Is there a function that returns $x$ with all repeated prime factors removed?

Asked Nov 08 '23 at 14:59

Active Nov 08 '23 at 19:36

Viewed 49 times

Let $x = f_1^{k_1} f_2^{k_2} \cdots f_n^{k_n}$ where $f_i \in \mathbb{P}^+$, $k_i \in \mathbb{Z}^+$ and $f_i \neq f_j \iff i \neq j$. I require a function that returns $x$ with all repeated prime factors removed.

Let us denote that function as $r$ such that $r(x) = f_1 f_2 \cdots f_n$.

I would like to know if there is a standardized function for this operation, or if it is something I need to define myself?

Thanks!

edited Nov 08 '23 at 19:36

Bill Dubuque

272,048

asked Nov 08 '23 at 14:59

Ryan Pierce Williams

5

This is known as $\operatorname{rad}(x)$ – TravorLZH Nov 08 '23 at 15:02
@TravorLZH Thanks :) – Ryan Pierce Williams Nov 08 '23 at 15:03
Not sure if this should be a comment, but it seems to me that you have already defined it. That is, decompose $x$ into its prime factorization, and let $r(x)$ be the product of all of $x$'s prime factors once each. Note that we require $x>1$ here – IAAW Nov 08 '23 at 15:03
@IAAW I'm working on writing up a proof, and I'd like to avoid redefining standard terms and functions – Ryan Pierce Williams Nov 08 '23 at 15:04
1

https://en.wikipedia.org/wiki/Radical_of_an_integer – user Nov 08 '23 at 15:08
@IAAW: You don't require $x>1$ if you adopt the standard convention that the product of the elements of an empty list is $1$ (just as the sum of the elements of an empty list is $0$). – TonyK Nov 08 '23 at 19:44

Is there a function that returns $x$ with all repeated prime factors removed?

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