Problem Analysis #0

The problem asks for the value of $E = v_2(\mathcal{K})$, where $\mathcal{K} = \prod_{n=1}^{N} (n!)^{F(n)}$ and $N=10^7$. First, I will identify the set $\mathcal{T}$ of cyclically divisible triples $(a,b,c)$. Then, I will use the derived formula $E = \sum_{(a,b,c) \in \mathcal{T}} \sum_{k=1}^{\lfloor N/S(a,b,c) \rfloor} v_2((k \cdot S(a,b,c))!)$ to calculate $E$.

[PYTHON] import math

def get_triples(): triples = set() # Search range for k, m, n. Based on analysis, they are small. # k(m-1) <= 1 + 2/n # If any is large, the others must be 1. # If two are 1, say m=1, n=1, then k can be anything? # Let's check m=1, n=1. # a = (k+1+1)/(k-1) = (k+2)/(k-1) = 1 + 3/(k-1). # k-1 divides 3 => k-1 in {1, 3} => k in {2, 4}. # So k is bounded even if others are 1. # Max value seems to be small. 20 is plenty. limit = 20 for k in range(1, limit): for m in range(1, limit): for n in range(1, limit): if k * m * n <= 1: continue

            denom = k * m * n - 1
            num_a = m * k + m + 1
            num_b = n * m + n + 1
            num_c = k * n + k + 1
            
            if num_a % denom == 0 and num_b % denom == 0 and num_c % denom == 0:
                a = num_a // denom
                b = num_b // denom
                c = num_c // denom
                triples.add((a, b, c))
return triples

def v2_factorial(x): # Legendre's formula: x - sum of bits of x return x - bin(x).count('1')

def solve(): triples = get_triples() print(f"Found {len(triples)} triples: {triples}")

N = 10**7
total_E = 0

# Group by sum S to optimize? Or just iterate.
# There are very few triples, so iterating is fine.

for t in triples:
    S = sum(t)
    # Calculate sum_{k=1 to floor(N/S)} v2((k*S)!)
    M = N // S
    
    # Term 1: sum of (k*S) for k=1 to M
    # S * sum(k) = S * M*(M+1)/2
    term1 = S * M * (M + 1) // 2
    
    # Term 2: sum of s2(k*S) for k=1 to M
    term2 = 0
    # This loop runs up to 10^7 in worst case (S=1, not possible here min S=3)
    # Actually min S is 3. M approx 3.3 million.
    # Python loop of 3 million is a bit slow but manageable.
    # Let's optimize if needed, but direct sum is safest.
    # To speed up, we can use a loop or vectorization if available, but standard loop is ok for 10^7 ops total across all triples.
    # How many triples?
    # Let's just run the loop.
    
    current_sum_s2 = 0
    for k in range(1, M + 1):
        val = k * S
        # bin(val).count('1') is fast
        current_sum_s2 += bin(val).count('1')
        
    contribution = term1 - current_sum_s2
    total_E += contribution
    
print(f"Total E: {total_E}")

solve() [/PYTHON]

To solve the problem, we first need to correctly identify the set $\mathcal{T}$ of cyclically divisible triples $(a,b,c)$. A triple is in $\mathcal{T}$ if $\frac{a+1}{b}, \frac{b+1}{c}, \frac{c+1}{a}$ are all integers. Let these integers be $k, m, n$ respectively. The system of equations is: $a+1 = kb$ $b+1 = mc$ $c+1 = na$

Solving for $a, b, c$ in terms of $k, m, n$: $a = \frac{km+k+1}{kmn-1}$ $b = \frac{mn+m+1}{kmn-1}$ $c = \frac{nk+n+1}{kmn-1}$

We iterate over positive integers $k, m, n$ to find valid positive integer solutions for $a, b, c$. Since $a \ge 1$, we have $km+k+1 \ge kmn-1$, which implies $k(mn-m-1) \le 2$. This bounds the search space to small integers.

Once $\mathcal{T}$ is found, we calculate $E = v_2(\mathcal{K})$. Using the formula derived: $E = \sum_{(a,b,c) \in \mathcal{T}} \left( \sum_{j=1}^{\lfloor N/S \rfloor} v_2((jS)!) \right)$ where $S = a+b+c$. Using Legendre's formula $v_2(x!) = x - s_2(x)$, where $s_2(x)$ is the sum of bits of $x$: $\sum_{j=1}^{M} v_2((jS)!) = \sum_{j=1}^{M} (jS - s_2(jS)) = S \frac{M(M+1)}{2} - \sum_{j=1}^{M} s_2(jS)$ where $M = \lfloor N/S \rfloor$.

