Let’s consider the reduced collatz problem.
Assuming there was an ever-growing sequence. Because of the unicity of trajectories, it cannot follow a periodic trajectory and thus it has to contain “down” steps.
How much, exactly? Could there be a sequence with a goggle up steps, one down step, a goggleplex of up steps, one down, etc?
I think I found a lower bound for the sparsity of down steps that forbid sequences that are that sparse. A better bound is surely possible and I’d gladly see yours!
So, if the sequence starts at X, the first down step has to happen before, at most, log_2(X). After this point, we go from k.2^N-1 to k.3^N-1, so our starting number has been multiplied by (k.3^N-1)/(k.2^N-1) which is approximately (3/2)^N as N gets big.
N<log_2(X), so (3/2)^N < (3/2)^(log_2(X)) = X^log_2(3/2) (approx X^0.58)
Then, we can apply the same reasoning as if the sequence started at Y below X*(3/2)^N/2, then Y < X*X^log_2(3/2) = X^(1+log_2(3/2)) (approx X^1.58).
That second step has to happen before log_2(Y) step, so at most log_2(X)+log_2(Y) = log_2(XY) < log_2(X^(2+log_2(3/2)) = log_2(X^(2+log_2(3/2)) = X^(2+log2(3/2)) (approx X^1.29)
For next one, Z<X*(3/2)^X^(1+log2(3/2))
…
I am pretty sure my exponent calculation is wrong, but you get the idea.