James Tanton asks in a series of tweets (which I’ve modified slightly): Write 20 numbers. Erase any two, and , and replace with , where some possible choices of are:
Repeat 19 times, until you get a single number. The final result will be independent of the choices of pairs made. Why?
For example, consider . And say we start with the numbers 2, 3, 6, 7. Then we could choose to do the replacement as follows. In each case the two bolded numbers are replaced by one.
- 2, 3, 6, 7 becomes 3, 7, 20
- 3, 7, 20 becomes 7, 83
- 7, 83 becomes 671
and so we’re left with 671. Or we could have
- 2, 3, 6, 7 becomes 2, 7, 27
- 2, 7, 27 becomes 23, 27
- 23, 27 becomes 671
and again we’re left with 671. What’s going on here? It’s not immediately apparent, but if you try this starting with lots of small integers, you often get results which are one less than some number with many small factors — in this case, 672. And in fact , which we can rewrite as
We can think of the whole process, starting with , as computing the product by combining two factors at a time; of course the order doesn’t matter.
Similarly with the function , if we start with we end up with . With , we get . The result with is a little harder to see but if we start with we eventually get . This is a bit easier to see if we realize that ; therefore applying conserves the sum of the reciprocals.
What others are there? Of course there are trivial examples like and ; iterating these functions will just give the sum or the product of the original numbers. But what other nontrivial examples are there? Can you say what all of them are?