You know when you create the sieve of Eratosthenes you use all the primes, as generators of sub-sieves, whose value is less than the square root of n? Well...
... Yes. Let's call q=sqrt(n) in the above. The numbers you're talking about are "numbers with no factors less than q" and crop up in all kinds of contexts. ...
Do the margins representing maxima and minima for the predictions as to the number of places where no sieve components fall, closely resemble those of the PNT...
Can you please reply in group, so that others can see the discussion. If you're using the web interface, then select 'reply to all' in the pulldown above the...
A: Because it messes up the order in which people normally read text. Q: Why is top-posting such a bad thing? A: Top-posting. Q: What is the most annoying...
If one could prove that the residue of mod(P) over all larger primes was equally likely to be 1,2,3...p(n-1), would that in any way prove or be equivalent to...
... Yes, Dirichlet's original formulation says just that. I can't find out when it was strengthened to include the density statement, but anyway, the density...
... This is an interesting hunt. It _appears_ that Dirichlet's theorem does prove that /if/ the primes along the AP have a natural density, then that natural ...
... Which I think is strong enough support for my original claim. However, I'm not familiar with what analytic density is, so shall have to do a bit more...
Hello all: I have obtained a limit that related The Riemann Zeta function Zeta(i) the prime gaps pi-p(i-1) where i is integer and the nth prime number pn. ...
Wow, I get two copies - I'm so lucky! ... Yes, it's trivial and uninteresting. ... Yes, it's trivial and uninteresting. Exactly the same type of relation holds...
... Speaking of converging to one, I have wondered if there is an exponent k such that (1/2)^k + (1/3)^k + (1/5)^k + (1/7)^k + ... (1/p)^k + ... has a limit...
... 1/2^k + 1/3^k + ... + 1/p^k + ... just equals to Sum[MoebiusMu[n]*Log[Zeta[n*k]]/n, {n, 1, +Infinity}] which converges rapidly; so it can be used to...
... 1.399433328726330318202807214745644327904727429484383941274765822888062492487247800233390475384227... (if you need more digits) Best, Andrey [Non-text...
As we know, the number of primes up to n is about n/log(n). Given this, it is easy to show that the number of primes between n^2 and (n+1)^2 is also about...
1 * 1 = 7 * 13 = 11 * 11 = 17 * 23 = 19 * 19 = 29 * 29 = 1 mod 30 For the integer k, If for every value of n, none of m1 = (k-n)/(30*n+1) m2 = ( k - 7 * n -...
Hi, Mike Oakes said: 2^11795+11795 137-PRP by PFGW (PRIMO certification would take c. 5 days) My computer says: [PRIMO - Task Report] Version=2.3.1 ...
... Just did some prime counting and so far it holds that the number of primes between n^2 and (n+1)^2 is within the range n/log(n) +/- sqrt(n). At least for n...
This is a very surprising result. It is over 14 million lower than the likely value that Gourdon found. Has it been checked by computing an offset value and...
In the same way one proves that the number of primes between cubes amounts to approximately: N = pi(n+1)^3-pi(n^3) ~ (n/ln(n))*(n+1) ~ pi(n)*(n+1), better ~...
Hi , ... primes ... prove or ... likelihood seems ... does prove ... natural ... the natural ... Phil , I apologize for being such a bad correspondent , I owe...
Walter Nissen
wnissen@...
Apr 13, 2007 2:38 am
18873
Why can't prime numbers be analyzed in terms of a sequential prime generating algorithm? Due to some very basic playing with the numbers, I can immediately see...
... Most of the properties of integers in general depend on their prime factorizations, so the study of primes is fundamental to the study of numbers in...
When you take your Sieve of Eratosthenes and consider values of n well beyond the square of the highest prime that you have selected to give a filter, does...
This amused me: http://www.xkcd.com/c247.html Of course the pedant in me makes me think that the representation is a mixed radix one, and that 2:53 is in fact...
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