Let amicable chain of $n$ numbers be (a$_{1}$, a$_{2}$, a$_{3},\dotsc$, a$_{n}$) such that for $i < n$ sum of proper divisors of a$_{i}$ is a$_{i+1}$. For sum of proper divisors of a$_{n}\rightarrow$ a$_{1}$. All numbers in the chain are different.

Case $n = 1$ has small solutions ($\{6\}, \{28\},\dotsc$).

Case $n=2$ has ($\{220, 284\},\{1184,1210\},\dotsc$) examples below $1000$.

What about $n=3$? It doesn't seem to have any examples below $364$ million. Are examples known?

Note that this is different from definition given in "Amicable Number Triples" L. E. Dickson.

Update: I checked up to Amicable$_{5}$ (up to $50$M) and up to Amicable$_{32}$ (up to $10$M). Here are some Amicable$_{n}$ with the smallest number in a cycle shown:

Amicable$_{12}$: $4256$ Amicable$_{5}$: $12496$ Amicable$_{28}$: $14316$

Amicable$_{4}$ are common:
$1264460$; $2115324$; $2784580$; $4938136$; $7169104$; $18048976$; $18656380$; $28158165$; $46722700$

Amicable$_{3}$ - no hits.

  • $\begingroup$ A script to verify such a property should not be too complex to write, I presume. What happens with larger numbers above one million? And as a side note, are there numbers for which your process "converges" to known cycles? $\endgroup$ Commented Nov 7, 2016 at 23:02
  • $\begingroup$ I just wrote an script (complexity O(n^2) - not very efficient) and checked cases up to 1M. I am going to improve it but I was curious if someone have already studied it. I will write an update once I have more data. $\endgroup$
    – Stepan
    Commented Nov 8, 2016 at 0:46
  • $\begingroup$ You can obtain $n^{3/2}$ without much trouble - just check the divisors up to $\sqrt n$. $\endgroup$ Commented Nov 8, 2016 at 10:02

1 Answer 1


You can read this link: https://www.jstor.org/stable/2973750. It is very informative on the generalisation of these numbers.

  • 1
    $\begingroup$ This is not relevant. Both a L. E. Dickson (my link) and Thomas E. Mason (your link) talk about numbers such that S(a1)=S(a2)=...=S(an)=a1+a2+...+an. I am looking at a case when S(a_(i))=a_(i+1) and S(a_n) = a_1. $\endgroup$
    – Stepan
    Commented Nov 8, 2016 at 16:44

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