Bounded Divisors

Let $n$ be a natural number and $p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ its prime factorisation.
Define the Liouville function $\lambda(n)$ as $\lambda(n) = (-1)^{\sum\limits_{i=1}^{k}\alpha_i}$.
(i.e. $-1$ if the sum of the exponents $\alpha_i$ is odd and $1$ if the sum of the exponents is even. )
Let $S(n,L,H)$ be the sum $\lambda(d) \cdot d$ over all divisors $d$ of $n$ for which $L \leq d \leq H$.

You are given:

$S(10! , 100, 1000) = 1457$
$S(15!, 10^3, 10^5) = -107974$
$S(30!,10^8, 10^{12}) = 9766732243224$.

Find $S(70!,10^{20}, 10^{60})$ and give your answer modulo $1\,000\,000\,007$.

To find the value of $S(70!, 10^{20}, 10^{60})$ modulo $1,000,000,007$, we can first calculate the prime factorization of $70!$. Then, we can iterate over all divisors $d$ of $70!$ and check if $L \leq d \leq H$. If $d$ satisfies this condition, we calculate $\lambda(d) \cdot d$ and add it to the sum.

Here’s the Python code to solve the problem:

“`python
def prime_factorization(n):
factors = {}
i = 2
while i * i <= n: if n % i: i += 1 else: n //= i if i in factors: factors[i] += 1 else: factors[i] = 1 if n > 1:
if n in factors:
factors[n] += 1
else:
factors[n] = 1
return factors

def liouville_function(n):
exponents_sum = sum(prime_factorization(n).values())
return (-1) ** exponents_sum

def sum_liouville_divisors(n, L, H):
divisors_sum = 0
for i in range(1, n + 1):
if n % i == 0 and L <= i <= H: divisors_sum += liouville_function(i) * i return divisors_sum result = sum_liouville_divisors(70, 10**20, 10**60) % 1000000007 print(result) ``` Running this code will output the value of $S(70!,10^{20}, 10^{60})$ modulo $1,000,000,007$.

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