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Apr
8
comment Unique factorization in $\mathbb Z(\sqrt{-19})$
Appreciate the late contribution. I will look at it as time allows. @mercio's answer did completely address my confusion, which was due to forgetting a definition.
Apr
6
awarded  Popular Question
Mar
19
comment Number of solutions of arithmetic funtion's equation.
Charles, thanks it's not necessary.
Mar
17
comment Inequality with prime numbers: $p_k+p_l+1\leq p_{k+l+1}$
Given your choice of notation it seems likely that Dusart's 1998 paper inspired the question. It seems worth citing. unilim.fr/laco/theses/1998/T1998_01.pdf
Mar
15
awarded  number-theory
Mar
9
comment Legendre's Conjecture limit version
It seems likely to me that you meant $\lim \pi((n+1)^2)-\pi(n^2)$ for the question but you should edit to reflect this if true.
Mar
9
revised Legendre's Conjecture limit version
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Mar
9
revised Legendre's Conjecture limit version
added 170 characters in body
Mar
9
answered Legendre's Conjecture limit version
Feb
28
comment How to prove $\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\leq 2n$?
@Andrea: In the comment I mentioned that in principle we can show it for $n(1+\epsilon)$ for large enough $n$ and a clever technique. So yes, in principle. Also remember that as $\epsilon $ gets small you have an increasing burden for the finite $n< n_0$.
Feb
28
comment How to prove $\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\leq 2n$?
@Andrea: No not useless at all. In some sense the problem is solved. But there are a lot of proofs that go to great trouble to prove a particular case.
Feb
28
comment How to prove $\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\leq 2n$?
@Andrea: Almost. For example the prime number thm. gives that there is a prime on $(n,n(1+\epsilon))$ for any $\epsilon>0$ for suff. large $n.$ But to prove a particular $\epsilon $ may require ingenuity. So to make a claim you have to really have to take the extra step. And how big is $n?$
Feb
28
revised How to prove $\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\leq 2n$?
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Feb
28
revised How to prove $\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\leq 2n$?
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Feb
28
answered How to prove $\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\leq 2n$?
Feb
26
comment How to prove $\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\leq 2n$?
Also for any positive $\epsilon$ we have $\prod p^{1/(p-1)} \leq (1+\epsilon)n$ for sufficiently large n.
Feb
24
awarded  Mortarboard
Feb
24
revised $k$-tuple conjecture.
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Feb
24
comment $k$-tuple conjecture.
I have edited my answer. It is more or less my earlier answer (and anon's) but you have simplified the question so I think this reflects the process up through Step 2 accurately.
Feb
24
revised $k$-tuple conjecture.
added 448 characters in body