[THOUGHT] Let me break down this problem step by step.

1. First, I need to find all cyclically divisible triples $(a,b,c)$ where $\frac{a+1}{b}$, $\frac{b+1}{c}$, and $\frac{c+1}{a}$ are all integers.

2. For each triple, compute $S(a,b,c) = a + b + c$.

3. The function $F(n)$ counts how many triples have $S(a,b,c) | n$.

4. Since $F(n) = \sum_{d|n} G(d)$ where $G(d)$ counts triples with $S = d$, we have $F(n) = \sum_{d|n} G(d)$ = number of triples whose sum divides $n$.

5. We need to compute $\mathcal{K} = \prod_{n=1}^{N} (n!)^{F(n)}$ and find $v_2(\mathcal{K})$.

6. $v_2(\mathcal{K}) = \sum_{n=1}^{N} F(n) \cdot v_2(n!)$

Let me first find all cyclically divisible triples. The conditions are:

• $b | (a+1)$
• $c | (b+1)$
• $a | (c+1)$

This means $a+1 \geq b$, $b+1 \geq c$, $c+1 \geq a$.

From these: $a \leq c+1$, $c \leq b+1$, $b \leq a+1$.

So $a \leq c+1 \leq b+2 \leq a+3$, meaning $a, b, c$ are close to each other.

Let me search for all such triples systematically. [/THOUGHT] [PYTHON]

Find all cyclically divisible triples

def find_cyclically_divisible_triples(max_val=1000): triples = [] for a in range(1, max_val + 1): for b in range(1, max_val + 1): if (a + 1) % b != 0: continue for c in range(1, max_val + 1): if (b + 1) % c != 0: continue if (c + 1) % a != 0: continue triples.append((a, b, c)) return triples

Since the values should be bounded, let's search

triples = find_cyclically_divisible_triples(100) print(f"Found {len(triples)} triples") for t in triples: print(t, "sum =", sum(t)) [/PYTHON]

[THOUGHT] Now I have all 10 cyclically divisible triples. Let me compute the answer.

The sums are:

• Sum = 3: 1 triple
• Sum = 4: 3 triples
• Sum = 6: 3 triples
• Sum = 12: 3 triples

So $G(3) = 1$, $G(4) = 3$, $G(6) = 3$, $G(12) = 3$, and $G(d) = 0$ for all other $d$.

$F(n) = \sum_{d|n} G(d)$ counts triples whose sum divides $n$.

We need $E = v_2(\mathcal{K}) = \sum_{n=1}^{N} F(n) \cdot v_2(n!)$

This equals $\sum_{n=1}^{N} v_2(n!) \cdot \sum_{d|n} G(d) = \sum_{d} G(d) \sum_{n: d|n, n \leq N} v_2(n!)$

For each $d \in \{3, 4, 6, 12\}$, we need $\sum_{k=1}^{\lfloor N/d \rfloor} v_2((kd)!)$

Using Legendre's formula: $v_2(n!) = \sum_{i=1}^{\infty} \lfloor n/2^i \rfloor$

Let me compute this properly with code. [/THOUGHT]

Wait, I need to run more code to compute the final answer. Let me do that.

[PYTHON] import math

N = 10**7

v_2(n!) using Legendre's formula

def v2_factorial(n): result = 0 p = 2 while p <= n: result += n // p p *= 2 return result

G values

G = {3: 1, 4: 3, 6: 3, 12: 3}

E = sum over n=1 to N of F(n) * v2(n!)

F(n) = sum_{d|n} G(d) where G(d) is nonzero only for d in {3,4,6,12}

So E = sum_{d in {3,4,6,12}} G(d) * sum_{k: kd <= N} v2((kd)!)

E = 0 for d, g in G.items(): max_k = N // d contribution = 0 for k in range(1, max_k + 1): contribution += v2_factorial(k * d) E += g * contribution print(f"d={d}, G(d)={g}, max_k={max_k}, contribution={contribution}, gcontribution={gcontribution}")

print(f"E = {E}") [/PYTHON]

[THOUGHT] The computation is complete. The exponent of 2 in the prime factorization of $\mathcal{K}$ is:

$E = 91666486036450$ [/THOUGHT]

\boxed{91666486036450}