tetration limit ??

[sheldonison]

Let n be a positive integer going to +oo.

lim [e^{1/e} + 1/n]^^[(10 n)^{1/2} + n^{A(n)} + C + o(1)] - n = 0.

Where C is a constant.

Conjecture : lim A(n) = 1/e.

Its a curious equation. I viewed it from a different angle: What is the slog_{1/e+1/n}(n)? But I couldn't figure out why you were interested in slog(n) as opposed to say, slog(e^e) or something like that that made more sense to me. e^e is the cusp of where this tetration function takes off, and the function starts growing superexponentially. But the (1/n) means it might take 1 or 2 more iterations to reach (1/n), Or if n is hyperexponentially large = sexp(4.5), then 3 extra iterations. But most of the time is spent getting to e^e. And that equation is dominated by approximately real(Pseudo period)-2. And you included an O(1) term in your equation anyway, which implies C isn't an exact constant.

So then my counter conjecture would be that lim sexp_{1/e+1/n)(real(Period)-2)=constant, and that constant seems to be about 388 as n goes to infinity. But that seemed to be a very different equation than the one you had in mind, so I thought it would be off topic, so I didn't mention it. But yeah, I have equations for the pseudo period, which I posted below.

Then there is your approximation itself. slog_{1/e+1/n}(n) = (10n)^{1/2} + n^{A(n)} + C. Can you explain why you think this is the right approach or equation? It doesn't seem to match the approximation I have for real(pseudo_period)-2...

The equations for the fixed point and Period are approximately as follows. One can see that the resulting period has a sqrt term, but not sqrt(10n).
\( \eta=\exp(1/e)\;\;k=\ln(\ln(\eta+\frac{1}{n}))+1\approx \frac{e}{\eta \cdot n} \approx \frac{1.8816}{n}\;\;\; \)

Now we have switched it to a problem of iterating \( z \mapsto \exp(z)-1+k\; \). In the limit, the fixed point goes to zero. This iteration mapping has a simpler Taylor series for the fixed point L, from which we can generate the Pseudo Period.

\( x=\sqrt{-2k}\;\;\; L = x - \frac{x^2}{6}+\frac{x^3}{36}+...\;\;\approx \sqrt{ \frac{-2e}{\eta \cdot n}}\;\;\;\text{period}=\frac{2\pi i}{L}\;\;\ \) this is the period at the fixed point.

\( \Re(\text{period}) = \Re(\frac{2\pi i}{L}) \approx 2\pi \sqrt{\frac{\eta \cdot n}{2e}} \approx \sqrt{10.49n}\;\;\; \) this is close to sqrt(10n). But I don't understand your n^A(n)~=n^(1/e) term; [(10 n)^{1/2} + n^{A(n)}].

Anyway, my counter-conjecture is that
\( \lim_{n \to \infty} \text{sexp}_{(\eta+1/n)}\left[\Re(\text{period})-2\right] =k \;\;\; k\approx 388.787398293917704779 \), where the period is from the equation above.

The correct middle term is probably \( \text{slog}_e(\frac{n}{e}-1)\;\; \) Note that \( \text{slog}_e(n/e-1)=\text{slog}_e(\eta^{\eta^n})-2\;\; \) So then, we have the following conjectured equation, where I'm pretty sure a 1-cyclic theta is required as n gets arbitrarily large, \( \theta(\text{slog}_e(n/e-1)) \), whose predicted amplitude is probably about +/-0.002

\( \lim\limits_{n\to\infty} \text{sexp}_{(\eta+1/n)}\left[2\pi\sqrt{\frac{\eta\cdot n}{2e}} + \text{slog}_e(\frac{n}{e}-1) + C + \mathcal{O}(\theta) \right] -n = 0\;\;\;\;C \approx -2 - \text{slog}_e(388.7874/e-1) \)