Number Theory Home

## $$\tau$$ and $$\sigma$$

A number-theoretic or arithmetic function is a function whose domain is the set of positive integers. A number-theoretic function $$f$$ is called multiplicative if $f(mn)=f(m)f(n),$ for all relatively prime positive integers $$m$$ and $$n$$ (i.e., $$\gcd(m,n)=1$$ ). If $$n=p_1^{a_1}\cdot p_2^{a_2}\cdots p_k^{a_k}$$ is the prime-power factorization of $$n>1$$, then by the multiplicativity of $$f$$ we have $f(n)=f(p_1^{a_1}\cdot p_2^{a_2}\cdots p_k^{a_k})=f(p_1^{a_1}) f(p_2^{a_2})\cdots f(p_k^{a_k}).$

Definition. For any positive integer $$n$$, $$\tau(n)$$ is the number of all positive divisors of $$n$$ and $$\sigma(n)$$ is the sum of all positive divisors of $$n$$.

Example. The positive divisors of $$n=12$$ are $$1,2,3,4,6,12$$. So $$\tau(12)=6$$ and $$\sigma(12)=1+2+3+4+6+12=28.$$

Note that $$\tau$$ and $$\sigma$$ can also be defined as sums as follows: $\tau(n)=\sum_{d\vert n}1 \text{ and } \sigma(n)=\sum_{d\vert n} d,$ where $$d$$ runs over all the positive divisors of $$n$$. This would be our standard notation throughout for the sums over the positive divisors of $$n$$.

Theorem. Let $$n=p_1^{a_1}\cdot p_2^{a_2}\cdots p_k^{a_k}$$ be the prime-power factorization of $$n>1$$. Then

1. $$\tau(n)=(a_1+1)(a_2+1)\cdots(a_k+1)$$ and
2. $$\sigma(n)=\displaystyle\frac{p_1^{a_1+1}-1}{p_1-1}\frac{p_2^{a_2+1}-1}{p_2-1}\cdots \frac{p_k^{a_k+1}-1}{p_k-1}$$.

Proof. (a) The positive divisors of $$n=p_1^{a_1}\cdot p_2^{a_2}\cdots p_k^{a_k}$$ are $$p_1^{b_1}\cdot p_2^{b_2}\cdots p_k^{b_k}$$ where $$0\leq b_i\leq a_i$$ for $$i=1,2,\ldots,k$$. There are $$a_i+1$$ choices for $$b_i$$ for $$i=1,2,\ldots,k$$. Thus $\tau(n)=(a_1+1)(a_2+1)\cdots(a_k+1).$
(b) Note that each positive divisor of $$n=p_1^{a_1}\cdot p_2^{a_2}\cdots p_k^{a_k}$$ appears exactly once as a term in the expansion of the product $(1+p_1+p_1^2+\cdots+p_1^{a_1})(1+p_2+p_2^2+\cdots+p_2^{a_2})\cdots(1+p_k+p_k^2+\cdots+p_k^{a_k}),$ where each term in the expansion of the product is a positive divisor of $$n$$. Thus $\sigma(n)=(1+p_1+\cdots+p_1^{a_1})(1+p_2+\cdots+p_2^{a_2})\cdots(1+p_k+\cdots+p_k^{a_k})=\displaystyle\frac{p_1^{a_1+1}-1}{p_1-1}\frac{p_2^{a_2+1}-1}{p_2-1}\cdots \frac{p_k^{a_k+1}-1}{p_k-1}. ◼$

Example. For $$1176=2^3\cdot 3^1\cdot 7^2$$, $$\tau(1176)=(3+1)(1+1)(2+1)=24$$ and $$\sigma(1176)=\frac{2^4-1}{2-1}\frac{3^2-1}{3-1}\frac{7^3-1}{7-1}=3420$$.

Theorem. $$\tau$$ and $$\sigma$$ are multiplicative functions.

Proof. Let $$m$$ and $$n$$ be two relatively prime positive integers. We show $$\tau(mn)=\tau(m)\tau(n)$$ and $$\sigma(mn)=\sigma(m)\sigma(n)$$. Let $$m=p_1^{a_1}\cdot p_2^{a_2}\cdots p_r^{a_r}$$ and $$n=q_1^{b_1}\cdot q_2^{b_2}\cdots q_s^{b_s}$$ be the prime-power factorizations of $$m$$ and $$n$$ respectively. Since $$\gcd(m,n)=1$$, $$p_i\neq q_j$$ for all $$i,j$$ and a prime factorization of $$mn$$ is $$mn=p_1^{a_1}\cdot p_2^{a_2}\cdots p_r^{a_r}q_1^{b_1}\cdot q_2^{b_2}\cdots q_s^{b_s}$$. Then $\tau(mn)=(a_1+1)(a_2+1)\cdots(a_r+1)(b_1+1)(b_2+1)\cdots(b_s+1)=\tau(m)\tau(n)$ and $\sigma(mn)=\displaystyle\frac{p_1^{a_1+1}-1}{p_1-1}\frac{p_2^{a_2+1}-1}{p_2-1}\cdots \frac{p_r^{a_r+1}-1}{p_r-1} \frac{q_1^{b_1+1}-1}{q_1-1}\frac{q_2^{b_2+1}-1}{q_2-1}\cdots \frac{q_s^{b_s+1}-1}{q_s-1}=\sigma(m)\sigma(n). ◼$

We have an alternative proof using the following result.

Theorem. Let $$f$$ be a multiplicative function and $$F$$ be an arithmetic function defined by $F(n)=\sum_{d\vert n} f(d).$ Then $$F$$ is also a multiplicative function.

Proof. Let $$m$$ and $$n$$ be two relatively prime positive integers. We show $F(mn)=\sum_{d\vert mn} f(d)=F(m)F(n).$ Note that since $$\gcd(m,n)=1$$, each positive divisor $$d$$ of $$mn$$ is a unique product of a divisor $$d_1$$ of $$m$$ and a divisor $$d_2$$ of $$n$$ where $$\gcd(d_1,d_2)=1$$. Thus $\begin{array}{ll}F(mn)&=\displaystyle\sum_{d\vert mn} f(d)\\ &=\displaystyle\sum_{\substack{d_1\vert m\\d_2\vert n}} f(d_1d_2)\\ &=\displaystyle\sum_{\substack{d_1\vert m\\d_2\vert n}} f(d_1)f(d_2) \text{ since }\gcd(d_1,d_2)=1\\ &=\displaystyle\sum_{d_1\vert m} f(d_1) \sum_{d_2\vert n}f(d_2)\\ &=F(m)F(n). \hfill ◼\\ \end{array}$

Note that the constant function $$f_1(n)=1$$ and the identity function $$f_2(n)=n$$ are multiplicative functions. Since $$\tau(n)=\sum_{d\vert n}1=\sum_{d\vert n} f_1(d)$$ and $$\sigma(n)=\sum_{d\vert n} d=\sum_{d\vert n} f_2(d)$$, $$\tau$$ and $$\sigma$$ are multiplicative functions by the preceding theorem.

Last edited