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November 24th, 2007, 06:28 PM   #1
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prime frequency

will there always be a prime between a given prime and the least prime greater than the square root of the first prime?
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November 24th, 2007, 06:56 PM   #2
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Yes, for p >= 7. By Bertrand's postulate we know that there is a prime strictly between n/2 and n for n > 2, and sqrt(n) <= n/2 for n >= 4. So there is a prime strictly between sqrt(p) and p for all p >= 4. Since you want at least two primes (since you want one strictly between nextprime(sqrt(p)) and p), you'll need the extended form of Bertrand's postulate, proved as I recall by Ramanujan, that shows that for large enough n there are at least k primes between n and 2n.

Edit: Here's Ramanujan's paper: see equation (18) at the bottom.
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November 24th, 2007, 08:02 PM   #3
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not what i meant

what i meant was

p<p_1<p+(sqrt(p) rounded up the the nearest prime)
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November 24th, 2007, 08:09 PM   #4
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In that case, you're essentially asking about Legendre's Conjecture or some variant on it. I don't think there's any hope for a quick solution. Even Schoenfeld's version of the Riemann hypothesis seems too weak to prove this.
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November 24th, 2007, 10:23 PM   #5
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not sure

but i dont think so,

I am doing this:

p_0<P_1<P_0+(Sqrt(p) rounded up to the nearest prime).

how is that the same as:
there is a prime between p^2 and (p+1)^2?
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November 25th, 2007, 03:18 AM   #6
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Re: not sure

Quote:
Originally Posted by soandos
but i dont think so,

I am doing this:

p_0<P_1<P_0+(Sqrt(p) rounded up to the nearest prime).

how is that the same as:
there is a prime between p^2 and (p+1)^2?
p_0=23

p_0+sqrt(p_0)= 23+4 (or 5) =27 or 28

p_1=29

???
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November 25th, 2007, 11:01 AM   #7
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Re: not sure

Quote:
Originally Posted by soandos
I am doing this:

p_0<P_1<P_0+(Sqrt(p) rounded up to the nearest prime).

how is that the same as:
there is a prime between p^2 and (p+1)^2?
(p+1)^2 = p^2 + 2p + 1 = p^2 + 2sqrt(p^2) + 1 ~= p^2 + 2sqrt(p^2)

The generalized Legendre conjecture is that there are some integers K, N where there is a prime between n and K sqrt(n) for all n > N. Legendre predicted in particular that it would hold for some N (probably 1) and K = 2. You're asking if it holds for K = 1.
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November 25th, 2007, 06:29 PM   #8
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not really

there is a difference as i am doing this:

between n and n+(sqrt(n) rounded up to the nearest prime).

not between n and k(sqrt(n))
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November 25th, 2007, 06:56 PM   #9
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Re: not really

Quote:
Originally Posted by soandos
there is a difference as i am doing this:

between n and n+(sqrt(n) rounded up to the nearest prime).

not between n and k(sqrt(n))
Not really, it just means you're looking for two primes instead of one.
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November 26th, 2007, 04:52 AM   #10
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no

i start out with a given p_0 2003 for example

i then take the square root ~44.75

i then round that up to the nearest prime 47

then i say that there is a prime between 2003 and 2003+47

there are:

2011
2017
2027
2029
2039

does make it clearer?
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