9 votes

If for the first $\|n\|$ primes $p_i, \left(\frac{p_i}n\right)=+1$, then $n$ is a square

This is not known. However, it may be provably false under the Generalized Riemann Hypothesis (GRH), depending on a constant calculated in a paper of Montgomery. Least Quadratic Non-Residue You are ...
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  • 69.9k
6 votes
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Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$?

We have \begin{align*} \prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } & = \exp \left( { - \sum\limits_{2<p \le n} {\log \left( {1 - \frac{2}{p}} \right)} } \right) \\ &...
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  • 20.4k
4 votes
Accepted

Do we have the following property about positive integer sequences?

The answer is no, and $P=\{$all primes$\}$ provides a tidy counterexample: we know that $\sum_{p\in P} \frac1p$ diverges. However, given any $U\ge1$, the number of integers in $[U,2U)$ that are the ...
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  • 64.3k
4 votes

Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$?

Asymptotically, $$g(n) = \prod_{2 < p < n}^{p \text{ prime}} 1-\frac2p \sim \frac{4C_2e^{-2\gamma}}{(\log n)^2} \approx \frac{0.83244}{(\log n)^2}$$ And presumably your expression is simply the ...
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  • 1,385
3 votes
Accepted

Show that $\sum_{n\le x}\max(n)=O(x)$

Miracles often happen when we turn this expression into a multiple sum. Let $p,q$ denote primes. Then we have \begin{aligned} \sum_{n\le x}\max(n) &=\sum_{t\le\log_2x}\sum_{\substack{n\le x\\\max(...
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  • 3,635
3 votes
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Mertens' theorem with the numerator other than $1$

That first asymptotic relation is incorrect. For $p \geq 3$, $$ \frac{(1-1/p)^2}{1-2/p} = \frac{1-2/p+ 1/p^2}{1-2/p} = 1 + \frac{1}{p(p-2)} $$ and $\prod_{p\geq 3} (1 + \frac{1}{p(p-2)})$ converges by ...
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  • 30.7k
2 votes

Inverse Laplace transform of the Riemann zeta function

$\mathcal{L}$ means the bilateral Laplace transform. For each $\sigma\ne 1$, $t\mapsto \zeta(\sigma+it)$ is a tempered distribution, so its inverse Fourier transform makes sense in the sense of ...
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  • 68.4k
2 votes
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The composition of an algebraic function and a transcendental function

Is it true that the result is transcendental if I precompose a transcendental function with a non-constant algebraic function? Yes, it is true. Statement: Let $X,Y\subseteq\mathbb{R}$ (or $\mathbb{C}$)...
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  • 4,208
2 votes
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Prove that $\sum_{d|n}\mu(d)\log^m d=0$

A way to finish the claim of the linked post is the following: The required sum $$\sum_{d \mid s}{\mu(d)\log^m(d)}+\sum_{d\mid s}{\mu(pd)\log^m(pd)}$$ can be combined as $$\sum_{d \mid s} \left[ {\mu(...
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  • 8,427
2 votes
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Difference between the usage of Big-Omega notation as used by Computer Scientists and Mathematicians.

The math definition is that there exists some infinite subsequence $I=\{n_i:i\geq 1\}$of the natural numbers and a positive constant $c$ such that for all $n\in I,$ $$ f(n)\geq c g(n). $$ One could ...
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  • 6,378
1 vote

An integral formula valid for $\zeta(s)$ for all $s\neq 1$

If you complete the Riemann zeta funcition by multiplying by the "local factor at infinity", you have the integral representation $$L(s) = \pi^{-\frac{s}{2}}\Gamma(\frac{s}{2}) \zeta(s) = \...
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  • 30.9k
1 vote

An integral formula valid for $\zeta(s)$ for all $s\neq 1$

Wikipedia (https://en.wikipedia.org/wiki/Riemann_zeta_function#Integral) says, $$\zeta(s)={1\over s-1}+{1\over2}+2\int_0^{\infty}{\sin(s\arctan t)\over(1+t^2)^{s/2}(e^{2\pi t}-1)}\,dt$$ for all ...
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1 vote
Accepted

Product form of Mobius Inversion formula: $g(n)=\prod_{d|n}f(d)^{a(n/d)}\iff f(n)=\prod_{d|n} g(d)^{b(n/d)}$

Wikipedia's form is a specific case. Defining the unit function $u(n) = 1$, and defining the product Dirichlet multiplication by $$\newcommand{\p}{\;\square\;} (f \p g)(n) := \prod_{d \mid n} f(d)^{g(...
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1 vote
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Titchmarsh divisor problem

For the reciprocal sum of $1/\varphi(d)$, partial summation shall give $$ \sum_{\substack{d\le x\\(d,a)=1}}{1\over\varphi(d)}=c(a)\log x+O(1), $$ which indicates that \begin{aligned} 2\operatorname{Li}...
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  • 3,635
1 vote
Accepted

Help needed in deducing an inequality in a lemma in the proof of linnik's theorem.

So if $h\in \Omega_p$ that means $h$ is a part of one of the excluded residue classes mod $p$. By definition of $S$, any $n\in S$ will not be a part of $\Omega_p$ and thus we can never have $n\equiv h\...
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  • 1,616

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