# 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 '14 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 '14 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 '14 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 '14 at 13:41

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.
Yes. This is partly because $\sigma(n)$ tends to be smoother than $n$ for the reasons that te Reile discusses. – deinst May 26 '14 at 1:57
Please help me out with te Riele's paper. He defines an amicable pair as $\sigma(m)=\sigma(n) = m+n, whereas I've understood it to be the sum of m's divisors equalling n, and n's equalling m. – Wanderlust May 26 '14 at 8:25$\sigma(m)=\sigma(n) = m+n was meant – Wanderlust May 26 '14 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 '14 at 14:24