Just looking for any and all cute closed/alternate forms for arithmetic functions.

Question

kainui · Answer

Here's one I found on my own: von Mangoldt function: $$\Lambda(n) = \ln[ rad(n) \delta(\omega(n)) ] $$ Functions used for the curious: rad(n) Basically makes all the exponents on the prime factorization =1. https://en.wikipedia.org/wiki/Radical_of_an_integer $\delta(n) = \lfloor \tfrac{1}{n} \rfloor $ (In other words, 1 if n=1, 0 otherwise for positive integers) $\omega (n)$ number of distinct prime factors http://mathworld.wolfram.com/DistinctPrimeFactors.html

amistre64 · Answer

Does this count?

amistre64 · Answer

forbidden script?  lol ... well that was unexpected

ganeshie8 · Answer

.

amistre64 · Answer

copy to notebook and save as html file ....

kainui · Answer

Hahahhaa @amistre64 
$$\huge \color{red} \heartsuit$$

kainui · Answer

Hahahhaa @amistre64 
$$\huge \color{red} \heartsuit$$

amistre64 · Answer

:)

kainui · Answer

Here's another one I just found, it's not particularly crazy or serious haha:

$$rad( n!) = n \#$$

$n \#$ is the primorial, it's like a factorial only it is the product of all primes less than or equal to n.

kainui · Answer

Just discovered a new thing pretty cute:

$$rad^2(n) = \sum_{d|n} \mu^2(d)\varphi(d) \sigma(d)$$

kainui · Answer

Just defined a new arithmetic function for fun:
for n defined this way:
$$n = \prod_{primes} p^{k}$$
My new multiplicative function is:
$$\chi(n) = \prod_{primes} p^{kp}$$
and taking the arithmetic derivative of this function seems to give:

$$\chi'(n) = \Omega(n)\chi(n)$$

Also it appears that $\chi(n)$ has an inverse,

$$\chi^{-1}(n) = \prod_{primes} p^{k/p}$$

So by that I mean $\chi^{-1}(\chi(n))=\chi(\chi^{-1}(n))=n$

(I haven't really proven this to myself yet so I don't know if these are all true, just thought they were fun)

ganeshie8 · Answer

Very interesting! do we have  :
$$\sum\limits_{d\mid n} \mu(d) \chi(d) = \prod\limits_{\text{primes}} (1-\chi(p))$$

kainui · Answer

I'll denote the Dirichlet convolution of something with itself n times like this:
$$f^{\star n}(a)$$
So for example:
$$f \star f \star f \star f = f^{\star 4}$$

It looks like this is true:

$$\mu^{\star n} (p^k) = (-1)^k\binom{n}{k}$$

And since it's multiplicative, we have not too much trouble just finding the general form of the nth mobius convolution with itself.

kainui · Answer

@ganeshie8 Ahhh cool I think we might I'm still trying to work it out, I am also curious what $\chi \star u$ will look like next as well.

So far I am getting a weird result though haha

$$(\mu \star \chi)(p^k) = \chi(p^k) ( 1 - \tfrac{1}{p} ) = \phi(\chi(p^k))$$

kainui · Answer

Oh I see I am calculating something slightly different than you, no wonder I am getting different results.

I was doing this:
$$(\mu \star \chi)(n) = \sum_{d|n} \mu(d) \chi ( \tfrac{n}{d})$$
You were doing this:
$$((\mu* \chi) \star u)(n) = \sum_{d|n} \mu(d) \chi (d)$$

ganeshie8 · Answer

yeah mine is not a straight convolution

ganeshie8 · Answer

More generally, for any multiplicative function $f(n)$ we have: 
$$\sum\limits_{d\mid n} \mu(d)f(d) = \prod\limits_{\text{primes}} (1-f(d))$$

kainui · Answer

Awesome, I got it! Actually this just made me realize the same thing in a slightly different way that you've just written,  really we can replace anywhere we find $N(n)=n$ with $\chi(n)$ and so it has all the same convolutions as it.

kainui · Answer

For fun I wanted to see what using the arithmetic derivative would give us for an upper bound under a certain peculiar circumstance...

$$\chi(n) = \prod_{primes} p^{kp}$$

$$\chi'(n) = \Omega(n)\chi(n)$$

Now two new relations:

$$\chi'(rad(n)) = \omega(n)\chi(rad(n))$$
$$\chi'(n\#) = \pi(n) \chi(n \#)$$
using the upper bound on the arithmetic derivative:
$$\chi'(n\#) \le  \chi(n \#)\frac{\log_2( \chi(n \#))}{2}$$
$$\pi(n)  \le  \frac{\log_2( \chi(n \#))}{2}$$
$$\pi(n)  \le  \frac{1}{2}\sum_{primes} p \log_2(p)$$

However that's not really too exciting since we already know

$$\pi(n)  \le  \sum_{primes} p$$

Hahaha at least I tried. xD