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For what positive x's the number of distinct values taken by x^x^...^x with parentheses inserted in all possible ways is not represented by the sequence A000081? Is it exactly the set of positive algebraic numbers? Is it a superset of positive algebraic numbers? Is it countable? Is $2^{\sqrt 2}$ or $\log_2 3$ in the set?

Vladimir Reshetnikov
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  • Off-hand, I'd be quite surprised if the generic number isn't attained in the case $x = 3$. Am I missing something? – Todd Trimble Oct 29 '11 at 02:34
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    The case $x=3$ gives http://oeis.org/A003018 which differs from http://oeis.org/A000081 starting from 7th term. – Vladimir Reshetnikov Oct 29 '11 at 02:53
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    Wow, that's really surprising! Can you tell me which two parenthesizations in the case $x = 3$ coincide? – Todd Trimble Oct 29 '11 at 03:07
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    3^(3^(3^3) * 3 * 3 * 3) = 3^(3^(3 * 3 * 3) * 3^3). [I've written these using products in the exponent, which of course can be rewritten into iterated exponentiations in various equivalent orders] – Sridhar Ramesh Oct 29 '11 at 05:50
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    The same phenomenon occurs for any natural number, of course: b^(b^(b^b) * the product of b many bs) = b^(b^(the product of b many bs) * b^b). So every natural number b fails to be generic for parenthesization of x^x^x... with 4 + b many copies of x. [Paraphrased from "The Nesting and Roosting Habits of The Laddered Parenthesis", by R. K. Guy and J. L. Selfridge] – Sridhar Ramesh Oct 29 '11 at 05:51
  • Ah, I see! Thanks for hunting that down, Sridhar! – Todd Trimble Oct 29 '11 at 05:52

1 Answers1

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The answer to the second question is "no". Consider the unique solution $x > 0$ to the equation $x^x = 3$. By the Gelfond-Schneider theorem, this number is transcendental. But we have

$$((x^x)^x)^x = x^{x^3} = x^{(x^{(x^x)})}$$

so that two of the parenthesizations coincide. So evidently this set contains transcendental numbers. Lots of other solutions can be similarly generated (e.g., solve $x^{(x^x)} = 4$).

Todd Trimble
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    A similar example I find interesting is to take the unique positive solution of $x^x=x+1$. Then you can easily check that $x^{(x^{(x^x)})} =(x^x)^{(x^x)}$ using the defining equation and the usual exponentiation rules. Although the transcendentality doesn't seem obvious anymore ... – Dejan Govc Nov 02 '11 at 21:30