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68 views

Around similar inequalities than an inequality due to Nicolas, that involve products of consecutive Ramanujan primes

This is cross-posted (and this post is a version to ask just around the veracity of Conjecture 1) as the post with identifier 3594907 and same title), that I've edited on Mathematics Stack Exchange ...
1 vote
1 answer
347 views

On equations with arithmetic functions [closed]

Is this good topic for research: equations with arithmetic functions, for example equations like $\varphi(n)=\sigma(n)$ or $\varphi(n)+\sigma(n)=d(n)$ ? If Anyone here have an advise please tell me ...
0 votes
0 answers
89 views

A similar inequality for the Dedekind psi function, than an inequality stated by Schinzel

I would like to ask about the next question that seems to me interesting. I know an article that was written by Andrzej Schinzel in which he stated Lemma 2. In this post we denote the Dedekind psi ...
3 votes
1 answer
134 views

Is it possible to find an estimate of $\sum_{k=1}^n\frac1{\varphi(k\cdot p_k)}$?

Is it possible to find an estimate of the summation $$s(n)=\sum_{k=1}^n\frac1{\varphi(k\cdot p_k)}$$ where $\varphi(n)$ is the totient function and $p_k$ the k-th prime? The corresponding series seems ...
7 votes
1 answer
231 views

The asymptotic of $|\{1\leq n\leq x|\gcd(n,S(n))=1\}|$, with $S(n)$ the sum of remainders, and get idea for other miscellany problem

Let $n\geq 1$ be an integer. In this post we denote the sum of remainders function as $$S(n)=\sum_{k=1}^n n \bmod k,$$ for example $S(1)=S(2)=0+0$ and $S(5)=0+1+2+1+0=4$. In the literature there are ...
8 votes
1 answer
427 views

Goldbach's conjecture for the Liouville function

Is it true that for every even integer $N > 2$, there exist positive integers $a,b$ such that $a + b = N$ and $\lambda(a) = \lambda(b) = -1$ ? Here $\lambda$ is the Liouville function.
5 votes
2 answers
314 views

Congruences for the non-divisors of Euler's $\phi(n)$

If $n$ is composite, then $\phi(n) < n-1$: hence, there is at least one number $d$ which does not divide $\phi(n)$ but divides$(n-1)$. We shall call $d$ the totient divisor of $n$. The purist will ...