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 $n=2^k{p}_1^{a_1}\dots p_r^{a_r}{q}_1^{b_1}\dots q_s^{b_s}$ , where the $p_i$ are distinct primes with $p_i\equiv 1(\mathrm{mod}4)$ and the $q_i$ are distinct primes with $q_j\equiv 3(\mathrm{mod}4)$. 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 $$\lceil \frac{1}{2}(a_1+1)(a_2+1)\dots (a_r+1)\rceil$$ 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:$${\sigma }_x(n)=\prod _{i=1}^r\sum _{j=0}^{a_i}{p}_i^{jx}=\prod _{i=1}^r(1+p_i^x+p_i^{2x}+\cdots +p_i^{a_{i}x}).$$ It follows (by setting $x=0$) that $d(n)$ is: $${\sigma }_0(n)=\prod _{i=1}^r(a_i+1).$$