A fun function involving primes

$\displaystyle \large H(x) = \sum_{p \, | \, x} \left(p*\log_p({\text{gcf}(x,p^{\lfloor \log_p x \rfloor})})\right)$ The idea for this function was adding together the largest $np$ such that $p^n \mid x$ and $p^{n+1}\nmid x$ for a given $x$ and all prime $p\le\sqrt{x}$ . For example, $H(28)=11$ since $28=7*2^2$ .

Here's a list of the first 30 $H(x)$ , starting from $x=2$ : $2,3,4,5,5,7,6,6,7 ,11,7 ,13,9 ,8 ,8 ,17,8 ,19,9 ,10,13,23,9 ,10,15,9 ,11,29,10,31,\dots$

I'm interested in learning more about this function. To start things off, here's some identities I've already found: $H(p_n)=p_n$ $H(ab)=H(a)+H(b)$ $H(x^b)=b*H(x)$ $H(x!)=\sum_{k=0}^x H(k)=\sum_{k=2}^x H(k)$

#NumberTheory

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

You don't mean all prime $p \le \sqrt{x},$ you mean all $p \le x,$ right?

The first two properties determine your function. If you look at $h(x) = 2^{H(x)},$ then $h$ is completely multiplicative.

There are some good links here. Apparently your function is sometimes called sopfr, for "sum of prime factors with repetition."

Patrick Corn - 3 years ago

In number theory there is a function $\log_{p}(x)=$ largest $n$ such that $p^{n}\mid x$ . Or in algebra the $p$ -adic evaluation delivers this $\nu_{p}(x)=n$ . In any case, we can do away with the multiple layers of superfluous terms in your function and write it simply as

$H(x):=\sum_{p\in\mathbb{P}}p\log_{p}(x)$

Noting that $\log_{p}(0)$ should be defined as $\infty$ , we have $H(0)=\infty$ , not $0$ (!!). I don’t know why you then say all prime $\leq\sqrt{x}$ that makes no sense. Then $H(p)$ would be $0$ for all $p\in\mathbb{P}$ .

Anyhow. Of interest would be, if you can extend $H$ to $\mathbb{Q}$ . Extending to $\mathbb{Z}$ is easy (just set $H(-n)=-H(n)$ or $H(-n)=H(n)$ however you like. Ideally, due to property 2, we would like to have $H(x/y)=H(x)-H(y)$ . It’s easy to see that this is in fact well defined, provided we choose $H(-n)=H(n)$ above $\ldots$ and provided you set $H(0)=\infty$ , otherwise you will run into problems.

Now set $|x|:=\exp(-H(x))$ for all $x\in\mathbb{Q}$ . Then $|x|=0$ iff $x=0$ and $|\cdot|$ is multiplicative. The operation is however not a norm, since $|5+-3|=|2|=\exp(-2)>\exp(-5)+\exp(-3)=|5|+|-3|$ .

R Mathe - 3 years ago

@Calvin Lin please take a look at this

Shivam Jadhav - 3 years ago

how do we write an interger

Kennedy Asiedu - 3 years ago

Another note: If we define a function $P_n(x)$ as the highest power of $n$ that divides $x$ , we can express this function as $\sum_{n=1}^{\pi(\sqrt{x})} p_n P_{p_n}(x)$ where $\pi(x)$ is the prime counting function.

William Crabbe - 3 years ago

I found some decent upper and lower bounds for this: $2k \le H(n) \le \frac{n}{2^{k-1}} + 2(k-1)$

Where $k$ is the number of prime factors of $n$ .

Levi Walker - 2 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$