To mock a mockingbird

We evaluate the $M$ -functions as follows $\begin{aligned} (M \circ M \circ M \circ M)(f) &= M(M(M(M(f)))) \\ &= M(M(M(f \circ f))) \\ &= M(M((f \circ f) \circ (f \circ f))) \\ &= M(M( f \circ f \circ f \circ f )) \\ &= M(( f \circ f \circ f \circ f ) \circ ( f \circ f \circ f \circ f )) \\ &= M( f \circ f \circ f \circ f \circ f \circ f \circ f \circ f ) \\ &= ( f \circ f \circ f \circ f \circ f \circ f \circ f \circ f ) \circ ( f \circ f \circ f \circ f \circ f \circ f \circ f \circ f ) \\ &= f^{16} \end{aligned}$ where $f^n$ denotes a $n$ -fold function composition. In our case, we have $f : x \mapsto x^2$ , so that $\begin{aligned} f^{16}(x) &= \underbrace{f(f( \dots f(}_{\text{16 times}} x) \dots )) \\ &=( \dots (x\underbrace{^2)^2 \dots )^2}_\text{16 times} \\ &= x^{2^{16}} \\ &= x^{65536} \end{aligned}$ Therefore, the result $y = 2^{65536}$ is a number too large to be explicitly stated here. But we can determine the number $d$ of digits by calculating the decadic logarithm of $y$ and rounding it up to the nearest integer: $\begin{aligned} d &= \lceil \log_{10} (y) \rceil \\ &= \left\lceil \frac{\log (2)}{\log(10)} \cdot 65536 \right\rceil \\ &= 19729 \end{aligned}$ The following Python program explicitly evaluates the number $y$ :

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from math import *

def square(x):
    return x*x

def M(f):
    def compose(x):
        return f(f(x))
    return compose

result = M(M(M(M(square))))(2)
digits = ceil(log(result)/log(10))

print(result)
print(digits)

$\begin{aligned} M (x \mapsto x^2) &= (x^2)^2 = x^{2^2} \\ \\ M \circ M (x \mapsto x^2) = M (x \mapsto x^{2^2}) &= (x^{2^2})^{2^2} = x^{2^4} \\ \\ M \circ M \circ M (x \mapsto x^2) = M (x \mapsto x^{2^4}) &= (x^{2^4})^{2^4} = x^{2^8} \\ \\ M \circ M \circ M \circ M (x \mapsto x^2) = M (x \mapsto x^{2^8}) &= (x^{2^8})^{2^8} = x^{2^{16}} \\ \\ &\text{so} \\ \\ M \circ M \circ M \circ M (x \mapsto x^2) (2) &= 2^{2^{16}} \end{aligned}$

We can determine the number of digits in any number by rounding down its base $10$ logarithm and adding $1$ . So the number of digits in $2^{2^{16}}$ is

$\left \lfloor \log_{10} (2^{2^{16}}) \right \rfloor + 1 = \left \lfloor 2^{16} \log_{10} 2 \right \rfloor + 1 = \boxed{19729}$