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It depends on what range of $\epsilon$ you are interested in. If the exponential sum is near its max possible size ($> \delta N$) then you are saying that your sequence correlates with an exponential …

Estimates on these quantities are used in Elsholtz and Tao's work on the Erdos-Straus conjecture. See their paper "Counting the number of solutions to the Erdos-Straus equation on unit fractions" and …

Zhang's strategy shows that any admissible tuple of size h contains at least 2 primes infinity often, as long as h is larger than some threshold (the primary theoretical thrust of the polymath project …

Updated answer: Here's an unconditional lower bound of exactly the right order $\sqrt{a}$, however I need $a$ to be comparable to $p$. It is known that (for any $a$) $ \frac{1}{p-1} \sum_{\chi} |S(\ …

See "On the number of primes $p$ for which $p+a$ has a large prime factor." (Goldfeld, Mathematika 16 1969 23--27.) Using Bombieri-Vinogradov he proves, for a fixed integer $a$, that $$\sum_{p \leq x …

For even integer exponents, say $p=2k$ and $p \geq2$, the quantity is just the $k$-order additive energy of the set $S \subset \mathbb{Z}$ of non-zero Fourier coefficients. It is easy to see that this …

Yes, Speiser's theorem is an if and only if. See Theorem 1 and "Corollary to Theorem 1" in Levinson and Montgomery's Zeros of the derivatives of the Riemann Zeta-function. Acta Math. 133 (1974), 49–65 …

This result (or at least the method) probably goes back to Vinogradov or Davenport. For an explicit statement/proof of this result you can see, for instance, Theorem 1 in: J. Liu, T. Zhan, Estimation …

J. Friedlander and H. Iwaniec, The polynomial $X^2+Y^4$ captures its primes. Ann. of Math. (2) 148 (1998), no. 3, 945--1040. see: http://www.ams.org/mathscinet-getitem?mr=1670069 http://www.jstor.o …

This might not be exactly what you have in mind, but it is an old result of Paley that for an infinite (but very sparse) set of real Dirichlet characters the inequality $max_{N} |\sum_{n}^{N} \chi(n) …

For a fixed a>0 it is known that there are infinitely many moduli q such that the least prime in the arithmetic progresion a mod q is at least >> q ln (q) lnln (q) ln ln ln ln (q) / (ln ln ln(q) )^2. …

Given such a $M$ note that $F(x)=M^2(x)$ will be (1) non-negative, (2) majorize $1_{[a,b]}$, and (3) has $\hat{F}$ supported in $[-2\delta,2\delta]$. Thus it will be a (possibly not optimal) solution …

As Joel mentions, this follows from the work of Tao and Ziegler. Alternatively, this can be directly deduced from the density of the primes and the known bounds on the Furstenberg–Sárközy theorem. Ind …

When I was a graduate student, I worked out a more simple direct variant of this argument avoiding any use of matrics, bilinear forms, etc. It's so simple I can't imagine I was the first to do this. h …

As Alexey has pointed out the problem can be reduced, via summation by parts, to understanding the asymptotic of Mertens' sum $$M(x) := \sum_{n\leq x} \mu(n).$$ Conditional on the Riemann hypothesis, …