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While tinkering with numbers, I found $n=f(x)$ where $n = \sigma(\sigma(n)-n)$ and $x \neq p$ where $p$ is a Mersenne prime exponent, but I need help input in regard to improving accuracy.

First off, I thought I'd try to explain how I got to this point, and please keep in mind that I am by no means a wizard at math, but I have a halfway decent understanding number theory and sequences due to my background as a programmer, so bear with me.

As previously stated, I was tinkering with numbers; Specifically Mersenne prime exponents, perfect numbers and powers of 2.

At some point I started to look at the relationship between perfect numbers and their binary representation, since it's proven that all perfect numbers are pernicious.

For example:

$ p=2:~~M_p=110_2\\\ p=3:~~M_p=11100_2\\\ p=5:~~M_p=111110000_2\\\ p=7:~~M_p=1111111000000_2\\\ ... $

Since all even perfect numbers are of this form, I started to investigate further, looking at the powers of two these numbers represent.

After making a myriad of different calculations, I came across a correlation between Mersenne prime exponents, Mersenne primes and perfect numbers that was previously unknown to me:

$$ B_n=\binom {B_{n \over 2},{{n \over 2} \mid 2}}{n+1} $$

Explanation of $B_n$:

$B_n$ is equal to $B_n \over 2$ while $n$ is divisible by $2$ else $B_n = n+1$

Expressed as a python snippet:

def B(n):
    if n%2==0:
        return B(n/2)
    else:
        return n+1

Based on this, I discovered that: If $n$ in $B_n$ is an even perfect number, I assume that:

  1. $B_n = \sigma(\sigma(B_n)-B_n)$, which implies that $B_n-1$ is a Mersenne prime
  2. $B_n$ is the product of two squares
  3. A $B_n$-gon could be made with a ruler and compass
  4. Is a $n$-th power of 2

Based on that assumption, it's safe to say that $B_n$ yields a perfect number, $P$, when $n$ of $B_n$ is a perfect number: $$ P = {(B_n-1)\over 2}\times B_n $$

In that case, assuming that $n$ is a Mersenne prime exponent, and $B_n=\binom {B_{n \over 2},{{n \over 2} \mid 2}}{n+1}$: $$ P_n = 2^{n-1} \times (2^n-1)\\\ P_n = {(B_{P_n}-1)\over 2}\times B_{P_n}\\\ B_{P_n} = 2^n = \sigma(\sigma(B_{P_n})-B_{P_n}) $$

With this in mind, one can see that:

$$ P_n = {(2^n-1)\over 2}\times 2^n $$

After generating the first 33 numbers of $\sigma(\sigma(B_{P_n})-B_{P_n})$ where $n$ is a Mersenne prime exponent, analyzed the sequence and I noticed that this sequence actually might be able to fit a formula.

So I decided to feed some processor cores with information with the help of number analysis software, and it seems as if there is a possible fit, in the form of a logistic curve...(?!)

I had no idea if I should get excited by this point, but it felt good seeing those numbers definable by a formula, so I felt pretty good!

So, here goes:

If: $$ f(x) = {a-b \over e^{-c (x-d)}+1}+b $$

where: $$ a = 1.36380\times 10^258716\\\ b = -1.97266\times 10^258709\\\ c = 10.0(+-)0.2\\\ d = 33.225(+-)0.004 $$ (Note: d $\approxeq$ size of sequence+.225(+-).005)

then: $$ G_n = ({a-b \over e^{-c (n-d)}+1}+b)-1 $$

If my assumption is correct, then $G_n$ is now the n-th Mersenne prime. $G_n$ should be $\approxeq$ to all Mersenne primes at least up to 33, which is the size of the sequence I based the formula on.

Is this a legitimate fit, or did I just get lucky?

Can anyone help me confirm/refute this?

Thanks!

share|improve this question
1  
See primes.utm.edu/notes/faq/NextMersenne.html and some of the linked pages; in particular there are conjectures about the log distributions of Mersennes and some supporting evidence. –  ARupinski Jan 23 '13 at 22:51
    
$33$ Mersenne primes? The first $42$ are now known, according to en.wikipedia.org/wiki/Mersenne_prime so you have some more data to check. –  Gerry Myerson Jan 23 '13 at 23:16
    
@ARupinski: Thanks for the information! It will surely come in handy! @Gerry Myerson: I am very aware of the fact that there are 42 known numbers/exponents, but computation $n > 33$ became very slow and memory intensive; And since I only have a sub-par laptop, I'd have to wait for some good amount of time in order to get the results. Further, computing a formula from $n > 33$ proved very expensive. –  JohnWO Jan 24 '13 at 2:01

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