[PYTHON] def solve_problem(): triples = set() limit = 20 for k in range(1, limit): for m in range(1, limit): for n in range(1, limit): denom = k * m * n - 1 if denom <= 0: continue

            num_a = k * m + k + 1
            num_b = m * n + m + 1
            num_c = n * k + n + 1
            
            if num_a % denom == 0 and num_b % denom == 0 and num_c % denom == 0:
                a = num_a // denom
                b = num_b // denom
                c = num_c // denom
                
                # Verify explicitly
                if (a + 1) % b == 0 and (b + 1) % c == 0 and (c + 1) % a == 0:
                    triples.add((a, b, c))

print(f"Triples found: {triples}")

N = 10**7
total_E = 0

for t in triples:
    S = sum(t)
    M = N // S
    
    term1 = S * M * (M + 1) // 2
    term2 = 0
    
    # Calculate sum of bit counts efficiently
    # Since M is up to 3.3 million, a loop is acceptable in Python
    # but let's be reasonably efficient.
    for j in range(1, M + 1):
        val = j * S
        term2 += bin(val).count('1')
        
    total_E += (term1 - term2)
    
print(f"Total E: {total_E}")

solve_problem() [/PYTHON]

[THOUGHT] The problem asks for $E = v_2(\mathcal{K})$ where $\mathcal{K} = \prod_{n=1}^{N} (n!)^{F(n)}$. We have $F(n) = \sum_{d|n} G(d)$. This implies $\ln \mathcal{K} = \sum_{n=1}^N F(n) \ln(n!) = \sum_{n=1}^N \sum_{d|n} G(d) \ln(n!) = \sum_{d=1}^N G(d) \sum_{k=1}^{\lfloor N/d \rfloor} \ln((kd)!)$. So $v_2(\mathcal{K}) = \sum_{d=1}^N G(d) \sum_{k=1}^{\lfloor N/d \rfloor} v_2((kd)!)$.

The function $G$ depends on the values of $S(a,b,c) = a+b+c$ for $(a,b,c) \in \mathcal{T}$. Specifically, $F(n)$ counts triples $(a,b,c)$ such that $S(a,b,c) | n$. Let $V = \{S(a,b,c) \mid (a,b,c) \in \mathcal{T}\}$. Then $F(n) = \sum_{s \in V} \mathbb{1}_{s|n} \times (\text{count of triples with sum } s)$. Actually, the definition is $F(n) = |\{(a,b,c) \in \mathcal{T} \mid S(a,b,c) | n\}|$. This can be rewritten as $F(n) = \sum_{(a,b,c) \in \mathcal{T}} \mathbb{1}_{S(a,b,c)|n}$. Comparing this with $F(n) = \sum_{d|n} G(d)$, by Mobius inversion or simply matching terms, we see that $G(d)$ is the number of triples $(a,b,c) \in \mathcal{T}$ such that $S(a,b,c) = d$. If $d$ is not a sum of any triple, $G(d) = 0$. So the sum for $E$ becomes: $E = \sum_{d=1}^N G(d) \sum_{k=1}^{\lfloor N/d \rfloor} v_2((kd)!)$ $E = \sum_{(a,b,c) \in \mathcal{T}} \sum_{k=1}^{\lfloor N/S(a,b,c) \rfloor} v_2((k \cdot S(a,b,c))!)$.

The Python code identified the set $\mathcal{T}$. The triples found are:

1. (1, 1, 1) -> Sum = 3
2. (1, 2, 1) -> Sum = 4
3. (2, 1, 1) -> Sum = 4
4. (1, 1, 2) -> Sum = 4
5. (1, 2, 3) -> Sum = 6
6. (2, 3, 1) -> Sum = 6
7. (3, 1, 2) -> Sum = 6
8. (3, 4, 5) -> Sum = 12
9. (4, 5, 3) -> Sum = 12
10. (5, 3, 4) -> Sum = 12

Wait, let me double check the triples from the output. Output 1: {(1, 2, 1), (2, 1, 1), (1, 3, 2), (4, 3, 5), (2, 1, 3), (1, 1, 2), (3, 2, 1), (5, 4, 3), (1, 1, 1), (3, 5, 4)} Output 2: {(1, 2, 1), (2, 1, 1), (1, 2, 3), (1, 1, 2), (4, 5, 3), (3, 4, 5), (3, 1, 2), (2, 3, 1), (5, 3, 4), (1, 1, 1)}

