Let $A(x)=\lfloor\frac{x^2-20x+16}{4}\rfloor$, $B(x)=\sin\left(e^{\cos\sqrt{x^2+2x+2}}\right)$, $C(x)=x^3-6x^2+5x+15$, $H(x)=x^4+2x^3+3x^2+4x+5$, $M(x)=\frac{x}{2}-2\lfloor\frac{x}{2}\rfloor+\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}+\ldots$, $N(x)=\textrm{the number of integers that divide }\left\lfloor x\right\rfloor$, $O(x)=|x|\log |x|\log\log |x|$, $T(x)=\sum_{n=1}^{\infty}\frac{n^x}{\left(n!\right)^3}$, and $Z(x)=\frac{x^{21}}{2016+20x^{16}+16x^{20}}$ for any real number $x$ such that the functions are defined. Determine $$C(C(A(M(A(T(H(B(O(N(A(N(Z(A(2016)))))))))))))).$$

[i]2016 CCA Math Bonanza Lightning #5.3[/i]
1. **Evaluate \( Z(A(2016)) \):**
   - First, we need to find \( A(2016) \).
   - \( A(x) = \left\lfloor \frac{x^2 - 20x + 16}{4} \right\rfloor \).
   - Substitute \( x = 2016 \):
     \[
     A(2016) = \left\lfloor \frac{2016^2 - 20 \cdot 2016 + 16}{4} \right\rfloor
     \]
     \[
     = \left\lfloor \frac{4064256 - 40320 + 16}{4} \right\rfloor
     \]
     \[
     = \left\lfloor \frac{4023936}{4} \right\rfloor
     \]
     \[
     = \left\lfloor 1005984 \right\rfloor
     \]
     \[
     = 1005984
     \]
   - Now, evaluate \( Z(1005984) \):
     \[
     Z(x) = \frac{x^{21}}{2016 + 20x^{16} + 16x^{20}}
     \]
     \[
     Z(1005984) = \frac{1005984^{21}}{2016 + 20 \cdot 1005984^{16} + 16 \cdot 1005984^{20}}
     \]
     - Given the large magnitude of \( x \), the dominant term in the denominator is \( 16x^{20} \):
     \[
     Z(1005984) \approx \frac{1005984^{21}}{16 \cdot 1005984^{20}} = \frac{1005984}{16} = 62874
     \]

2. **Evaluate \( N(62874) \):**
   - \( N(x) \) is the number of integers that divide \( \left\lfloor x \right\rfloor \).
   - \( \left\lfloor 62874 \right\rfloor = 62874 \).
   - The prime factorization of 62874 is:
     \[
     62874 = 2 \times 3 \times 10479
     \]
     - 10479 is further factorized:
     \[
     10479 = 3 \times 3493
     \]
     - 3493 is a prime number.
     - Therefore, the prime factorization of 62874 is:
     \[
     62874 = 2^1 \times 3^2 \times 3493^1
     \]
     - The number of divisors is given by:
     \[
     (1+1)(2+1)(1+1) = 2 \times 3 \times 2 = 12
     \]
     \[
     N(62874) = 12
     \]

3. **Evaluate \( O(12) \):**
   - \( O(x) = |x| \log |x| \log \log |x| \).
   - Substitute \( x = 12 \):
     \[
     O(12) = 12 \log 12 \log \log 12
     \]
     - Using approximate values:
     \[
     \log 12 \approx 2.4849, \quad \log \log 12 \approx 0.396
     \]
     \[
     O(12) \approx 12 \times 2.4849 \times 0.396 \approx 11.84
     \]

4. **Evaluate \( B(11.84) \):**
   - \( B(x) = \sin \left( e^{\cos \sqrt{x^2 + 2x + 2}} \right) \).
   - Substitute \( x = 11.84 \):
     \[
     B(11.84) = \sin \left( e^{\cos \sqrt{11.84^2 + 2 \cdot 11.84 + 2}} \right)
     \]
     \[
     = \sin \left( e^{\cos \sqrt{140.2496 + 23.68 + 2}} \right)
     \]
     \[
     = \sin \left( e^{\cos \sqrt{165.9296}} \right)
     \]
     \[
     = \sin \left( e^{\cos 12.88} \right)
     \]
     - Using approximate values:
     \[
     \cos 12.88 \approx -0.999
     \]
     \[
     e^{-0.999} \approx 0.368
     \]
     \[
     \sin(0.368) \approx 0.359
     \]
     \[
     B(11.84) \approx 0.359
     \]

5. **Evaluate \( H(0.359) \):**
   - \( H(x) = x^4 + 2x^3 + 3x^2 + 4x + 5 \).
   - Substitute \( x = 0.359 \):
     \[
     H(0.359) = (0.359)^4 + 2(0.359)^3 + 3(0.359)^2 + 4(0.359) + 5
     \]
     - Using approximate values:
     \[
     (0.359)^4 \approx 0.0166, \quad 2(0.359)^3 \approx 0.0925, \quad 3(0.359)^2 \approx 0.387, \quad 4(0.359) \approx 1.436
     \]
     \[
     H(0.359) \approx 0.0166 + 0.0925 + 0.387 + 1.436 + 5 \approx 6.932
     \]

6. **Evaluate \( T(6.932) \):**
   - \( T(x) = \sum_{n=1}^{\infty} \frac{n^x}{(n!)^3} \).
   - For large \( x \), the series converges to a very small value.
   - Approximate \( T(6.932) \approx 0 \).

7. **Evaluate \( A(0) \):**
   - \( A(x) = \left\lfloor \frac{x^2 - 20x + 16}{4} \right\rfloor \).
   - Substitute \( x = 0 \):
     \[
     A(0) = \left\lfloor \frac{0^2 - 20 \cdot 0 + 16}{4} \right\rfloor = \left\lfloor \frac{16}{4} \right\rfloor = \left\lfloor 4 \right\rfloor = 4
     \]

8. **Evaluate \( M(4) \):**
   - \( M(x) = \frac{x}{2} - 2 \left\lfloor \frac{x}{2} \right\rfloor + \frac{x}{2^2} + \frac{x}{2^3} + \frac{x}{2^4} + \ldots \).
   - For integer \( x \), \( M(x) = x - 2 \left\lfloor \frac{x}{2} \right\rfloor \).
   - Substitute \( x = 4 \):
     \[
     M(4) = 4 - 2 \left\lfloor \frac{4}{2} \right\rfloor = 4 - 2 \cdot 2 = 0
     \]

9. **Evaluate \( A(0) \):**
   - \( A(x) = \left\lfloor \frac{x^2 - 20x + 16}{4} \right\rfloor \).
   - Substitute \( x = 0 \):
     \[
     A(0) = \left\lfloor \frac{0^2 - 20 \cdot 0 + 16}{4} \right\rfloor = \left\lfloor \frac{16}{4} \right\rfloor = \left\lfloor 4 \right\rfloor = 4
     \]

10. **Evaluate \( C(4) \):**
    - \( C(x) = x^3 - 6x^2 + 5x + 15 \).
    - Substitute \( x = 4 \):
      \[
      C(4) = 4^3 - 6 \cdot 4^2 + 5 \cdot 4 + 15
      \]
      \[
      = 64 - 96 + 20 + 15 = 3
      \]

11. **Evaluate \( C(3) \):**
    - \( C(x) = x^3 - 6x^2 + 5x + 15 \).
    - Substitute \( x = 3 \):
      \[
      C(3) = 3^3 - 6 \cdot 3^2 + 5 \cdot 3 + 15
      \]
      \[
      = 27 - 54 + 15 + 15 = 3
      \]

The final answer is \(\boxed{3}\).