Numbers that are generic w.r.t. exponentiation This is a follow-up to my old question Number of distinct values taken by $x\hat{\phantom{\hat{}}}x\hat{\phantom{\hat{}}}\dots\hat{\phantom{\hat{}}}x$ with parentheses inserted in all possible ways.
In the following let us assume $n$ to be a positive integer, and all other variables to be positive reals. Let $a\hat{\phantom{\hat{}}}b$ denote exponentiation $a^b$. 
The number of distinct $\mathbb{R}^+\to\mathbb{R}^+$ functions obtained from the expression 
$$\underbrace{x\hat{\phantom{\hat{}}}x\hat{\phantom{\hat{}}}\dots\hat{\phantom{\hat{}}}x}_{n\text{ occurences of }x}\tag1$$ 
by inserting parentheses in all possible ways depends on $n$ and is given by the OEIS sequence A000081. Note that different parenthesizations can result in the same function, e.g.
$$(x\hat{\phantom{\hat{}}}x)\hat{\phantom{\hat{}}}(x\hat{\phantom{\hat{}}}x)=(x\hat{\phantom{\hat{}}}(x\hat{\phantom{\hat{}}}x))\hat{\phantom{\hat{}}}x.\tag2$$
If instead of considering functions, we fix some value of $x$, and ask about the number of distinct numeric outcomes of the expression $(1)$ for all possible parenthesizations, then, depending on the value of $x$ we fixed, the result can be either A000081 (in this case we call the value of $x$ generic), or a slower growing sequence. 
For example, the number $2$ is not generic, because the corresponding sequence is A002845 due to some identities specific to the number $2$, e.g.
$$2\hat{\phantom{\hat{}}}(2\hat{\phantom{\hat{}}}2)=(2\hat{\phantom{\hat{}}}2)\hat{\phantom{\hat{}}}2.\tag3$$
Actually, it is not difficult to see that no positive integer is generic. Likewise, $\sqrt2$ is not generic. Furthermore, it can be proved that no positive algebraic number is generic.
Questions:

  
*
  
*Can we prove that $2^{\sqrt2}$ is generic?
  
*Can we find an explicit$^*$ computable generic number?
  

One might think that a plausible candidate for a generic number could be $\pi$, but, unfortunately, we do not even know yet if $\pi^{\pi^{\pi^\pi}}$ is an integer.

$^*$ By explicit I mean something that can be constructed from algebraic numbers and known $^{**}$ constants, elementary and known special functions, or an isolated root of an equation constructed from those constants and functions.
$^{**}$ known means they appeared in published books or reviewed papers.

References:


*

*R. K. Guy and J. L. Selfridge, The nesting and roosting habits of the laddered parenthesis. Amer. Math. Monthly 80 (1973), 868-876. ᵖᵈᶠ

*F. Göbel and R. P. Nederpelt, The number of numerical outcomes of iterated powers, Amer. Math. Monthly, 80 (1971), 1097-1103.  ᵖᵈᶠ
 A: Your first question can be taken with several senses, from weak to
strong. Let us understand your "explicit" terminology to mean
"computable".


*

*(weak version) Is there a computable generic value?

*(medium) Can we give a specific algorithm for computing a generic
value?

