As a next application of arithmetic functions we provide a general formula for the number of positive divisors \(\tau(n)\).
Let \(n > 1 \) be a natural number with the factorization. \[n=\prod_{i=1}^\infty p_i^{e_i}.\]
Then, the number of positive divisors \(\tau(a)\) can be calculated as
\[\tau(a)=\prod_{i=1}^\infty (e_i+1).\]
\(n=877800\) has \(192\) positive divisors, since
$$n=877800=2^3\cdot 3^1\cdot 5^2\cdot 7^1\cdot 11^1\cdot 13^0\cdot 17^0\cdot 19^1\cdot\prod_{i=9}^\infty p_i^0$$ and \[\tau(877800)=4\cdot 2\cdot 3\cdot 2\cdot 2\cdot 1\cdot 1\cdot 2\cdot\prod_{i=9}^\infty 1=192.\]
Proofs: 1
Proofs: 1
