Recent content by flouran

I was wondering if one of the consequences of the Elliott-Halberstam conjecture would imply the Riemann Hypothesis (RH) or the Generalized Riemann Hypothesis (GRH)? Or at least if there is a connection between the Elliott-Halberstam conjecture and RH or GRH? I ask because the...

No. Managed to prove it.

I was looking through some tables of Laplace Transforms on f(t) the other day, and I noticed that in all cases, as s \to \infty, F(s) \to 0. A question that I have been trying to prove is that if \lim_{s\to\infty}F(s) = 0, then does that necessitate whether F(s) can undergo an inverse Laplace...

The residue representation of -1 in Z_{42} is (-1,-1,-1).

I was wondering what the residue number representation of -1 was? (Or for negative numbers in general) Thanks, flouran

[SIZE="7"]BUMP Anyone?

Let k and n \le X be large positive integers, and p is a prime. Define F(X,n) := \sum_{\substack{k^2+p = n\\X/2\le p<X\\\sqrt{X}/2 \le k < \sqrt{X}}}\log p Q(n) := \sum_{k^2+p = n}\log p.Note that in Q(n), the ranges of k and p are unrestricted. My question is: I know that F(X,n) and Q(n) can...

There are no limits on the value of n or m. Nice to see you on the forums, by the way Charles!

Would d(p) be different if it was instead the number of solutions to: F(n) \equiv 0 \pmod p? Here F(n) = n - q where q = m^2 is a perfect square where m is a natural number.

No, it's O. Look up "Vinogradov symbol" and you'll see that I am right.

f(x) >> g(x) translates as: f(x) = \Omega(g(x)). Although according to Knuth, using an equality in front of a Landau symbol is supposedly abuse of notation (apparently \in is preferred).

I have a rather simple question which requires a direct answer: We have two functions, f(x) and g(x). I know that f(x) << g(x) is the same as f(x) = O(g(x)). But if f(x) >> g(x), how can I write f(x) in terms of g(x) using the one of the four Landau symbols (\Omega, \omega, o, or O)? I...

I can be clearer: Let F(n) be a polynomial of degree g => 1 with integer coefficients. Let d(p) denote the number of solutions to the congruency F(n) \equiv 0 \pmod p for all primes p (and suppose that d(p) < p for all p). We may take F(n) = N-n^2, where N is an integer greater than (or equal...

I am an idiot. Hint: quadratic residue

I have a trivial upper bound for d(p). That is, d(p) < p-1 for p\nmid N. I think that suffices for my usages of d(p) for now.