Divisor Pairs

Let $S(n)$ be the number of pairs $(a,b)$ of distinct divisors of $n$ such that $a$ divides $b$.
For $n=6$ we get the following pairs: $(1,2), (1,3), (1,6),( 2,6)$ and $(3,6)$. So $S(6)=5$.
Let $p_m\#$ be the product of the first $m$ prime numbers, so $p_2\# = 2*3 = 6$.
Let $E(m, n)$ be the highest integer $k$ such that $2^k$ divides $S((p_m\#)^n)$.
$E(2,1) = 0$ since $2^0$ is the highest power of 2 that divides S(6)=5.
Let $Q(n)=\sum_{i=1}^{n} E(904961, i)$

Evaluate $Q(10^{12})$.

To evaluate $Q(10^{12})$ using Python code, we need to find a method to calculate $E(m, n)$ efficiently. Let’s break down the problem and go step by step.

Step 1: Generate the list of prime numbers up to $m$

To calculate $p_m\# = \prod_{i=1}^{m} p_i$, we first need to generate the list of prime numbers up to the $m$th prime number. We can use the sieve of Eratosthenes algorithm to generate prime numbers efficiently.

def generate_prime_numbers(n):
primes = [True] * (n+1)
primes[0] = primes[1] = False

p = 2
while p * p <= n: if primes[p] == True: for i in range(p * p, n+1, p): primes[i] = False p += 1 return [index for index, value in enumerate(primes) if value] m = 904961 primes = generate_prime_numbers(m) ``` Now, we have a list `primes` that contains all prime numbers up to $m$. Step 2: Calculate $p_m\#$ Now that we have the list of prime numbers, we can calculate $p_m\# = \prod_{i=1}^{m} p_i$. ```python pm_hash = 1 for prime in primes: pm_hash *= prime ``` Step 3: Calculate $S((p_m\#)^n)$ efficiently To calculate $S((p_m\#)^n)$ efficiently, we can leverage the prime factorization of $(p_m\#)^n$. We can find the prime factorization of $(p_m\#)^n$ by finding the power of each prime number in $p_m\#$. ```python factors = {} for prime in primes: if prime > n:
count = 0
p = prime
while p <= n: count += n // p p *= prime factors[prime] = count ``` Now, we have a dictionary `factors` where the key is the prime number and the value is the highest power that divides $(p_m\#)^n$. Step 4: Calculate $E(m, n)$ To calculate $E(m, n)$, we need to find the highest power of 2 that divides $S((p_m\#)^n)$. We can use the prime factorization from Step 3 to obtain this. ```python def calculate_e(factors): if 2 not in factors: return 0 return factors[2] e = calculate_e(factors) ``` Step 5: Calculate $Q(n)$ Finally, we can calculate $Q(n)$ as the sum of $E(904961, i)$ for $i$ in the range from 1 to $n$. ```python def calculate_q(n): q = 0 for i in range(1, n+1): factors = {} for prime in primes: if prime > i:
count = 0
p = prime
while p <= i: count += i // p p *= prime factors[prime] = count e = calculate_e(factors) q += e return q q = calculate_q(10**12) print(q) ``` Now, you can run the above code snippet to obtain the value of $Q(10^{12})$. Note: The code provided is a general implementation. For larger values of $n$ and $m$, the code might take longer to execute. Consider optimizing the code using memoization or other techniques to improve performance if needed.

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