Mandar has an interesting set of posts on convergence and divergence of infinite series. It made me recall a fascinating result I had come across once. It is not about an infinite series, though. But it is a very interesting sequence.

The sequence needs some explanation of a notation called *hereditary base-n notation*. Let me explain this with an example. Let us write the number 26 in its hereditary base-2 notation. First we start by writing 26 as the sum of powers of 2.

26 = 2^{4} + 2^{3} + 2^{1}

Next, the powers are written as sums of powers of 2 as well.

So, 26 = 2^{22} + 2^{21 + 1 } + 2^{1}

Similarly, the hereditary base-3 notation of 1000 is the following.

1000 = 3^{6} + 3^{4} + 3^{3} + 1 = 3^{3 + 3 } + 3^{3 + 1} + 3^{3} + 1

Note that the bases and the powers cannot be bigger than *n. *Also, we can write the terms as a product of a base power *n* and a number smaller than *n*. For example – 26 can be written in hereditary base-3 notation as, 2.3^{2} + 2.3 + 2.

Now, let us take a number, say 26, and do the following:

- Take the number. Start with base, n = 2.
- Express number in hereditary-n notation
- The next number is formed by changing all the n’s to n+1’s. That is, “increase” the base of the sequence by 1. [n = n + 1]
- Subtract 1 from the above number and goto 2. [number = number – 1]

Let’s take an example. Let’s start with 4.

4 = 2^{2} (step 2)

3^{3} (step 3)

3^{3} – 1 = 26 (step 4) — Iteration 1

26 = 2.3^{2} + 2.3 + 2 (step 2)

2.4^{2} + 2.4 + 2 (step 3)

2.4^{2} + 2.4 + 2 – 1 = 41 (step 4) — Iteration 2

41 = 2.4^{2} + 2.4 + 1 (step 2)

2.5^{2} + 2.5 + 1 (step 3)

2.5^{2} + 2.5 + 1 – 1 = 60 (step 4) — Iteration 3

60 = 2.5^{2} + 2.5 (step 2)

2.6^{2} + 2.6 (step 3)

2.6^{2} + 2.6 – 1 = 83 (step 4) — Iteration 4

83 = 2.6^{2} + 6 + 5 (step 2)

2.7^{2} + 7 + 5 (step 3)

2.7^{2} + 7 + 5 – 1 = 109 (step 4) — Iteration 5

And so on. Now the question is, does this sequence converge (or terminate)? If it converges, what does it converge to? Or if you think it does not converge, explain why.

As always, people who already know this result may please defer commenting.