(x^x)^(x^x), x^(x^(x^x)), etc...
Meeussen Wouter (bkarnd)
wouter.meeussen at vandemoortele.com
Tue Apr 29 11:11:26 CEST 2003
specialising x to Sqrt[2], only these remain:
1,2,4,8,17,38,88,206,498 ...
Wouter.
-----Original Message-----
From: Edwin Clark [mailto:eclark at math.usf.edu]
Sent: maandag 28 april 2003 17:30
To: seqfan
Cc: math-fun at mailman.xmission.com
Subject: (x^x)^(x^x), x^(x^(x^x)), etc...
Consider the following five functions defined on positive reals:
f1(x) = ((x^x)^x)^x;
f2(x) = (x^(x^x))^x;
f3(x) = x^((x^x)^x);
f4(x) = x^(x^(x^x ));
f5(x) = (x^x)^(x^x);
Note that f2 = f5. There are really only 4 different
functions of this type. We know the Catalan number C(n-1)
gives the number of ways to parenthesize a product of
n variables. If we use x^x for the binary operation and identify
products that give equal functions when x is positive,
how many products do we get?
If the number of x's is n let e(n) be the number of distinct
functions so obtained. I used Maple to simplify the set
of C(n-1) expressions and got the following cardinalities
for the simplified sets for n from 1 to 11:
(*) 1,1,2,4,9,20,48,115,286,719,1842
Except for some of the smaller values, I cannot be sure that
Maple has not failed to identify some equal functions. So
this may not be the sequence e(n).
The sequence (*) coincides with A000081: rooted trees with n
nodes --- at least as far as I have computed. Is this known?
If not can anyone prove that the sequences coincide?
Maple undoubtedly uses standard laws of exponents to simplify
these expressions. This in itself raises a number of questions.
Edwin Clark
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