All Questions
7 questions
0
votes
0
answers
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 ...
- 31
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 ...
- 1,008
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.
- 11.3k
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 ...
- 5,470