Number of Divisors

I came across this formula for calculating the number of divisors of a given number: $$\sigma_0(n)=\prod_{i=1}^r(a^i+1)$$ where $r$ is the number of distinct prime factors and $a^i$ are their coefficients. For example, $$\sigma_0(24)=\prod_{i=1}^2(a^i+1)=(3+1)(1+1)=8$$ where $r$ = 2 because there are two distinct prime factors (2 and 3) with indices of 3 and 1 respectively.

There are indeed eight factors of $24: 1, 2, 4, 8, 3, 6, 12, 24$ so this is a quick and easy way to calculate the number of factors based on prime factors of a number and the degrees to which each of them are raised. It's similar to the formula for calculating the number of ways in which a number can be written as the sum of two squares viz.:

The fundamental theorem on sums of two squares is:

Let $n=2^kp_1^{a_1} \dots p_r^{a_r}q_1^{b_1} \dots q_s^{b_s}$ , where the $p_i$ are distinct primes with $p_i \equiv 1 \pmod{4}$ and the $q_i$ are distinct primes with $q_j \equiv 3 \pmod{4}$. Then $n$ is the sum of two squares if and only if all the $b_j$ are even. In that case, the number of distinct solutions is $$\lceil \frac{\scriptstyle{1}}{\scriptstyle{2}} (a_1+1)(a_2+1) \dots (a_r+1) \rceil$$ where $\lceil x \rceil$ is the ceiling function, the smallest integer greater than or equal to $x$.

Of course by this rule, $24$ cannot be written as a sum of two squares because it has a $4k+3$ factor ($3$), raised to an odd power ($1$). However, $25=5^2$ has no $4k+3$ factors and its $4k+1$ factor ($5$) is raised to the $2nd$ power. This means that it can be written as the sum of two squares in two distinct ways because $\lceil \frac{1}{2}(2+1) \rceil =2$. The two ways are $5^2+0^2$ and $4^2+3^2$.

Getting back to the number of divisors however, the formula quoted is actually part of a more general formula (as explained in Wikipedia):

The divisor function is multiplicative, but not completely multiplicative.  The consequence of this is that, if we write $$n = \prod_{i=1}^r p_i^{a_i}$$ where $r=\omega (n)$ is the number of distinct prime factors of $n$, $p_i$ is the $i$th prime factor, and $a_i$ is the maximum power of $p_i$ by which $n$ is divisible, then we have $$\sigma_x(n) = \prod_{i=1}^{r} \frac{p_{i}^{(a_{i}+1)x}-1}{p_{i}^x-1}$$ which is equivalent to the useful formula: $$\sigma_x(n) = \prod_{i=1}^r \sum_{j=0}^{a_i} p_i^{j x} = \prod_{i=1}^r (1 + p_i^x + p_i^{2x} + \cdots + p_i^{a_i x}).$$ It follows (by setting $x = 0$) that $d(n)$ is: $$\sigma_0(n)=\prod_{i=1}^r (a_i+1).$$