Timeline for Number of distinct values taken by x^x^...^x with parentheses inserted in all possible ways
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 31, 2011 at 4:16 | vote | accept | Vladimir Reshetnikov | ||
| Oct 30, 2011 at 3:05 | history | edited | Vladimir Reshetnikov | CC BY-SA 3.0 |
added 115 characters in body; edited tags
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| Oct 29, 2011 at 5:52 | comment | added | Todd Trimble | Ah, I see! Thanks for hunting that down, Sridhar! | |
| Oct 29, 2011 at 5:51 | comment | added | Sridhar Ramesh | 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] | |
| Oct 29, 2011 at 5:50 | comment | added | Sridhar Ramesh | 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] | |
| Oct 29, 2011 at 3:07 | comment | added | Todd Trimble | Wow, that's really surprising! Can you tell me which two parenthesizations in the case $x = 3$ coincide? | |
| Oct 29, 2011 at 3:01 | answer | added | Todd Trimble | timeline score: 16 | |
| Oct 29, 2011 at 2:53 | comment | added | Vladimir Reshetnikov | The case $x=3$ gives oeis.org/A003018 which differs from oeis.org/A000081 starting from 7th term. | |
| Oct 29, 2011 at 2:34 | comment | added | Todd Trimble | Off-hand, I'd be quite surprised if the generic number isn't attained in the case $x = 3$. Am I missing something? | |
| Oct 29, 2011 at 1:21 | history | asked | Vladimir Reshetnikov | CC BY-SA 3.0 |