Question

Oh I am curious about perfect numbers if you know anything interesting, please share. My other question is, similar to how they satisfy $$\sigma(n)=2n $$ is there a name for numbers that satisfy $$\sigma(n)=3n$$ or other integer multiples?

Answer 1

For people who are also curious, a perfect number is equal to the sum of its divisors. So for example here's a simple perfect number \[\Large 1+2+3=6\] since 1,2, and 3 are all the divisors of 6 and add to 6.

Answer 2

@rational @ikram002p

Answer 3

\( (2^{p-1})(2^{p}-1)\) will be an even perfect number as long as p is prime.

Answer 4

That is the Euclid-Euelr theorem fyi :P

Answer 5

Oh that's interesting, I wonder why that works

Answer 6

One good question is, are there any odd perfect numbers?

Because none have been found.

Answer 7

I guess that means that there exists or there might exist some perfect numbers that aren't of that Euler form, since those are all going to be even.

Answer 8

It also seems to be breaking a kind of symmetry, since the number they have is \[\Large \frac{2^p(2^p-1)}{2}\] but what about \[\Large \frac{(2^p+1)2^p}{2}\] I wonder if there's any relationship here to triangular numbers or if there are other numbers with a relationship to sum of squares or cubes... hmm

Answer 9

If \(\sigma(n) = kn\) then \(n\) is called a \(k\)-perfect number
example : \(\sigma(30240) = 4*30240\)

there is a conjecture that for each \(k\ge 3\) there can exist only finitely many \(k\)-perfect numbers

Answer 10

Interesting! That's exactly what I wanted to hear about. Now I've really been interested in Mersenne primes ever since we were talking about how \(2^p-1\) can only be prime if p is prime and writing it in binary. Actually I'm trying to remember which thing you asked a little while back where I had tried to prove it because I had mistakenly believed it was proven there was infinitely many Mersenne primes...

Specifically while we're on the topic of Mersenne primes, I was wondering about this recursion of: \[\Large f(n)=2^n-1\] Since it seems the more we do it the more _likely_ it will be for it to become a prime number in a way.\[\Large f(f(f(...)))\]

Answer 11

fixed typo :
If \(2^p-1\) is prime, then \(2^{p-1}(2^p-1)\) is a perfect number and every even perfect number is of this form.


\(2^p-1\) is called mersenne prime. we don't know whether there are infinitely many mersenne primes. one interesting thing to notice is that solving mersenne primes problem solves the even perfect numbers problem too because they both go hand in hand. above proposition tells us that a perfect number can be constructed from mersenne prime.

Answer 12

are you looking at numbers of form \(\large 2^{2^{2^{\vdots}}}-1\) ? these are trivially composites right ?

Answer 13

Nooo, cause there's still the -1 on each component.
\(\large 2^{2^{2^{\vdots}-1}-1}-1\)

Answer 14

ohkay i see..

Answer 15

Yeah, because if \(2^p-1\) is prime then p is prime we know if p is composite that \(2^p-1\) is always composite. So by making \(2^{2^p-1 }-1\) we are in a sense "more likely" for it to be prime. I don't know, just an interesting thing I was playing with.

Answer 16

Nice :) that rapidly becomes a very huge number because of tetration...

there is a nice way to test primality of mersenne primes
http://math.stackexchange.com/questions/1209599/factorization-of-huge-integer/1209633#1209633

this reduces the effort by a factor of 2p+1 which is much better than checking all prime divosors because we look at very large primes in our search

Answer 17

thats my fav tool for checking primality of mersenne numbers

Answer 18

there is another interesting number form which looks similar to mersenne numbers but has entirely different properties : \(\large 2^k+1\)

Answer 19

i have been thinking of them as you have mentioned about that form in the start of this thread

Answer 20

Interesting! I'm working through what you put on MSE but I am interested in these as well, if you want to share, I'd like to hear!

What I was really interested in from the start is I realized that \[\Large n=\frac{p(p+1)}{2}\] is a perfect number if p is a Mersenne prime so I was wondering if there was an analogue like \[\Large n=\frac{p(p+1)(2p+1)}{6}\] where n was a 3-perfect number if \[\Large p=3^p-2\] was kind of like the "3 Mersenne prime" sorta analogue kinda thing or something like this existed. Basically where it generalized to sums of cubes, etc... and higher as well.

Answer 21

But I mean this is all related so I'm not really going in any particular direction on this problem just sorta feeling it out, I just wanted to get that off my chest as my motivation for starting this, I'm really interested in these 2^k+1 primes at the moment.

Answer 22

\[2^{p-1}(2^p-1) = \dfrac{(2^p-1)2^p}{2} = T_{2^p-1}\]

I agree this cannot be a coincidence

Answer 23

Yeah, it's part of this sum of divisors expression, I think that is clear. I really should read up on how Euler connected the two, it's just shortly after this I discovered something fascinating in the Collatz Conjecture and took a turn down that road haha.

Answer 24

that proof is pretty straightforward and im sure its too easy for you because sigma is ur fav function

Answer 25

Haha well actually I like the tau function more but... I'm still just a baby at these, but they keep having these surprising properties to them. It's kind of like when someone shows you \[\Large 10^2+11^2+12^2=13^2+14^2\] for the first time. Just fun. =D

Answer 26

^ that is interesting. Haha :)

Answer 27

Indeed it is! for some reason it reminds me of identity \[(a^2+b^2)(c^2+d^2) = (ac+bd)^2+(ad-bc)^2\]

i remember we worked its proof many years ago..

Answer 28

you had used complex numbers or some other weird stuff i think..

Answer 29

Too much to read , sort of busy hope to read this sometime

Answer 30

that comes from complex numbers