Difference between revisions of "Divisor function"

Latest revision as of 05:58, 21 May 2009

The divisor function is denoted $\sigma_k(n)$ and is defined as the sum of the $k$ th powers of the divisors of $n$ . Thus $\sigma_k(n) = \sum_{d|n}d^k = d_1^k + d_2^k + \cdots + d_r^k$ where the $d_i$ are the positive divisors of $n$ .

Counting divisors

Note that $\sigma_0(n) = d_1^0 + d_2^0 + \ldots + d_r^0 = 1 + 1 + \ldots + 1 = r$ , the number of divisors of $n$ . Thus $\sigma_0(n) = d(n)$ is simply the number of divisors of $n$ .

Example Problems

Demonstration

Consider the task of counting the divisors of 72.

First, we find the prime factorization of 72: $72=2^{3} \cdot 3^{2}.$

Since each divisor of 72 can have a power of 2, and since this power can be 0, 1, 2, or 3, we have 4 possibilities. Likewise, since each divisor can have a power of 3, and since this power can be 0, 1, or 2, we have 3 possibilities. By an elementary counting principle, we have $3\cdot 4=12$ divisors.

We can now generalize. Let the prime factorization of $n$ be $p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ . Any divisor of $n$ must be of the form $p_1^{f_1}p_2^{f_2} \cdots p_k^{e_k}$ where the $f_i$ are integers such that $0\le f_i \le e_i$ for $i = 1,2,\ldots, k$ . Thus, the number of divisors of $n$ is $\sigma_0(n) = (e_1+1)(e_2+1)\cdots (e_k+1)$ .

Introductory Problems

2005 AMC 10A Problem 15

Sum of divisors

The sum of the divisors, or $\sigma_1(n)$ , is given by

$\sigma_1(n) = (1 + p_1 + p_1^2 +\cdots p_1^{e_1})(1 + p_2 + p_2^2 + \cdots + p_2^{e_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{e_k}).$

There will be $(e_1+1)(e_2+1)(e_3+1)\cdots (e_k+1)$ products formed by taking one number from each sum, which is the number of divisors of $n$ . Clearly all possible products are divisors of $n$ . Furthermore, all of those products are unique since each positive integer has a unique prime factorization.

Since all of these products are added together, we can conclude this gives us the sum of the divisors.

Sum of kth Powers of Divisors

Inspired by the example of the sum of divisors, we can easily see that the sum of the $k^\text{th}$ powers of the divisors is given by $\begin{align*} \sigma_k(n) &= (1+p_1^k+p_1^{2k}+\cdots +p_1^{e_1k})(1+p_2^k+p_2^{2k}+\cdots +p_2^{e_2k})\cdots (1+p_i^k+p_i^{2k}+\cdots +p_i^{e_ik}) \\ &= \prod_{a=1}^{i}\left(\sum_{b=0}^{e_a}p_a^{bk}\right) \end{align*}$ where $p_1,p_2,...,p_i$ are the distinct prime divisors of $n$ .

This is proven in a very similar way to the $\sigma_1$ case.

@@ Line 22: / Line 22: @@
 == Sum of divisors ==
-The sum of the divisors, or <math>\sigma_1(n)</math>, is given by <center><math> \sigma_1(n) = (1 + p_1 + p_1^2 +\cdots p_1^{e_1})(1 + p_2 + p_2^2 + \cdots + p_2^{e_2}) \cdots (1 + p_n + p_n^2 + \cdots + p_n^{e_n}).</math> </center>
+The sum of the divisors, or <math>\sigma_1(n)</math>, is given by <center><math> \sigma_1(n) = (1 + p_1 + p_1^2 +\cdots p_1^{e_1})(1 + p_2 + p_2^2 + \cdots + p_2^{e_2}) \cdots (1 + p_k + p_k^2 + \cdots + p_k^{e_k}).</math> </center>
+There will be <math>(e_1+1)(e_2+1)(e_3+1)\cdots (e_k+1)</math> products formed by taking one number from each sum, which is the number of divisors of <math>n</math>.  Clearly all possible products are divisors of <math>n</math>.  Furthermore, all of those products are unique since each positive integer has a unique prime factorization.
+Since all of these products are added together, we can conclude this gives us the sum of the divisors.
+== Sum of kth Powers of Divisors ==
+Inspired by the example of the sum of divisors, we can easily see that the sum of the <math>k^\text{th}</math> powers of the divisors is given by
+<cmath>\begin{align*}
+\sigma_k(n) &= (1+p_1^k+p_1^{2k}+\cdots +p_1^{e_1k})(1+p_2^k+p_2^{2k}+\cdots +p_2^{e_2k})\cdots (1+p_i^k+p_i^{2k}+\cdots +p_i^{e_ik}) \\
+&= \prod_{a=1}^{i}\left(\sum_{b=0}^{e_a}p_a^{bk}\right)
+\end{align*}</cmath>
+where <math>p_1,p_2,...,p_i</math> are the distinct prime divisors of <math>n</math>.
+This is proven in a very similar way to the <math>\sigma_1</math> case.
 == See also ==

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