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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?

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    $\begingroup$ 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. $\endgroup$
    – 2'5 9'2
    Sep 27, 2013 at 2:01
  • $\begingroup$ 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? $\endgroup$
    – jessica
    Sep 27, 2013 at 2:07
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    $\begingroup$ Another pattern: $(\{17,18,22,27,38,73\} \bmod 6)=\{5,0,4,3,2,1\}$ $\endgroup$
    – Fred Daniel Kline
    Sep 27, 2013 at 3:03
  • $\begingroup$ 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. $\endgroup$
    – Maesumi
    Sep 27, 2013 at 12:14

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