I have a question.
I have recently discovered that if you get the golden ratio $φ$, which is equal to $\frac{1 + \sqrt 5}{2}$, then if you raise this to the power of $2^{n - 1}(2^n - 1)$, the higher the value of $n$, the closer this comes to an integer. It does not converge to a particular integer, but just gets very close to some integer. Let this integer be $y$.
I know that the expression $2^{n - 1}(2^n - 1)$ attains perfect numbers, which are numbers that are the sum of all its divisors. And if $2^n - 1$ is prime, then $2^n - 1$ is a Mersenne prime.
Let's call this formula $x$. By testing $φ^x$ with the value $n \leq 30$ (on a $2000$ digit calculator), I have seen that it gets closer and closer to some integer $y$. I have also seen that if there is a Mersenne prime that divides into $x$ then $y$ is divisible by it as well.
However, I cannot prove that this is true. Just by looking at some equation and seeing similar results does not mean I have proven it works for all results (which I can't test because $n$ can tend to infinity), so how do I prove/disprove that $φ^x$ gets closer to an integer $y$ the higher the value of $n?$
I would appreciate it if you showed me step by step and not skip all the steps and possibly show a formula that proves/disproves this if there is one. Thanks.
Any connection with Mersenne numbers is a red herring. We have $\left(\frac {1+\sqrt 5}2\right)^n+ \left(\frac {1-\sqrt 5}2\right)^n$ is exactly an integer for even $n$ because all the terms with odd powers of $\sqrt 5$ cancel. Because $\left|\frac {1-\sqrt 5}2\right|\lt 1$ when we raise it to a large power it becomes very small, so $\left(\frac {1+\sqrt 5}2\right)^n$ becomes very close to an integer for large even $n$. It also doesn't depend on the base being $\varphi$. Any quadratic surd with conjugate smaller than $1$ in absolute value shows the same behavior.