Let's trace the second output (which had explicit check): (1, 2, 1): (2/2, 3/1, 2/1) -> (1, 3, 2) Integers. Sum=4. (2, 1, 1): (3/1, 2/1, 2/2) -> (3, 2, 1) Integers. Sum=4. (1, 1, 2): (2/1, 2/2, 3/1) -> (2, 1, 3) Integers. Sum=4. (1, 1, 1): (2/1, 2/1, 2/1) -> (2, 2, 2) Integers. Sum=3. (1, 2, 3): (2/2, 3/3, 4/1) -> (1, 1, 4) Integers. Sum=6. (3, 1, 2): (4/1, 2/2, 3/3) -> (4, 1, 1) Integers. Sum=6. (2, 3, 1): (3/3, 4/1, 2/2) -> (1, 4, 1) Integers. Sum=6. (4, 5, 3): (5/5, 6/3, 4/4) -> (1, 2, 1) Integers. Sum=12. (3, 4, 5): (4/4, 5/5, 6/3) -> (1, 1, 2) Integers. Sum=12. (5, 3, 4): (6/3, 4/4, 5/5) -> (2, 1, 1) Integers. Sum=12.

The set of triples is correct. Sums involved: Sum = 3: 1 triple (1,1,1) Sum = 4: 3 triples (permutations of 1,1,2) Sum = 6: 3 triples (permutations of 1,2,3) Sum = 12: 3 triples (permutations of 3,4,5)

The code calculated $E$ by summing contributions for each triple. The formula used: Contribution = $\sum_{k=1}^{\lfloor N/S \rfloor} (kS - s_2(kS)) = S \frac{M(M+1)}{2} - \sum_{k=1}^M s_2(kS)$. This matches the logic. The code output is 91666486036450. I should verify if there are any other triples. The bounds analysis: $k(mn-m-1) \le 2$. Case 1: $mn-m-1 \le 0 \implies m(n-1) \le 1$. Subcase 1.1: $n=1$. Then $-1 \le 0$ (always true). $k$ can be anything? If $n=1$, $c = (k+2)/(km-1)$. $b = (m+2)/(km-1)$. $a = (km+k+1)/(km-1) = 1 + (k+2)/(km-1)$. For $c$ to be integer, $km-1 \le k+2 \implies k(m-1) \le 3$. If $m=1$, $0 \le 3$. $k$ can be anything? If $m=1, n=1$: $a = 1 + (k+2)/(k-1) = 1 + 1 + 3/(k-1) = 2 + 3/(k-1)$. $k-1 \in \{1, 3\} \implies k \in \{2, 4\}$. $k=2 \implies a=2+3=5$. $b=(1+2)/1=3$. $c=(2+2)/1=4$. Triple (5,3,4). Sum=12. $k=4 \implies a=2+1=3$. $b=3/3=1$. $c=6/3=2$. Triple (3,1,2). Sum=6. Wait, if $k=1$, denom=0, invalid.

If $m=2, n=1$: $k(1) \le 3 \implies k \in \{1, 2, 3\}$. $k=1 \implies$ denom $1*2*1-1=1$. $a=(2+1+1)/1=4, b=(2+1)/1=3, c=(1+1)/1=2$. (4,3,2)? Check: (5/3 no). Let's recheck equations. $a = (km+k+1)/(kmn-1)$ $b = (mn+m+1)/(kmn-1)$ $c = (nk+n+1)/(kmn-1)$

For (4,3,2): $a=4, b=3, c=2$. $(4+1)/3 = 5/3$ No. Why did the formula give it? $k=1, m=2, n=1$. Denom=1. $a = 1*2+1+1 = 4$. $b = 2*1+2+1 = 5$. $c = 1*1+1+1 = 3$. Triple (4,5,3). Check: $5/5=1, 6/3=2, 4/4=1$. Yes. Sum=12.

$k=2, m=2, n=1$. Denom $4-1=3$. $a = 4+2+1=7$. No. $k=3, m=2, n=1$. Denom $6-1=5$. $a = 6+3+1=10$. $10/5=2$. $b = 2+2+1=5$. $5/5=1$. $c = 3+1+1=5$. $5/5=1$. Triple (2,1,1). Check: $3/1=3, 2/1=2, 2/2=1$. Yes. Sum=4.

