Crazy conjecture connecting the sqrt(e) and tetrations!

[tommy1729]

e^(1/e) = 1.444...
Let d = 1/e

Set infinity to be some arbitrarily high number, e.g. 9.99e10000000

I can further generalize this conjecture:

if d= 1/c, for any constant > 1

the infinite tetration of e^(1/e) + d, will reach "infinity" after 1/sqrt© iterations. I can post the data if anyone is interested:

For example, when d=1/10 we reach "infinity" after:
12, 34, 104, 325, 1024 iterations for d=(1/10,1/100,1/10^3,1/10^4)

This series grows at sqrt(10) for each iteration approximately.

Ryan

Hm, so what you are saying is that
\( \lim_{y\to\infty} \frac{\operatorname{slog}_{\eta+1/c^{n+1}}(y)}{\operatorname{slog}_{\eta+1/c^n}(y)}\to \sqrt{c} \)
Or at least
\( \lim_{n\to\infty}\lim_{y\to\infty} \frac{\operatorname{slog}_{\eta+1/c^{n+1}}(y)}{\operatorname{slog}_{\eta+1/c^n}(y)}\to \sqrt{c} \)
where \( \eta=e^{1/e} \) and \( \operatorname{slog}_b \) is the inverse function of \( f(z)=\exp_b^{\circ z}(1) \)

Sounds really interesting, however I have no idea how to tackle.

i noticed that too , very long ago.

perhaps the count till 'oo' is the confusing part.

what if we replace d with -d and count until we reach 'e' (instead of 'oo')

then would the limit also give sqrt© ?

if so , i think we are close to a proof.

or at least arrive at showing these limits depend on earlier conjectured limits ( such as the limit by gottfried )

regards

tommy1729