Sigma, Sigh

Since $p$ prime $\sigma_{2}(p)=p^{2}+1$ Also using the property $\sigma(a_{1}\times a_{2} \cdots\times a_{n})=\sigma(a_{1})\times \sigma(a_{2} )\cdots \times\sigma(a_{n})$

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26

from functools import reduce
from operator import mul
numprimes = 1000
count = 0
n = 2
def product(iterable):
    return reduce(mul, iterable, 1)

def isprime(num):
    if num == 2:return True

    elif num < 2 or not num % 2: return False

    for i in range(3, int(num ** .5 + 1), 2):
        if not num % i: return False
    return True
L=[]
while count < int(numprimes):
    if isprime(n):
        L.append(n)
        count += 1
    n += 1

for i in range(len(L)):
    L[i] =((L[i]**2)+1)
print (len(str(product(L))))

Notice that $\sigma_2(\omega(n))=\prod_{i=1}^{n}(1+p_i^2)$ Hence, $\log_{10}(\sigma_2(\omega(n)))=\sum_{i=1}^{n} \log_{10}(1+p_i^2)$ Which can be readily computed from a list of first 1000 primes . The required answer is the ceiling of this sum.