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Not assuming PNT, what is $a$ in $$(p\#_x)^a=(2^{1/2})(3^{1/3})(5^{1/5})...$$

where $p\#$ is primorial till $x$, and r.h.s is over primes.

Also answer can be asymptotic !

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Assuming the rhs stops at $x$, clearly $a \to 0$. When $x$ increments to a new prime, the right side is multiplied by something very close to $1$, while the base on the left is multiplied by $x$. So $a=\frac{\log [(2^{1/2})(3^{1/3})(5^{1/5})\dots]}{\log p\#\_x}$ will have the denominator increase by $\log x$ and the numerator hardly increase at all.

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  • $\begingroup$ I don't know how you can if you don't know roughly how many primes there are. With PNT you have $\log p\#\_x \sim x$ I suppose you could prove that directly, but that seems like another way of stating PNT. $\endgroup$ – Ross Millikan Jan 21 '14 at 23:24
  • $\begingroup$ No, it is the base on the left (before the power $a$) that is about $e^x$ $\endgroup$ – Ross Millikan Jan 21 '14 at 23:29
  • $\begingroup$ The thing is, I had shown that the RHS is ~ x, but I am unable to show p#_n ~ e^x from that. But nevertheless thanks for help. $\endgroup$ – user3058477 Jan 21 '14 at 23:31
  • $\begingroup$ I get that the right side at $x=59$ is only about $18.4$, much less than $x$ and each new prime is only raising it $7\%$ here. $\endgroup$ – Ross Millikan Jan 21 '14 at 23:34
  • $\begingroup$ Well, Its asymptotic ! also I used Sigma_(log p)/p ~ log x (also plz see my comment just after the question) $\endgroup$ – user3058477 Jan 21 '14 at 23:37

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