Well done to Ken from Blessed Robert Johnson Catholic School who sent us his solution to this problem:

After a while of playing with the numbers on a spreadsheet I have discovered that the formula to find the "limiting value" for 2 starting numbers is:

x+2y
3
where x is the first number chosen and y is the second number chosen.



The formula to find the ``limiting value'' for 3 starting numbers will be:
x+2y+3z
6
Terence from Brumby Engineering College has made a prediction for what the limit will be when there are n starting numbers. He thinks it will be

a1+2a2+3a3+4a4+¼+(n-1)an-1+n an
1
2
 n(n+1)
where a1, a2, ..., an-1 and an are the n starting numbers.
Can you see why Terence has made this prediction?
As mathematicians, we like to prove our answers, rather than just predict. Here is an outline of how we might prove the formula when there are two starting numbers.
Suppose that the first two numbers are a and b. Let's work out the next few terms of the sequence. We have


a, b,  a+b
2
 ,  a+3b
4
 ,  3a+5b
8
 ,  5a+11b
16
 ,  11a+21b
32
 ,  21a+43b
64
and so on. We'd like to show that in the limit the coefficient of b is twice the coefficient of a, and also that the sum of these coefficients is the denominator. Can you see why this will prove the formula?

The nth term in the sequence is of the form
an a+bn b
2n-2

(for n ³ 2). We can work out an in terms of an-1 and an-2. Can you see how?

We have an=an-1+2an-2 (for n ³ 4). Similarly, we have bn=bn-1+2bn-2 (for n ³ 4).

It turns out that we can use these formulae to prove that bn=2an +(-1)n (for n ³ 2). Test this formula on the above examples to see that it works for these cases.

We'd like to show that as n tends to infinity, bn becomes twice an. Let's think about bn/an. From the above formula, this is 2+(-1)n/an. But an is getting bigger and bigger, so the (-1)n/an part of this gets smaller and smaller. In fact, it tends to 0, and so the limit of the whole expression is 2.

It is also possible to use the formulae for an and bn to show that an+bn=2n-2 (for n ³ 3) (that is, the coefficients of a and b sum to the number on the bottom of the fraction). Again, test this formula on the examples above.

Combining the last two paragraphs, we find that the limit of the sequence is
a+2b
3
as predicted.