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