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

franktaw at netscape.net franktaw at netscape.net
Tue Nov 8 05:39:54 CET 2011

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


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 

I doubt that the first sequence continues to be zeros. More terms, 

Franklin T. Adams-Watters

-----Original Message-----
From: franktaw <franktaw at netscape.net>
To: seqfan <seqfan at list.seqfan.eu>
Sent: Thu, Nov 3, 2011 3:39 am
Subject: [seqfan] Re: Intersections of x^x^...^x

It's perhaps a subtle point, but he asked for intersections of the
functions, not of the expressions. Expressions that produce the same
function are not considered different.

Franklin T. Adams-Watters

-----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
> 0<x<1? And for x>1?

for n>3, it is infinite...

(x^x)^(x^x) = (x^(x^x))^x

ln ((x^x)^(x^x)) = ln ((x^(x^x))^x)

(x^x) * ln (x^x) = x * ln (x^(x^x))

x*(x^x) * ln (x) = x*(x^x) * ln (x)




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