# Amicable pair sums - intriguing sexagesimal relationships

Some amicable pair sums show intriguing relationships, for example:

1) The sums of the two numbers in each of the first five pairs have a gcd of 126. 12600 is the sum of those in the fifth pair, which I designate Am(5)

2) The sum of Am(6) is 21600, and that of both Am(32) and Am(35) is 1296000. 216 = 6^3, while 1296 = 6^4. Is there an Am(n) whose sum is equal to some other power of 6 multiplied by a power of 10?

3) The pair sum of Am(26), 756000, is equal to 60 times 12600, Am(5). Sum Am(32)and(35) is 60 times 21600, Am(6).

Apologies if I haven't formalised these equalities more clearly.

(See OEIS, A180164, "The sum of the two numbers in an amicable pair". "List of amicable numbers from 1 to 20,000,000", www.vaxasoftware.com)

Any explanation or rule?

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You present a small number of isolated facts --- I don't know what would count as an explanation, or a rule. –  Gerry Myerson Feb 23 at 9:19
I know they're isolated. I was hoping for someone who could spot any kind of connection and bigger picture. –  Wanderlust Feb 23 at 9:32
You know that there are formulas that generate amicable pairs. Have you checked whether those formulas give any insights? –  Gerry Myerson Feb 23 at 9:35
Thanks Gerry. I'd already looked at Wikipedia, and looked again just now, but they didn't have anything under "Rules of generation" that revealed anything to me, and/or seemed to result in more than half a dozen pairs. I'd be very grateful if you could have a look too. –  Wanderlust Feb 23 at 13:41

As (2) is really the only question, I will attempt to address it along with what I think is your fundamental observation, which is that the pair sums of amicable pairs are really smooth. This stems from the fact that if a number is smooth then there are likely lots of numbers whose sum of divisors equals that number, and if there are enough such numbers then it becomes likely that the sum of two such numbers has the same value. H. J. J. te Riele used this observation to create "A new method for finding Amicable Pairs".

I am too lazy to attempt to find more pairs of the form $6^m 10^n$, but I would not be surprised if many exist.

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Most the pair sums I've looked at in response to your answer so far are smoother than either of the two numbers in the pair concerned, or at least have a greatest prime factor smaller than the average gpf of the two numbers. That is, if amicable sum m + n is k smooth, then k < (j+l)/2 iff m is j smooth and n is l smooth. Summing AP's seems to be a way of smoothing out, reducing the incidence of big lumpy primes. Getting lots of small ones instead: 2's, 3's, 5's, makes it less surprising to get multiples and powers of 6 and 10 so frequently. –  Wanderlust May 25 at 15:48
Yes. This is partly because $\sigma(n)$ tends to be smoother than $n$ for the reasons that te Reile discusses. –  deinst May 26 at 1:57
Please help me out with te Riele's paper. He defines an amicable pair as $\sigma(m)=\sigma(n) = m+n, whereas I've understood it to be the sum of m's divisors equalling n, and n's equalling m. – Wanderlust May 26 at 8:25$\sigma(m)=\sigma(n) = m+n was meant –  Wanderlust May 26 at 8:29
These two descriptions are the same. Your sum of divisors function does not include the number itself, while $\sigma$ does. The definition of $\sigma$ is a more natural definition because it is multiplicative, i.e. $\sigma(p_1^{e_1} p_2^{e_2}\cdots p_n^{e_n}) = \sigma(p_1^{e_1})\sigma(p_2^{e_2})\cdots\sigma(p_n^{e_n})$ where the $p_i$ are distinct primes and $e_1$ are their exponents. Furthermore $\sigma(p^e)=(p^(e+1)-1)/(p-1)$ when $p$ is a prime. It would be well worth your while to get an elementary number theory text (there are probably many suggestions elsewhere on math.stackexchange) –  deinst May 26 at 14:24