daniel
Reputation
4,692
Top tag
Next privilege 5,000 Rep.
Approve tag wiki edits
 Sep 6 asked Approximate zeros of a (hypothetical) analog of $\zeta(s)$ Sep 5 comment A (possibly) easier version of Bertrand's Postulate Bertrand implies a prime on p, 2p. Choose p(n) max less than a non-prime n. Then there is a prime on n, 2p(n) which implies a prime on n,2n (Bertrand). So I think the two are equivalent. You don't need case 1, for the reason you give. Sep 5 comment I made an observation on prime numbers, want to check if any conjecture already exist or not? @ColmBhandal: Yes, just an example of something along these lines that works. The OP conjecture is a special case of Opperman (is implied by it without any work) but I'm not sure why the reverse would seem possible. May 8 awarded Popular Question 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 added 12 characters in body 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 $\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\leq 2n$? added 5 characters in body Feb 28 revised How to prove $\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\leq 2n$? added 5 characters in body 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.