Averaging problem

[Rainer Rosenthal]

David Wilson wrote:

> At n = 4, however, there are fewer than 3! = 6 possible values. Starting 
> with a sequence of 4 elements, if we average the first pair, then the last 
> pair, then the only pair, we end up with the same value as when we average 
> the last pair, then the first pair, then the only pair. This means there are 
> at most 5 distinct final values (in fact, there are sequences of 4 elements 
> with 5 final values, e.g, (8, 16, 32, 64), which has final values 20, 26, 
> 30, 40, and 43.
> 
> The question is, for a sequence of n elements, what is the largest possible 
> number final values? The sequence begins 1, 1, 2, 5, ... for n = 1, 2, 3, 4, 
> ... 

This reminds me of "The Book of Numbers", p. 270, where there are
ordinal numbers added together. Wouldn't it be nice if your next
sequence members were ..., 13, 33, 81, ...?

Cheers,
Rainer Rosenthal
r.rosenthal at web.de