Let's consider the following variant of Collatz $(3n+1) : $
If $n$ is odd then $n \to n^2-1.$
$1\to 0.$
$3\to 8\to 1\to 0.$
$5\to 24\to 3\to 0.$
$7\to 48\to 3\to 0.$
$9\to 80\to 5\to 0.$
$11\to 120\to 15\to 224\to 7\to 0.$
$\color{red}{13\to 168\to 21\to 440\to 55\to 3024\to 189\to 35720\to 4465\to 19936224\to 623007\to\ldots\ ?}\ $
Does the sequence starting with $13$ go off to infinity? If yes, what is a proof? If no, is there a starting number whose sequence does go off to infinity, and how do we prove either that such a number must exist, or even better that a specific starting number goes off to infinity?
Here is some Python code I ran, which suggests that the numbers in the sequence starting at $13$ quickly become large:
n=13
num_loops=0
print(n)
while n!=0:
if n%2==0:
n//=2
else: n=n**2-1
print('\n', n)
num_loops+=1
if num_loops==70:
print("too many loops")
break
About Eric Syder's comment: suppose $ x^2 - 1 = 2^k y$ with $x,y$ odd. We wish to investigate what happens when $y \leq x \; . \; \;$ Note that $\gcd(x+1, x-1) = 2$ because $x \equiv 1,3 \pmod 4.$ One of the $x \pm 1 $ is $\equiv 2 \pmod 4. $
Let us make the name $ \delta = \pm 1.$ Then we may demand $$ x + \delta \equiv 2 \pmod 4$$ This tells us that the integer $ \frac{x + \delta}{2} $ is odd. We also have $$ \frac{x + \delta}{2} \frac{x - \delta}{2} = 2^{k-2}y$$ By repeated division by $2$ it follows that $$ \frac{x - \delta}{2^{k-1}} = w $$ is an odd positive integer, with $$\frac{x + \delta}{2} \; \frac{x - \delta}{2^{k-1}} = y $$ IF WE ASSUME $$ w = \frac{x - \delta}{2^{k-1}} \geq 3, $$ we find $$ y = w \frac{x + \delta}{2} \geq 3 \frac{x + \delta}{2} \geq \frac{3x - 3}{2}$$ The assumption $ w \neq 1$ has led us to $y \geq \frac{3x - 3}{2}$ The hypothesis that $x \geq y$ now says $x \geq \frac{3x - 3}{2},$ or $ 2x \geq 3x - 3,$ or $3 \geq x$
If $ x \geq y$ and $ x \geq 5$ in $ x^2 - 1 = 2^k y,$ we find that $w=1$ in $ \frac{x - \delta}{2^{k-1}} = w .$ So that $ x - \delta = 2^{k-1} .$ or $$ x = \pm 1 + 2^{k=1}$$