# Ore numbers: how do they grow?

Posted by: Gary Ernest Davis on: November 15, 2010

Ã˜ystein Ore

An Ore number (also known as a harmonic divisor number) is a number whose harmonic mean of its divisors is an integer.

The harmonic mean of a set of numbers $\{x_1,\ldots x_k\}$ is $\frac{n}{\frac{1}{x_1}+\ldots +\frac{1}{x_k}}$.

When the $x_1, \ldots , x_k$ are the whole number divisors of the number n, then we get the harmonic mean of the divisors of n.

These numbers were named for Ã˜ystein Ore who proved that all perfect numbers are Ore numbers. Recall a whole number is perfect if the sum of the divisors of the number is exactly twice the number. 6 and 28 are well-known perfect numbers.

For example, the divisors of 3 are 1 and 3, so the harmonic mean of these divisors is $\frac{2}{\frac{1}{1}+\frac{1}{3}}=\frac{3}{2}$ which is not an integer, so 3 is not an Ore number.

More generally, if $p > 1$ is a prime number then the harmonic mean of its divisors is $\frac{2}{1+\frac{1}{p}}=\frac{2p}{p+1}$ which is not an integer, so no prime number is an Ore number.

On the other hand, the divisors of 6 are 1, 2, 3 and 6 so the harmonic mean of these divisors is $\frac{4}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=\frac{24}{6+3+2+1}=2$ so 6 is an Ore number.

Using a computational language such as Maple, Mathematica, MATLAB or Python one can calculate a few Ore numbers.

Here is a list of the Ore numbers up to 10,000,000:

1, 6, 28, 140, 270, 496, 672, 1638, 2970, 6200, 8128, 8190, 18600, 18620, 27846, 30240, 32760, 55860, 105664, 117800, 167400, 173600, 237510, 242060, 332640, 360360, 539400, 695520, 726180, 753480, 950976, 1089270, 1421280, 1539720, 2178540, 2229500, 2290260, 2457000, 2845800, 4358600, 4713984, 4754880.

How fast do these Ore numbers grow?

Let’s write $O(n)$ for the $n^{th}$ Ore number.

So, $O(1) = 1, O(2) = 6, O(3) = 28, O(4) = 140$ and so on.

What does a plot of $O(n)$ versus $n$ look like?

A regression indicates that $O(n)$ grows as a power of n, at least in the range $1 < n < 10000000$

A plot of $\log (O(n)) \textrm{ versus } \log (n)$ should therefore look like a straight line, and it does:

This suggest that there are constant $A, B$ such that as n increases, $O(n) \approx An^B$.

If this is so, then $\frac{O(n+1)}{O(n)} \approx \frac{A\times (n+1)^B}{A\times n^B}=(1+\frac{1}{n})^B \to 1 \textrm{ as } n \to \infty$.

From the data we have for the Ore numbers less than 10000000 the plot of $\frac{O(n+1)}{O(n)} \textrm{ versus } n$ looks as follows:

So,Â  does $\frac{O(n+1)}{O(n)} \to 1 \textrm{ as } n \to \infty$ ?

This is a weaker question than: “is $O(n) \approx An^B$ for some constants $A, B$“.

#### Postscript

• Here are the Ore numbers up to 1 billion ( = 1,000 million):

1, 6, 28, 140, 270, 496, 672, 1638, 2970, 6200, 8128, 8190, 18600, 18620, 27846, 30240, 32760, 55860, 105664, 117800, 167400, 173600, 237510, 242060, 332640, 360360, 539400, 695520, 726180, 753480, 950976, 1089270, 1421280, 1539720, 2178540, 2229500, 2290260, 2457000, 2845800, 4358600, 4713984, 4754880, 5772200, 6051500, 8506400, 8872200, 11981970, 14303520, 15495480, 16166592, 17428320, 18154500, 23088800, 23569920, 23963940, 27027000, 29410290, 32997888, 33550336, 37035180, 44660070, 45532800, 46683000, 50401728, 52141320, 56511000, 69266400, 71253000, 75038600, 80832960, 81695250, 90409410, 108421632, 110583200, 115048440, 115462620, 137891520, 142990848, 144963000, 163390500, 164989440, 191711520, 221557248, 233103780, 255428096, 287425800, 300154400, 301953024, 318177800, 318729600, 326781000, 400851360, 407386980, 423184320, 428972544, 447828480, 459818240, 481572000, 499974930, 500860800, 513513000, 526480500, 540277920, 559903400, 623397600, 644271264, 675347400, 714954240, 758951424, 766284288, 819131040, 825120800, 886402440, 900463200, 995248800