Divisor function ($\tau (n)$)

The divisor function $\tau (n)={\sum }_{d\mid n}1$ counts the positive integer divisors of a positive integer $n$. In particular $\tau (1)=1$, because $1$ has just one positive integer divisor. More generally the notation ${\sigma }_k(n)={\sum }_{d\mid n}{d}^k$ gives $\tau ={\sigma }_0$ and the sum-of-divisors function ${\sigma }_1$. Thus the divisor function counts the positive integer divisors, while ${\sigma }_1$ adds their sizes. ^[ref]

If the prime factorization is $n={\prod }_{k=1}^r{p}_k^{a_k}$ with distinct prime numbers $p_k$ and positive exponents $a_k$, thenThe short proof is a count: each positive integer divisor has a unique form ${\prod }_{k=1}^r{p}_k^{b_k}$, where $0\le b_k\le a_k$. There are $a_k+1$ choices at each prime number, and choosing them gives every positive integer divisor exactly once. For $n=1$ there are no prime factors and the product is $1$.

For $12=2^2\cdot 3$, choose $i\in \{0,1,2\}$ and $j\in \{0,1\}$ to obtain $2^i{3}^j$. The six values are $1,2,4,3,6,12$; sorted, the positive integer divisors are $1,2,3,4,6,12$. Therefore $\tau (12)=(2+1)(1+1)=6$. The diagram makes the independent exponent choices visible.

This count is useful because a known prime factorization gives $\tau (n)$ without testing every smaller positive integer for divisibility. It also proves the divisor function is a multiplicative function: coprime integers have disjoint prime factors, so their exponent choices combine. For example $\tau (6)=4=\tau (2)\tau (3)$, but $\tau (4)=3\ne 4=\tau (2{)}^2$, so it is not a completely multiplicative function. Equivalently, $\tau =1\ast 1$ under Dirichlet convolution, counting ordered pairs of positive integers whose product is $n$.

For each $\eta >0$, the divisor function satisfies the displayed bound. Write $n=\prod p^{a_p}$, so $\tau (n)=\prod (a_p+1)$. For all sufficiently large primes, $a+1\le 2^a\le p^{\eta a}$ for every $a\ge 0$. For each of the finitely many smaller primes, ${\sup }_{a\ge 0}(a+1)p^{-\eta a}$ is finite. Multiplying these finitely many constants proves the bound. Products of a fixed number of divisor functions obey the same conclusion after reducing $\eta$.

The divisor function satisfies ${\sum }_{n\le X}d(n{)}^2\ll X{\log }^3(2X)$. Partial summation consequently gives ${\sum }_{n\le X}d(n{)}^2/n\ll {\log }^4(2X)$.