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Is there any nice characterization of the class of polynomials can be written with the following formula for some $c_i , d_i in mathbbN$? Alternatively, where can I read more about these? do they have a name?
$$c_1 + left( c_2 + left( dots (c_k + x^d_k) dots right)^d_2 right)^d_1$$

For instance, it is not possible to write $1 + x + x^2$ in this way, but it is possible to write $1 + 2x + x^2$ or $0 + x^3$.

For some context: two actions on the set of polynomials $A times mathbbN[x] to mathbbN[x]$, and $B times mathbbN[x] to mathbbN[x]$ can be combined into a single one $left<A,Bright> times mathbbN[x] to mathbbN[x]$ that takes a word of elements on $A$ and $B$ and applies the multiple actions in order. In the case of multiplication and exponential, we can see that the class of polynomials
$$c_1 left( c_2 left( dots (c_k x^d_k) dots right)^d_2 right)^d_1$$
can be just described as the polynomials of the form $cx^d$. I do not expect such a simple characterization in the case of sums and exponentials, but I would like to know if this class of polynomials has been described or studied somewhere.

Some partial elementary observations that might lead to necessary conditions, probably not to a characterization.

If the nonconstant polynomial $p(x)$ has this form then the leading coefficient must be $1$ and you can make the constant term anything you like.

The quadratics are precisely the ones where the coefficient of $x$ is even.

The cubics are the ones of the form
$$
c + 3t^2x + 3tx^2 + x^3 .
$$
That is, those where the coefficient of $x^2$ is a multiple $t$ of $3$ and the coefficient of $x$ is $t$ times the coefficient of $x^2$.

For degree $n$, must the coefficient of $x^n-1$ be a multiple of $n$? How will it restrict some lower order coefficients?

You can figure out if a given polynomial is of that form by "going backwards". Let:
$$(a_n x^n + a_n-1 x^n-1 + ...a_1 x^1 + a_0) = c_1 + left( c_2 + left( dots (c_k + x^d_k) dots right)^d_2 right)^d_1$$
A valid polynomial will always be of the form:
$$ c_1 + left( c_2 + left( dots (c_k + x^d_k) dots right)^d_2 right)^d_1 = x^n + (fracn c_kd_k) x^n-d_k + ... = P(x)$$
where there are no non-zero intermediate powers between $x^n$ and $x^n-d_k$, and $c_k neq 0$ except in the cases where the polynomial is $x^d_1$. We solve it by substituting $x = (y-c_k)^frac1d_k$ in our polynomial from the computed values, checking if $P((y-c_k)^frac1d_k)$ is still an integer polynomial and $c_k$ is an integer, then solving recursively for :
$$c_1 + left( c_2 + left( dots (c_k-1 + x^d_k-1) dots right)^d_2 right)^d_1 = P((y-c_k)^frac1d_k)$$ until all pairs $(c_k,d_k)$ are computed.

Eg: $x^2 + x + 1$ gives $d_k = 1$ and $frac2c_k1 = 1$ which is non-integer $c_k$ so there are no solutions.

Eg: $P(x) = x^3 + 6x^1 $ gives $d_k = 3-1 = 2$ and $frac3c_k2 = 6$ so $c_k = 4$. We then calculate $P((y-4)^frac12) = (y-4)^frac32 + 3 (y-4)^frac12 $. We see that the exponents are not natural so this cannot be further expanded. There are no solutions.

Eg: $P(x) = x^20 + 12 x^15 + 54 x^10 + 108 x^5 + 83$. We see $d_k = 20-15 = 5 $. Also $frac20c_k5 = 12$ so $c_k = 3$. We compute $ P((y-3)^frac15) = y^4 + 2 = P_2(y)$. Applying the same procedure to $P_2$ we get $d_k-1 = 4-0 = 4 $. and $frac4c_k-14=2$ so $c_k-1 = 2$. Substituting again we see that $P_2((z-2)^frac14) = z$ . So our final result is: $((x^5+3)^4 + 2)$ .