RE: Tetration operator in functional programming

[Peter Douglass]

Christian Sievers wrote:
 
> This looks strange, as (b^b)^b = b^(b^2), or generally
> n-->b = b^(b^(n-1)). A quick web search showed me tetration as
> 
>   b ^^ 1     = b
>   b ^^ (n+1) = b ^ (b ^^ n)
> 
> so b^^3 = b^(b^b); clearly a more interesting (and faster 
> growing) definition.

Faster growing if and only if b > e ^ (1/e).  For example if b = sqrt(2),
b^^n < 2 for all finite n>0.
--PeterD