Is it possible to give an estimation (maybe an approximation) of the size of the $k$-th prime number?
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6Yes, google prime-number-theorem – Peter Sep 12 '17 at 22:32
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1https://en.wikipedia.org/wiki/Prime_number_theorem $p_n \sim n \ln n $ – Donald Splutterwit Sep 12 '17 at 22:35
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$$p_n\approx n(\ln(n)+\ln(\ln(n)))$$ is a rough estimation – Peter Sep 12 '17 at 22:45
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1The OP obviously meant the $k$-th smallest prime number. – Mark Fischler Sep 12 '17 at 22:46
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are we talking about the size in terms or quantity of digits in base $10$? i.e. the $10^{th}$ prime number is $31$ and its length is $2$ in base $10$. In that case, it is near to $\lfloor\log_{10}(n \cdot ln(n))\rfloor+1$ as per this answer: https://math.stackexchange.com/questions/231742/proof-how-many-digits-does-a-number-have-lfloor-log-10-n-rfloor-1 – iadvd Sep 13 '17 at 00:02
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@iadvd Thank you for your comments. I am interested by the binary size of primes. – Adam54 Sep 13 '17 at 12:49
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@Adam54 then if I am not wrong, the lenght in binary should be bounded by $\lfloor log_2 (n \cdot ln(n)) \rfloor + 1$ assuming that we are bounding the $n^{th}$ prime by $n \cdot ln(n)$ (it is a rough estimation). – iadvd Sep 13 '17 at 14:33