Symmetry in the Sum of Factors

The factors of 72 are 1,2,3,4,6,8,9,12,18,24,36,72. The factors can be organized as such $$m=1,2,3,4,6,8$$ $$n=72,36,24,18,12,9$$

Knowing the symmetric properties and that the 12 factors exist. Pair the factors by symmetry. $$(1,72),(2,36),(3,24),(4,18),(6,12),(8,9)$$ The sum of each pair is (arranged in ascending order), $$17,18,22,27,38,73$$

My question: Is there a pattern am I not seeing? Or is the sum of the pair just random gibberish? Assuming we are only working with positive $m$ and $n$, there is an upper bound on the sum of any two multiplied positive numbers, in our example it is $n=1$ and $m=72$ and the lower bound is (assuming we have the whole real line) $n=\sqrt72$ and $m=\sqrt72$. So is there a pattern here?

• You have whole number outputs of $x+\frac{72}{x}$ where $x$ is a whole number too. Double these numbers and you have perimeters of rectangles of area 72. These are some "patterns" but maybe not what you are looking for. Sep 27, 2013 at 2:01
• Alex, that is the formula I used to compute those values. Excluding this formula, is their a natural sequence in these numbers? Purely on the virtue of the numbers alone? Sep 27, 2013 at 2:07
• Another pattern: $(\{17,18,22,27,38,73\} \bmod 6)=\{5,0,4,3,2,1\}$ Sep 27, 2013 at 3:03
• Once you start with the prime factorization then you can easily write the pairs, but ordering them from small to largest is not easy. Unless the number is $p^n$ in which case you can easily order the sum of pairs. For $p^nq^m$ the ordering seems much more complex. So ordering of sum of pairs of divisors of $2^n3^m$ does not seem trivial. Sep 27, 2013 at 12:14