A fun function involving primes

H(x)=px(plogp(gcf(x,plogpx)))\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 npnp such that pnxp^n \mid x and pn+1xp^{n+1}\nmid x for a given xx and all prime pxp\le\sqrt{x}. For example, H(28)=11H(28)=11 since 28=72228=7*2^2.

Here's a list of the first 30 H(x)H(x), starting from x=2x=2: 2,3,4,5,5,7,6,6,7,11,7,13,9,8,8,17,8,19,9,10,13,23,9,10,15,9,11,29,10,31,2,3,4,5,5,7,6,6,7 ,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(pn)=pnH(p_n)=p_n H(ab)=H(a)+H(b)H(ab)=H(a)+H(b) H(xb)=bH(x)H(x^b)=b*H(x) H(x!)=k=0xH(k)=k=2xH(k)H(x!)=\sum_{k=0}^x H(k)=\sum_{k=2}^x H(k)

Note by William Crabbe
3 years, 2 months ago

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

The first two properties determine your function. If you look at h(x)=2H(x),h(x) = 2^{H(x)}, then hh 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, 2 months ago

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In number theory there is a function logp(x)=\log_{p}(x)= largest nn such that pnxp^{n}\mid x. Or in algebra the pp-adic evaluation delivers this νp(x)=n\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 logp(0)\log_{p}(0) should be defined as \infty, we have H(0)=H(0)=\infty, not 00(!!). I don’t know why you then say all prime x\leq\sqrt{x} that makes no sense. Then H(p)H(p) would be 00 for all pPp\in\mathbb{P}.

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

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

R Mathe - 3 years, 1 month ago

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@Calvin Lin please take a look at this

Shivam Jadhav - 3 years, 2 months ago

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how do we write an interger

Kennedy Asiedu - 3 years, 2 months ago

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

William Crabbe - 3 years, 1 month ago

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I found some decent upper and lower bounds for this: 2kH(n)n2k1+2(k1) 2k \le H(n) \le \frac{n}{2^{k-1}} + 2(k-1)

Where kk is the number of prime factors of nn.

Levi Walker - 2 years, 8 months ago

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