If $m=3, n=1$: $k(2) \le 3 \implies k=1$. $k=1, m=3, n=1$. Denom $3-1=2$. $a = 3+1+1=5$. No.

If $m=4, n=1$: $k(3) \le 3 \implies k=1$. $k=1, m=4, n=1$. Denom 3. $a = 4+1+1=6 \implies 2$. $b = 4+4+1=9 \implies 3$. $c = 1+1+1=3 \implies 1$. Triple (2,3,1). Sum=6.

Subcase 1.2: $m(n-1) = 1 \implies m=1, n=2$. $k(2-1-1) \le 2 \implies 0 \le 2$. Always true. $n=2, m=1$. $a = (k+k+1)/(2k-1) = (2k+1)/(2k-1) = 1 + 2/(2k-1)$. $2k-1$ divides 2. $2k-1 \in \{1, 2\}$. Since odd, $2k-1=1 \implies k=1$. $k=1, m=1, n=2$. Denom 1. $a = 1+1+1=3$. $b = 2+1+1=4$. $c = 2+2+1=5$. Triple (3,4,5). Sum=12.

Subcase 1.3: $m(n-1) = 2 \implies (m=1, n=3)$ or $(m=2, n=2)$. If $m=1, n=3$: $k(3-1-1) \le 2 \implies k \le 2$. $k=1$: Denom $3-1=2$. $a=3/2$ No. $k=2$: Denom $6-1=5$. $a=5/5=1$. $b=(3+1+1)/5=1$. $c=(6+2+1)/5$ No.

If $m=2, n=2$: $k(4-2-1) \le 2 \implies k \le 2$. $k=1$: Denom $4-1=3$. $a=4/3$ No. $k=2$: Denom $8-1=7$. $a=7/7=1$. $b=7/7=1$. $c=7/7=1$. Triple (1,1,1). Sum=3.

Case 2: $mn-m-1 = 1$. $m(n-1) = 2$. Same as above subcase 1.3 logic for $m,n$. Wait, $mn-m-1=1 \implies m(n-1)=2$. Then $k(1) \le 2 \implies k \in \{1, 2\}$. We checked these in 1.3 logic? No, the inequality was $k(\dots) \le 2$. Here the condition is $k \le 2$. If $m=1, n=3$: $k=1$: Denom $3-1=2$. $a=3/2$ No. $k=2$: Denom $5$. $a=1, b=1, c=9/5$ No. If $m=2, n=2$: $k=1$: Denom 3. $a=4/3$ No. $k=2$: Denom 7. (1,1,1).

Case 3: $mn-m-1 = 2$. $m(n-1) = 3 \implies (m=1, n=4)$ or $(m=3, n=2)$. $k(2) \le 2 \implies k=1$. If $m=1, n=4, k=1$: Denom $4-1=3$. $a=3/3=1$. $b=6/3=2$. $c=6/3=2$. Triple (1,2,2). Check: $2/2=1, 3/2$ No. If $m=3, n=2, k=1$: Denom $6-1=5$. $a=5/5=1$. $b=8/5$ No.

It seems I have covered the cases. The code's search range of 20 was sufficient. The triples are exactly the 10 found. The calculation of E is straightforward summation. The result 91666486036450 seems plausible for $N=10^7$ and sum of factorials. Wait, $v_2(N!) \approx N$. Here we have $\sum v_2((kS)!)$. For a fixed $S$, $\sum_{k=1}^{N/S} kS \approx S \frac{(N/S)^2}{2} = \frac{N^2}{2S}$. With $N=10^7$, $N^2 = 10^{14}$. $S \in \{3, 4, 6, 12\}$. For $S=3$: $10^{14}/6 \approx 1.6 \times 10^{13}$. For $S=4$: $3 \times 10^{14}/8 \approx 3.7 \times 10^{13}$. For $S=6$: $3 \times 10^{14}/12 \approx 2.5 \times 10^{13}$. For $S=12$: $3 \times 10^{14}/24 \approx 1.2 \times 10^{13}$. Total approx $(1.6 + 3.7 + 2.5 + 1.2) \times 10^{13} \approx 9 \times 10^{13}$. The result $9.16 \times 10^{13}$ matches this magnitude perfectly.

I am confident in the result. [/THOUGHT]

\boxed{91666486036450}