The sequence satisfying $a_2^{a_3^{\dots^{a_n}}} = n $

Question

$a_2 = 2$

$a_2^{a_3 } = 3$

So $a_3= \ln(3)/\ln(2)$.

I wonder about all solutions $a_n$ such that

$a_2^{a_3^{\dots^{a_n}}} = n$

For all $n$.

How does $a_n$ behave? What are the best asymptotics?

Of course $a_n$ goes quickly towards values between $\exp(1/e)$ and $1$ that is trivial.

But I am not even sure If $a_n$ is strictly decreasing or if its limit exists.

Also If it is strictly decreasing with limit $A$ , I do not know the value of $A$.

I assume $A=1$.

---

I assume $A=1$ Because $a_2 = 2$. We could consider other starting values and define ( assuming a limit ) $A(a_2)$ as a function. If that function is even continuous or $C^1$ is another matter.

Answer 1

We can't have $a_n\to a<\exp(1/e)$. If we did, then there must exist some $N$ such that $a_n<\exp(1/e)$ for all $n>N$, and then we'd have

$$a_2^{a_3^{\dots^{a_n^{a_{n+1}^{\dots}}}}}<a_2^{a_3^{\dots^{a_n^{\exp(1/e)^{\dots}}}}}=a_2^{a_3^{\dots^{a_n^e}}}$$

which is bounded, while $n$ is unbounded.

We also can't have $a_n\to a>\exp(1/e)$, since then the power tower would have a lower bound that grows asymptotically faster than $n$ ($x^{x^{\dots}}$ diverges for $x>\exp(1/e)$).

Hence, we conclude that we must have $a_n\to\exp(1/e)$, if it approaches any limit at all. We also deduce that we must have $a_n\ge\exp(1/e)$ for infinitely many $n$, since $x^{x^{\dots}}$ converges for $x=\exp(1/e)$. So if the sequence is monotone, it must be decreasing.