*(strong) Can we identify a computable generic value that we can
also understand in a simple way, apart from the property of it being a
generic value?
Probably you had something like the strong version in mind when
asking your question. But in order to make some small progress,
let me point out that we can get affirmative answers to the weak
and medium versions of the question. This amounts, in effect, to a
pure existence proof that there is an explicit generic value. Let
us say that $x$ is fully generic if it is generic with respect
to your expressions for all values of $n$ (not just expressions of the same length), so that $f(x)\neq g(x)$ for any two of your functions, provided $f\neq g$. 
Theorem. There is a computable fully generic value.
Proof. Observe that functions corresponding to any of your
expressions is continuous, and furthermore, they have computable
moduli of continuity (that is, for any of them, at any rational
input value, we can compute sufficient $\delta$ from
$\epsilon=\frac 1n$ for continuity).
Next, I believe that distinct functions arising from your
expressions never agree on an interval of positive real input
values $x$. (Please let me know if this is wrong; I will defer to
experts.)
It follows that there is a computable procedure to enumerate the
pairs of your expressions that give rise to distinct functions:
simply evaluate on more and more rational number inputs, until the
inequality is detected. (Perhaps one would hope to computably
decide equality of expressions as well, but I don't need this.)
Now, we construct a computable $x$ in stages. At any stage $k$, we
have made a promise to a certain rational approximation $r_k$ to
the value $x$ we are computing, with a certain promise of accuracy
$\delta_k$, so that $x$ will be within $\delta_k$ of $r_k$. In the
background, we have been running the computable algorithm to
enumerate the pairs of expressions corresponding to distinct
functions. We now take the $k^{th}$ such pair that we have found,
$f$ and $g$. Since they are distinct, they will disagree on some
rational value within $\delta_k$ of $r_k$, and we can
computably find such a value. Using the $\epsilon$ value revealing
the difference between $f(r_{k+1})$ and $g(r_{k+1})$, we can
compute a new accuracy $\delta_{k+1}$ that will ensure $f(x)\neq
g(x)$ for any $x$ within $\delta_{k+1}$ of $r_{k+1}$. This is the
new approximation to $x$, and we proceed.
Thus, we diagonalize against all the pairs of distinct
expressions, and thereby compute rational approximations to a
value $x$ that will resolve all the expressions as distinct,
provided that those expressions in fact correspond to distinct
functions. So $x$ is computable and fully generic. QED
I know this answer is not really what you want, which is a
specific number that you already knew about in some simple way,
like the candidate $2^{\sqrt{2}}$ that you mentioned. My
reply to this objection is to point out that there is a
widespread phenomenon in computability theory — some call it
the "If you build it, they will come" phenomenon — that
if one wants to prove that there is an explicit example of
something, then often you've got to just build the thing to order.
Meanwhile, I will hope along with you that someone comes through with a solution to the strong formulation of the question.
Let me also add that the theorem obviously generalizes to cover genericity with respect to much larger collections of computable functions. For any computable listing of computable functions on $\mathbb{R}$ (in the sense of computable analysis) with computable moduli of continuity, such that distinct functions are revealed as distinct on rationals in any interval, then there will be a computable real $x$ resolving them all as different.
Update. Following the ideas in the comments:
Theorem. Every non-computable real number is fully generic.
Proof. This is a consequence of the fact that the points of agreement between two of your expressions, if they are not everywhere in agreement, are isolated. And since these are computable functions with computable moduli of continuity, it follows that we can compute  these points of agreement. (See the question Intermediate value theorem on computable reals for further discussion of this and similar issues.) So any violation of full genericity occurs only at a computable real. QED
So any non-computable real is fully generic, and many such reals qualify under your definition of "explicit". In particular:


*

*The number $0'$, which is the binary sequence encoding the halting problem.

*Kleene's $\mathcal{O}$, which can be viewed as a binary sequence corresponding to representations of the computable ordinals.

*The number $0^{(\omega)}$, which is  the binary sequence encoding the true theorems of arithmetic.

*The number $0^{\triangledown}$, which is the binary sequence encoding the halting problem for infinite time Turing machines.

*The number $0^\sharp$.

*The number $0^\dagger$.
None of these is computable and so each of them is fully generic, as well as explicit in the sense you described, since indeed, many of them have their own Wikipedia pages. And there are many more that would seem to qualify, at least formally, as known constants under your  definition. The field of computability theory has thousands of published examples of explicit constructions of binary sequences that are not computable, but which are the limit of a computably enumerable sequence of rational approximations from below, and all of these will be fully generic.
Probably the right response to this, as you hinted in the comments, is that you only want to consider computable constants in your definition of explicit.
A: Two observations:


*

*Not every transcendental is generic. Indeed the real solution $r$ of the equation $x^x=2$ is not generic because $r$ satisfies 
$$x^{x^{x^x}}=\left( x^x\right)^x,$$
and $r$ is transcendental by the Gelfond-Schneider theorem.

*Here is an argument showing that no positive real algebraic is generic (by which I mean that every positive real algebraic satisfies an equation between two exponential towers that have the same number of $x$'s and  that do not define the same functions on $\mathbb{R}^+$.) This is mentioned in the original problem but maybe it would be nice to put up a proof here. It seems quite significant to me that 'generic' is a strengthening of 'transcendental'.
Rather than give a formal proof of the claim, I'll present an example that should make the general argument clear. Let $r$ be a solution of the polynomial equation $x^2-3x-2=0$. To prove that $r$ is not generic, first write the equation as
$$xx=x+x+x+1+1.$$
(In general, construct an equation made from additions and multiplications,  all of whose coefficients are 1.)
Equating $x$ raised to the left-hand-side and $x$ raised to the right-hand-side, we obtain the following non-identity (also satisfied by $r$)
$$(x^x)^x=(x^x)(x^x)(x^x)xx.$$
Doing the same thing with the last equation, we get
$$x^{(x^x)^x}=\left(\left(\left(\left(x^{(x^x)}\right)^{(x^x)}\right)^{(x^x)}\right)^x\right)^x.$$
Now call the left and right hand sides of the last equation $a$ and $b$. Then $r$ satisfies the equation $a^b=b^a$, and both $a^b$ and $b^a$ have the same number of $x$'s. Furthermore $a^b$ and $b^a$ cannot define the same function. This follows from the calculus problem to the effect that if $u$ and $v$ are greater than $e$ (the base of the natural logarithm) and if $u^v=v^u$, then $u=v$.
