# [seqfan] Re: Intersections of x^x^...^x

Alois Heinz heinz at hs-heilbronn.de
Wed Nov 9 05:01:56 CET 2011

```I looked at the interval (0,1).

n=5, 9 distinct functions,
((x^(x^x))^x)^x, (x^((x^x)^x))^x, (x^(x^(x^x)))^x, (x^(x^x))^(x^x),
x^(((x^x)^x)^x), x^((x^(x^x))^x), x^(x^((x^x)^x)), x^(x^(x^(x^x))),
(((x^x)^x)^x)^x,
one intersection, between (x^(x^x))^(x^x) and x^(x^((x^x)^x)) at
x= 0.42801103796472992390204...

n=6, 20 distinct functions, 8 intersections

n=7, 48 distinct functions, 84 intersections

so the first sequence is 0,0,0,0,1,8,84,...

Alois

Am 08.11.2011 05:39, schrieb franktaw at netscape.net:
> So, getting back to the original question:
>
> Through n=2, there is only one function, so no intersections.
>
> For n=3, there are two functions: x^(x^x) and (x^x)^x. Setting these
> equal, we get x^x  =  x^2, and hence x = 2, so there is 1 intersection
> with x>1, and none for 0<x<1.
>
> For n=4, there are four functions: x^(x^(x^x)), (x^x)^(x^x) [which
> equals (x^(x^x))^x], x^((x^x)^x), and ((x^x)^x)^x. Take log base x
> twice, these become:
>
> x^x, x+1, x^2, and 3
>
> Then:
>
> x^x = x+1: 1 solution > 1 (A124930).
> x^x = x^2: 1 solution > 1 (2).
> x^x = 3: 1 solution > 1 (A173158).
> x+1=x^2: 1 solution > 1 (phi = A001622).
> x+1=3: 1 solution > 1 (2)
> x^2=3: 1 solution > 1 (sqrt(3) = A002194).
>
> So, through n=4, the first sequence is 0,0,0,0, and the second is
> 0,0,1,6.
>
> I doubt that the first sequence continues to be zeros. More terms,
> anybody?
>
>

> -----Original Message-----
> From: Andrew Weimholt <andrew.weimholt at gmail.com>
>
> On Wed, Nov 2, 2011 at 7:10 PM, Vladimir Reshetnikov
> <v.reshetnikov at gmail.com> wrote:
>> Consider a set of functions obtained by all possible
> parenthesizations of
>> x^x^...^x (with n x's). How many pairwise intersections do they have
> for
>> 0<x<1? And for x>1?
>>
>

```