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*p>n? set prime 
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#1
Sep1904, 09:16 PM

P: 1

is it true that for any set S:={1,....,n}
2*p>n , where p is the largest prime in S? Thanks in advance 


#2
Sep1904, 09:52 PM

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lets try it. {1} does not seem to work. {1,2} ok
{1,2,3,4} ok, it looks likely from now on, but why don't we use the theorem behind "mills constant"? (I just heard of this today, on this site.) i.e. there is a constant K such that for any prime p, the next prime is closer than K p^(5/8), in particular it is closer than Kp. so once p gets larger than K, we have that the next prime is closer than p^2. hence the next prime is smaller than p+p^2 = p(1+p), which is a lot smaller than 2^p. so this says that in any sequence of integers, {1,2,....,p,....,n} where p is the alrgest prime, then n is smaller than the next larger prime hence n is smaller than 2^p. so this is certainly true eventually, i.e. for large n. but it is probably easy to prove it is always true. i just do not see it right now. 


#3
Sep2004, 12:21 AM

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#4
Sep2004, 01:36 AM

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*p>n? set prime
i think this problem might relate to the matter of whether there is always a prime between any positive integer X in the interval X^2 and (X+1)^2. Call such a prime, p. If so then 2p, which at the smallest would be 2*(X^2+1), and assumed the only prime in that interval, would be bounded by one less than the smallest prime q, being also assumed the largest prime, in the interval (X+1)^2, (X+2)^2, which would have size at the most of (X+1)^2 + 2X+2.
This is going to be true when 2X^2+2 > (X+1)^2 +2X+2, or X^2>4X+1 or X>4. This may be a good start, but while this proposition seems obvious, it is not know that any such prime exists! This indicates just how difficult this problem might be. See http://nrich.maths.org/discus/messag...tml?1095166598 


#5
Sep2004, 02:51 AM

P: n/a

While you're at it, how about generalising the conjecture, that between x^n and (x+1)^n, there would always be a prime number?



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