smaller than $2^n$. – Tejas parashar Dec 16 '24 at 14:48
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That I why a asked weather we can substitute $2^n$ by something else to increase efficiency, which Willan wasn't concerned about. – Tejas parashar Dec 16 '24 at 14:50
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That video did clarify 1 part of the question. Thank you. – Tejas parashar Dec 16 '24 at 14:53
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2Better bounds for $n$-th prime (including uppers and lowers) were found by multiple authors, as you can see here: https://math.stackexchange.com/questions/1270814/bounds-for-n-th-prime – Duong Ngo Dec 16 '24 at 15:03