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 2017-02-08, 01:28 #1 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 23×3×79 Posts Bonse's inequality Hi all, https://en.m.wikipedia.org/wiki/Bonse%27s_inequality It seems to me that the inequality can be true for higher powers (if not any given higher power), for an appropriately higher (lower) bound for "n". Any thoughts, proofs, counter proofs our insights are appreciated. Thank you in advance. Last fiddled with by a1call on 2017-02-08 at 01:32
 2017-02-08, 01:43 #2 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 23×3×79 Posts In particular I am interested in estimates (or preferably lower bound) for n for: (p-1)# < p^n
2017-02-08, 01:48   #3
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by a1call Hi all, https://en.m.wikipedia.org/wiki/Bonse%27s_inequality It seems to me that the inequality can be true for higher powers (if not any given higher power), for an appropriately higher (lower) bound for "n". Any thoughts, proofs, counter proofs our insights are appreciated. Thank you in advance.
http://math.stackexchange.com/questi...first-n-primes suggest a proof could be as simple as noting things like there's always a prime between n and 2*n and then the fact that 2*...*p(n) has 1 product we can take away 2*p(n) we know this is larger than the base if we can then prove that 1/2*pn*1/4*pn*.....3 > p(n+1) we can prove that the statement is always true. that p(n+1)^2 < 2*...*p(n)

2017-02-08, 06:14   #4
CRGreathouse

Aug 2006

31·191 Posts

Quote:
 Originally Posted by a1call In particular I am interested in estimates (or preferably lower bound) for n for: (p-1)# < p^n
(p-1)# and p# are around e^p, so n would need to be around p/log p for that to work. In particular, for any fixed n, (p-1)# is larger than p^n for large enough p.

Last fiddled with by CRGreathouse on 2017-02-08 at 06:15

 2017-02-08, 09:25 #5 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 23×3×79 Posts Thank you very much for the replies. In particular Mr Greathouse for answering my particular question.

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