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At first, we count the number of residues $a$ for which both $a$ and $a+1$ are coprime with $n$. Let $q=\prod p_i^{k_i}$ be a factorization of $q$. For any $p_i$, there exist $p_i-2$ admissible remain …

As Will Sawin observes, there is no such function supported on $[1,2]$. For every larger interval $[a, b]$, $a<1<2<b$, you may take your favourite function $f$ supported on $[a, b]$ and positive on $( …

UPD. GH's approach is, of course, better, the reason is that it uses a stronger asymptotical estimate: asymptotics $\sum_{p\leqslant x} \log p/p=\log x+O(1)$ implies $\sum_{p\leqslant x} 1/p=\log\log …

$\DeclareMathOperator{\ap}{ap}$ $\DeclareMathOperator{\rad}{rad}$ I think so. Fix $\rad(m)=p_1\ldots p_k=:P$. Denote by $\Omega$ the set of positive integers with all prime divisors in $\{p_1,\ldots,p …

Denote $f_p(n)=\log(p)\cdot {\textbf 1}_{p|n}$. Your event is $\sum_{p^2<x} f_p<\log x$. The random variables $f_p$ are independent, so this is a sort of law of large numbers for not identically distr …

For large $n$ this product behaves like $(e+o(1))^{p_n}$, this is equivalent form of Prime Number Theorem. I do not know, however, whether $e^{p_n}$ is always an upper bound.

Asymptotics of such integral may be obtained by applying appropriate Tauberian theorems. If we denote $x_i=e^{t_i}$, then your integral is $F(e^{\alpha k})$, where $F(T)$ is a volume of the set $\{t_i …

Quick observation. Choose $\alpha$ close to 1 and consider the number for which $x\leqslant n^{\alpha}=:N$, $p(x)\leqslant n^{1-\alpha}=N^{(1-\alpha)/\alpha}$. The number of such $x$ grows as $\rho(\f …

It is problem number 10 of IMC 2018, you may find the solution on the official site.

Denote $\pi(x)=M\sim x/\log x$. Then $j$ varies between 1 and $M$, $p_j=j\log j+o(M\log M)$, and for $j_1,\ldots j_k$, denoting $j_i=Mt_i$ we have $$p_{j_1}+\ldots+p_{j_k}=\sum j_i\log j_i+o(M\log M)= …

Yes, and you get what should be expected (that is, $\sum_{\substack{|D_1D_2|\le x \\ D_1,D_2 \in A}}1\sim \frac{x\log x}{\zeta(2)^2}$.) For proving this you may choose $\rho>1$ (close to 1) and partit …

No $x$ less then $1/2$ may satisfy $S_N=o(n^x)$. Indeed, denote $f(x)=\chi_{[0,\pi]}-\chi_{[\pi,2\pi]}$, then we are interested in $|f(\theta)+f(2\theta)+\dots+f(n\theta)|$ for specific value of $\the …

This is false. Take $n$ equal to the product of first $k$ odd primes, or 3 times larger (so that $n-1$ is divisible by 4). For large $k$, $\varphi(n)/n=(1-1/3)(1-1/5) \ldots (1-1/p_{k+1})$ tends to 0. …

There is no known non-trivial (less than 1) bound for real parts of Zeta zeros (I guess, it is even called "weak Riemann conjecture" to find such a bound). So, your result is very-very interesting, ma …

As stated the answer is negative. Denote $N=lcm \{p_i-1|1\leq i \leq k\}$. Then $t^{N+1}-t$ is always divisible by $n$. But it may appear that $\sum (p_i-1)=N+N/2+\dots+N/k$, so it may be much greater …