Superbase 2 notation is power-of-2 notation in which each power is also written in base 2, and each power of a power as well, etc. For example: In base 2, 41 = 2^5 + 2^3 + 1 In superbase 2, 41 = 2^2^2+1 + 2^2+1 + 1. So, in superbase 2, only 1's and 2's appear.

Consider the following sequence of integers, from a given starting value n:

n_{1}= n;

For n_{2}, write n_{1}in superbase 2 notation and replace each 2 by a 3

For n_{3}, subtract 1 from n_{2}, write the result in superbase 3 notation, and replace each 3 by a 4

For n_{4}, subtract 1 from n_{3}, write the result in superbase 4 notation, and replace each 4 by a 5

and so on.It is an amazing fact that, for any n

_{1}, this sequence becomes 0 eventually!

Problem: Let n_{1}= 4. How many steps until the sequence zeros out?I would be pleased to see answers (not necessarily complete solutions!) on this one to see if they agree with what I think the answer is.

To see that you understand the definition and to underscore the surprising nature of the result, here are a few steps for n_{1}= 41.41 nAgain: The problem is: How many steps (approximately) until the result is 0?_{1}= 41 2^(2^2+1) + 2^(2+1) + 1 3^(3^3+1) + 3^(3+1) + 1 n_{2}= 22876792455043 3^(3^3+1) + 3^(3+1) 4^(4^4+1) + 4^(4+1) So n_{3}= 53631231719770388398296099992823384509917463\ 2823695735108942457748870561202941879072074\ 9719266761371076012743274594420341501553124\ 7786279785734596024337408 4^(4^4+1) + 4^(4+1) - 1 4^(4^4+1) + 3*4^4 + 3*4^3 + 3*4^2 + 3*4 + 3 5^(5^5+1) + 3*5^5 + 3*5^3 + 3*5^2 + 3*5 + 3 So n_{5}is about 10^2184 5^(5^5+1) + 3*5^5 + 3*5^3 + 3*5^2 + 3*5 + 2 6^(6^6+1) + 3*6^6 + 3*6^3 + 3*6^2 + 3*6 + 2 So n_{6}is about 10^36305© Copyright 1996 Stan Wagon. Reproduced with permission.

2 October 1998