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Number of Primes Between n and 2n 
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#1
Jul1911, 12:20 PM

P: 1,280

As we travel further up the number line, primes become more scarce.
But, as n grows larger and larger, the range of numbers between n and 2n grows larger ad larger. Do these two counteract each other? Does this cause the number of primes between n and 2n to stay relatively consistant, increase, or just shrink regardless? 


#2
Jul1911, 03:40 PM

Sci Advisor
P: 6,031

Using the prime number theorem, this can be estimated.
π(n) ~ n/ln(n), so π(2n)  π(n) ~ 2n/ln(2n)  n/ln(n) which is approximately n/ln(n). 


#3
Jul2211, 10:37 PM

P: 828

Chebyshev (and Erdos) said it and I'll say it again: there's always a prime between n and 2n.



#4
Jul2311, 04:01 AM

P: 460

Number of Primes Between n and 2n



#5
Jul2311, 04:02 PM

P: 828

Yes, in the sense that n/logn > infinity as n > infinity. (Just use L'Hospital's to see this.)



#6
Jul2311, 04:59 PM

P: 460

I don't really like the quote by chebyshev and erdos because it is bit misleading. 


#7
Jul2511, 07:48 PM

P: 828

Well, the quote is by Erdos. See, Chebyshev proved that there was always a prime between n an 2n using a rather technical argument. Erdos proved it (I think when he was 19 or so) using a combinatorial argument, so he said "Chebyshev said it I'll say it again, there's always a prime between n and 2n."
But what is misleading about it? 


#8
Jul2611, 12:15 AM

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P: 16,099




#9
Jul2611, 07:37 AM

P: 460




#10
Jul2611, 12:23 PM

P: 828

Well sure, what we said was a sharper result than Bertrand's Postulate. I still don't see how this is misleading (though I will agree it really doesn't have much to do with the the thread.) The fact that there is ALWAYS a prime between n and 2n for all n is interesting on its own, espicialy when you consider that I can give you an arbirarily long list of consecutive composite integers.
I just posted the quote as a joke in the first place. 


#11
Jul2611, 12:57 PM

P: 460

so in a simillar fashion r/n goes to zero as r goes to infinity. Remember that r is the number of consecutive composite integers and n is the integer where this sequence starts. Is my gut mistaken? It's all good, you don't have to see it my way, my opinions are not 'etched in stone' 


#12
Jul2711, 05:19 AM

P: 828

You're correct; its huge. The sequence begins at (n+1)!.
But that's not really the point. The fact that PNT, in a way, implies Chebyshev's theorem does not, in any way, mean that Chebyshev's Theorem is "misleading" or unimportant. And it certainly does not take away from the fact that Erdos came up with a magnificent proof of it. PNT also implies that there are infinite primes, but this does not take anything away from Eucilid's proof of the same. 


#13
Aug2811, 06:25 AM

P: 460




#14
Aug2811, 07:49 AM

P: 828

I was speaking of constructing a sequence like this: (n+1)! + 2, (n+1)! + 3, ..., (n+1)! + n + 1 This is a list of n consecutive composites. When you say "not necessarily" do you mean that you know of another way to construct an arbitrarily long list of composite numbers? If so, I'd be interested. Or, do you mean that not every list of m consecutive composites has to be constructed the way I describe? If this is the case, then I don't really care as I never claimed that the ONLY way to construct a list of n composites was the construction I mentioned above. Perhaps what I wrote in my last post was confusing, but that's only because I thought we understood that I was referring to creating a list of size n for a general n, not a specific n. Certainly for "most" n, there are much lower sequences of n consecutive composites; but this doesn't tell us anything about a general n. 


#15
Aug2811, 08:28 AM

P: 894




#16
Sep511, 11:23 AM

P: 96

From my 'numerical department', the number of primes between n and 2 n is:
100 < #21 > 200 ... 1000 < #135 > 2000 ... 10000 < #1033 > 20000 ... 100000 < #8392 > 200000 


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