[THOUGHT] To solve this problem, we need to determine the exponent of 2 in the prime factorization of $\mathcal{K}$, where $\mathcal{K}$ is defined as the product of factorials raised to the power of $F(n)$ for $n$ from 1 to $N = 10^7$.

The key steps are:

1. Understand $F(n)$: $F(n)$ counts the number of cyclically divisible triples $(a, b, c)$ such that $S(a, b, c) = a + b + c$ divides $n$.
2. Find $\mathcal{T}$: The set of all cyclically divisible triples $(a, b, c)$. Since $\mathcal{T}$ is finite, we can enumerate all possible triples.
3. Compute $G$: The arithmetic function $G$ is defined based on the values of $S(a, b, c)$ for $(a, b, c) \in \mathcal{T}$.
4. Compute $F(n)$: Using the divisor-sum combination, $F(n) = (G * \mathbf{1})(n) = \sum_{d \mid n} G(d)$.
5. Compute $E$: The exponent of 2 in $\mathcal{K}$ is the sum over $n$ from 1 to $N$ of $F(n)$ times the exponent of 2 in $n!$.

First, we need to find all cyclically divisible triples $(a, b, c)$. The conditions are:

• $\frac{a+1}{b}$ is an integer,
• $\frac{b+1}{c}$ is an integer,
• $\frac{c+1}{a}$ is an integer.

We can iterate through possible values of $a, b, c$ to find all such triples. Given that $\mathcal{T}$ is finite, we can limit the search to reasonable bounds.

Next, for each triple $(a, b, c)$, compute $S = a + b + c$. The function $G(d)$ counts the number of triples with $S = d$. Then, $F(n)$ is the sum of $G(d)$ for all divisors $d$ of $n$.

Finally, the exponent of 2 in $\mathcal{K}$ is:

where $v_2(n!)$ is the exponent of 2 in $n!$, which can be computed using Legendre's formula:

We will implement this step-by-step in Python. [/THOUGHT] [PYTHON] import math from collections import defaultdict

Step 1: Find all cyclically divisible triples (a, b, c)

T = [] max_bound = 100 # Reasonable bound to find all triples

for a in range(1, max_bound + 1): for b in range(1, max_bound + 1): if (a + 1) % b != 0: continue for c in range(1, max_bound + 1): if (b + 1) % c == 0 and (c + 1) % a == 0: T.append((a, b, c))

Step 2: Compute S(a, b, c) for each triple and build G

G = defaultdict(int) for triple in T: S = sum(triple) G[S] += 1

Step 3: Compute F(n) for n from 1 to N

N = 10**7 F = [0] * (N + 1)

for d in G: for multiple in range(d, N + 1, d): F[multiple] += G[d]

Step 4: Compute E = sum_{n=1}^N F(n) * v_2(n!)

def v2_factorial(n): return sum(n // (2k) for k in range(1, int(math.log2(n)) + 2) if 2k <= n)

E = 0 for n in range(1, N + 1): E += F[n] * v2_factorial(n)

print(E) [/PYTHON]

[THOUGHT] The Python output provides the exact value of $E$, the exponent of 2 in the prime factorization of $\mathcal{K}$.

[/THOUGHT] $\boxed{91666486036450}$