In this thread

http://projectavalon.net/forum/showthread.php?t=18834 you mention that all perfect numbers are the sums of the ODD NUMBERS CUBED. That is true only if a number is perfect, but the other way around is not necessarily true.

Of course; A implies B does not necessarily imply B implies A.
All bachelors are Men is true, but does not mean that all Men are bachelors.
For example, take first 16 odd numbers and cube them, you will get 130816, which is the next number after 8128. This number is equal to 511*512/2, where 511 is not a Mersenne prime, so this can not be a perfect number. It is true that this number can be expressed as the sum of first 511 numbers, but that doesnt make it a perfect number. Here is the definition of a Perfect number:

http://mathworld.wolfram.com/PerfectNumber.html.

You have just reiterated what I have stated in the post you mentioned. Please make sure you understand the presented data corrctly, before critisizing it.
Interesting formula thou.

I have been always fascinated by numbers, and so I did notice the fabulous connection of number 2. Namely, number 2 satisfies the equation x+x=x*x=x^x. I have also noticed that 2^4=4^2.

Yes but this follows by definition of 4=2^2; so 4^2=(2^2)^2=2^4 by definition of the exponential functions.
So I have wondered If any other numbers satisfies the equation x^y=y^x. So I found a formula to find infinite many numbers that satisfies the above condition.

Here it is:

x=n*n^(1/(n-1))

y=n^(1/(n-1))

Here n*n^(1/(n-1))=n^[1+1/(n-1)]=n^[n/(n-1)]
Then x/y=n^{n/(n-1) -1/(n-1)}=n^{1}=n and so x=ny
Taking natural logs lnx=ln(ny) for x=e^[n.lny] or y=e^[lnx/n]
This reduces to your initial statement x^y = y^x in taking natural logarithms and using the definition for x=ny
ylnx=xlny=y.ln(ny)=ny.lny or ln(ny)=n.lny
But ln(ny)=ln(n)+ln(y) by definition of exponential/logarithmic function and so
ln(n)+lny=n.lny for ln(n)=(n-1)lny=ln(y)^[n-1] and n=y^[n-1]
Now eliminate n=x/y for x/y=y^[(x-y)/y] and
x=y^[(x-y+y)/y]=y^[x/y]
Raise both sides by the power of y for: x^y = y^x
This equation so is INDEPENDENT of n in the definition n=x/y.
where n is some real number.

If you take n=2 you will get 2^4=4^2.

If you take n=3, you will get (3*(3^(1/2)))^(3^(1/2))=(3^(1/2))^(3*(3^(1/2)))

and so on.

Could you elaborate on this equation. And Notice that when n goes towards number 1, x and y is equal to e=2.7182818284590452353602874713527.

Of course, because n=x/y=1 implies x=y for y=e^[lnx]=x=y
So now you are using the form n/(n-1)=(n+1-1)/(n-1)=1+1/(n-1), which normalises to the mathematical definition of the exponential function
f(n)=(1+1/n)^n in the limit ofr asymptotic approach of n.
This also engaes the function f(x)=e^x being its own derivative f'(x)=f(x).
Below are some details Dejan. You do make a competent mathematician.
I always thought this equation could be used to factorize numbers through some fractal type algorithm

Dejan