Divisor

A divisor of a number $N$ is a number $d$ which Divides $N$, also called a Factor. The total number of divisors for a given number $N$ can be found as follows. Write a number in terms of its Prime Factorization

$$N={p_1}^{\alpha_1}{p_2}^{\alpha_2}\cdots{p_r}^{\alpha_r}.$$ (1)

For any divisor $d$ of $N$, $N=dd'$ where

$$d={p_1}^{\delta_1}{p_2}^{\delta_2}\cdots {p_r}^{\delta_r},$$ (2)

so

$$d'={p_1}^{\alpha_1-\delta_1}{p_2}^{\alpha_2-\delta_2}\cdots {p_r}^{\alpha_r-\delta_r}.$$ (3)

Now, $\delta_1 = 0, 1, \ldots, \alpha_1$, so there are $\alpha_1+1$ possible values. Similarly, for $\delta_n$, there are $\alpha_n+1$ possible values, so the total number of divisors $\nu(N)$ of $N$ is given by

$$\nu(N)=\prod_{n=1}^r (\alpha_n+1).$$ (4)

The function $\nu(N)$ is also sometimes denoted $d(N)$ or $\sigma_0(N)$. The product of divisors can be found by writing the number $N$ in terms of all possible products

$$N=\cases{d^{(1)}d'^{(1)}\cr \vdots\cr d^{(\nu)}d'^{(\nu)}\cr},$$ (5)

so

$$N^{\nu(N)} = [d^{(1)}\cdots d^{(\nu)}][d'^{(1)}d'^{(\nu)}]$$

$$= \prod_{i=1}^\nu d_i \prod_{i=1}^\nu {d_i}' = \left({\prod d}\right)^2,$$ (6)

and

$$\prod d = N^{\nu(N)/2}.$$ (7)

The Geometric Mean of divisors is

$$G\equiv \left({\prod d}\right)^{1/\nu(N)} = [N^{\nu(n)/2}]^{1/\nu(N)} = \sqrt{N}.$$ (8)

The sum of the divisors can be found as follows. Let $N\equiv ab$ with $a\not=b$ and $(a,b)=1$. For any divisor $d$ of $N$, $d=a_ib_i$, where $a_i$ is a divisor of $a$ and $b_i$ is a divisor of $b$. The divisors of $a$ are 1, $a_1$, $a_2$, ..., and $a$. The divisors of $b$ are 1, $b_1$, $b_2$, ..., $b$. The sums of the divisors are then

$$\sigma(a)=1+a_1+a_2+\ldots+a$$ (9)

$$\sigma(b)=1+b_1+b_2+\ldots+b.$$ (10)

For a given $a_i$,

$$a_i(1+b_1+b_2+\ldots+b)=a_i\sigma(b).$$ (11)

Summing over all $a_i$,

$$(1+a_1+a_2+\ldots+a)\sigma(b)=\sigma(a)\sigma(b),$$ (12)

so $\sigma(N)=\sigma(ab)=\sigma(a)\sigma(b)$. Splitting $a$ and $b$ into prime factors,

$$\sigma(N)=\sigma({p_1}^{\alpha_1})\sigma({p_2}^{\alpha_2})\cdots\sigma({p_r}^{\alpha_r}).$$ (13)

For a prime Power ${p_i}^{\alpha_i}$, the divisors are 1, $p_i$, ${p_i}^2$, ..., ${p_i}^{\alpha_i}$, so

$$\sigma({p_i}^{\alpha_i})=1+p_i+{p_i}^2+\ldots+{p_i}^{\alpha_i} ={{p_i}^{\alpha_i+1}-1\over p_i-1}.$$ (14)

For $N$, therefore,

$$\sigma(N) = \prod_{i=1}^r {{p_i}^{\alpha_i+1}-1\over p_i-1}.$$ (15)

For the special case of $N$ a Prime, (15) simplifies to

$$\sigma(p)={p^2-1\over p-1}=p+1.$$ (16)

For $N$ a Power of two, (15) simplifies to

$$\sigma(2^\alpha) = {2^{\alpha+1}-1\over 2-1} = 2^{\alpha +1}-1.$$ (17)

The Arithmetic Mean is

$$A(N)\equiv {\sigma(N)\over \nu(N)}.$$ (18)

The Harmonic Mean is

$${1\over H}\equiv {1\over N}\left({\sum {1\over d}}\right).$$ (19)

But $N=dd'$, so $1/d=d'/N$ and

$$\sum {1\over d}={1\over N}\sum d' = {1\over N} \sum d = {\sigma(N)\over N},$$ (20)

and we have

$${1\over H(N)} = {1\over \nu(N)} {\sigma(N)\over N} = {A(N)\over N}$$ (21)

$$N=A(N)H(N).$$ (22)

Given three Integers chosen at random, the probability that no common factor will divide them all is

$$[\zeta(3)]^{-1}\approx 1.20206^{-1} \approx 0.831907,$$ (23)

where $\zeta(3)$ is Apéry's Constant.

Let $f(n)$ be the number of elements in the greatest subset of $[1,n]$ such that none of its elements are divisible by two others. For $n$ sufficiently large,

$$0.6725\ldots \leq {f(n)\over n} \leq 0.673\ldots$$ (24)

(Le Lionnais 1983, Lebensold 1976/1977).

See also Aliquant Divisor, Aliquot Divisor, Aliquot Sequence, Dirichlet Divisor Problem, Divisor Function, e-Divisor, Exponential Divisor, Greatest Common Divisor, Infinary Divisor, k-ary Divisor, Perfect Number, Proper Divisor, Unitary Divisor

References

Guy, R. K. ``Solutions of $d(n)=d(n+1)$.'' §B18 in Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 73-75, 1994.

Le Lionnais, F. Les nombres remarquables. Paris: Hermann, p. 43, 1983.

Lebensold, K. ``A Divisibility Problem.'' Studies Appl. Math. 56, 291-294, 1976/1977.