(09/04/2017, 03:04 AM)sheldonison Wrote:(09/03/2017, 11:52 PM)JmsNxn Wrote: There is a result posted on here about how the "eta constants" converge to 2 and the "euler constants" converge to 4. The n'th eta constant is the sup of the x'th n'th hyperoperator root of x. And the "euler constants" are the actual values x_n such that x_n'th n'th hyperoperator root of x_n = n'th eta constant...
How interesting! I haven't seen this conjecture before. This is base 1.84, for which septation almost has an upper fixed point, that would be somewhere near 3.7 So octation for base=1.84 wouldn't quite be bounded, but oct(45)~=4.05, so it would take awhile before octation escapes; I can only calculate integer values for octation, until I do a theta mapping. Why would do the "euler constants" converge to 4, as n gets arbitrarily large for the "nth" hyperoperator for bases approaching 2?
I'll be a bit clearer, I was in a rush before. I wish I could find the post (i think it was by jaydfox). I can't remember the finesse involved. This is all based off foggy memories, but I think it went something like this.
\( h_n(x) \uparrow^n x = x \)
then \( h'_n(e_n) = 0 \)
and \( h_n(e_n) = \eta_n \)
then
\( e_n \to 4 \)
\( \eta_n \to 2 \)
But these constants are not unique (depend on which tetration, pentation, hexation, we're using), it simply follows that they converge to 4 or 2. (Maybe it was just a conjecture though...).
Again, I can't remember how to prove the following but, \( b > \eta_n \) if and only if \( b \uparrow^n x \to \infty \) as \( x \to \infty \). Which are the sequence of constants you're talking about. I'm pretty sure they're equivalent. They are for \( \uparrow^1 \) and \( \uparow^0 = \times \).
These two facts would claim if \( b<2 \) then \( b \uparrow^n x \) is bounded as \( n \) grows.
Again, I can't remember the full details, I may be forgetting something very important. Maybe it'll come to me in the morning, been a long day.
