Kainui
  • Kainui
Just looking for any and all cute closed/alternate forms for arithmetic functions.
Mathematics
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SOLVED
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chestercat
  • chestercat
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Kainui
  • Kainui
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
  • amistre64
Does this count?
amistre64
  • amistre64
forbidden script? lol ... well that was unexpected

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ganeshie8
  • ganeshie8
.
amistre64
  • amistre64
copy to notebook and save as html file ....
Kainui
  • Kainui
Hahahhaa @amistre64 \[\huge \color{red} \heartsuit\]
Kainui
  • Kainui
Hahahhaa @amistre64 \[\huge \color{red} \heartsuit\]
amistre64
  • amistre64
:)
Kainui
  • Kainui
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
  • Kainui
Just discovered a new thing pretty cute: \[rad^2(n) = \sum_{d|n} \mu^2(d)\varphi(d) \sigma(d)\]
Kainui
  • Kainui
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
  • ganeshie8
Very interesting! do we have : \[\sum\limits_{d\mid n} \mu(d) \chi(d) = \prod\limits_{\text{primes}} (1-\chi(p))\]
Kainui
  • Kainui
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
  • Kainui
@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
  • Kainui
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
  • ganeshie8
yeah mine is not a straight convolution
ganeshie8
  • ganeshie8
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
  • Kainui
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
  • Kainui
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

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