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: \[H(p_n)=p_n\] \[H(ab)=H(a)+H(b)\] \[H(x^b)=b*H(x)\] \[H(x!)=\sum_{k=0}^x H(k)=\sum_{k=2}^x H(k)\]

Note by William Crabbe
3 weeks, 5 days ago

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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 - 2 weeks, 6 days ago

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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


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 - 2 weeks, 1 day ago

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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 - 2 weeks, 1 day ago

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How is that fun?????

Annie Li - 2 weeks, 4 days ago

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how do we write an interger

Kennedy Asiedu - 2 weeks, 6 days ago

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@Calvin Lin please take a look at this

Shivam Jadhav - 3 weeks, 3 days ago

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