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On an inequality involving the radical of an integer and its greatest prime factor
Let $n\geq 1$ an integer, in this post I denote the greatest prime dividing $n$ as $\operatorname{gpf}(n)$, and the product of the distinct prime numbers dividing $n$ as $$\operatorname{rad}(n)=\prod_{...
user243301
1
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1
answer
243
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Square-free integers in the sequence $n^{\operatorname{rad}(n)}+\operatorname{rad}(n)+1$
For integers $n\geq 1$ in this post we denote the square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing an ...
user243301
5
votes
1
answer
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What about sequences $\{\sum_{k=1}^n (\operatorname{rad}(k))^p\}_{n\geq 1}$ containing an infinitude of prime numbers, where $p\geq 1$ is integer?
We denote the radical of the integer $n> 1$ as
$$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\ p\text{ prime}}}p,$$
taking $\operatorname{rad}(1)=1$ that is this definition from Wikipedia.
In ...
user243301
2
votes
1
answer
105
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Square-free integers in the sequence $\lambda+\prod_{k=1}^n(\varphi(k)+1)$, where $\lambda\neq 0$ is integer
While I was exploring the squares in the sequence defined for integers $n\geq 1$
$$\prod_{k=1}^n(\varphi(k)+1),\tag{1}$$
where $\varphi(m)$ denotes the Euler's totient function I wondered a different ...
user243301
2
votes
0
answers
48
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Compare $\sum_{k=1}^n k^{\operatorname{rad}\left(\lfloor\frac{n}{k}\rfloor\right)}$ and $\sum_{k\mid n}k^{\operatorname{rad}\left(\frac{n}{k}\right)}$
I would like to know how do a comparison between the sizes of these functions defined for integers $n\geq 1$, when $n$ is large
$$f(n):=\sum_{k=1}^n k^{\operatorname{rad}\left(\lfloor\frac{n}{k}\...
user243301