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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

\end{displaymath} (1)

For any divisor $d$ of $N$, $N=dd'$ where
d={p_1}^{\delta_1}{p_2}^{\delta_2}\cdots {p_r}^{\delta_r},
\end{displaymath} (2)

d'={p_1}^{\alpha_1-\delta_1}{p_2}^{\alpha_2-\delta_2}\cdots {p_r}^{\alpha_r-\delta_r}.
\end{displaymath} (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).
\end{displaymath} (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},
\end{displaymath} (5)

$\displaystyle N^{\nu(N)}$ $\textstyle =$ $\displaystyle [d^{(1)}\cdots d^{(\nu)}][d'^{(1)}d'^{(\nu)}]$  
  $\textstyle =$ $\displaystyle \prod_{i=1}^\nu d_i \prod_{i=1}^\nu {d_i}' = \left({\prod d}\right)^2,$ (6)

\prod d = N^{\nu(N)/2}.
\end{displaymath} (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}.
\end{displaymath} (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
\end{displaymath} (9)

\end{displaymath} (10)

For a given $a_i$,
\end{displaymath} (11)

Summing over all $a_i$,
\end{displaymath} (12)

so $\sigma(N)=\sigma(ab)=\sigma(a)\sigma(b)$. Splitting $a$ and $b$ into prime factors,
\end{displaymath} (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}.
\end{displaymath} (14)

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

For the special case of $N$ a Prime, (15) simplifies to
\sigma(p)={p^2-1\over p-1}=p+1.
\end{displaymath} (16)

For $N$ a Power of two, (15) simplifies to
\sigma(2^\alpha) = {2^{\alpha+1}-1\over 2-1} = 2^{\alpha +1}-1.
\end{displaymath} (17)

The Arithmetic Mean is
A(N)\equiv {\sigma(N)\over \nu(N)}.
\end{displaymath} (18)

The Harmonic Mean is
{1\over H}\equiv {1\over N}\left({\sum {1\over d}}\right).
\end{displaymath} (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},
\end{displaymath} (20)

and we have
{1\over H(N)} = {1\over \nu(N)} {\sigma(N)\over N} = {A(N)\over N}
\end{displaymath} (21)

\end{displaymath} (22)

Given three Integers chosen at random, the probability that no common factor will divide them all is
\begin{displaymath}[\zeta(3)]^{-1}\approx 1.20206^{-1} \approx 0.831907,
\end{displaymath} (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
\end{displaymath} (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


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.

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© 1996-9 Eric W. Weisstein