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:
- $B_n = \sigma(\sigma(B_n)-B_n)$, which implies that $B_n-1$ is a Mersenne prime
- $B_n$ is the product of two squares
- A $B_n$-gon could be made with a ruler and compass
- 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!