Timeline for The number of possible numeric outcomes of parenthesizations of 2^2^...^2
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
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| Dec 8 at 10:51 | answer | added | 138 Aspen | timeline score: 1 | |
| Jul 29, 2013 at 15:15 | answer | added | Peter Taylor | timeline score: 5 | |
| S Jul 29, 2013 at 11:02 | history | bounty ended | Peter Taylor | ||
| S Jul 29, 2013 at 11:02 | history | notice removed | Peter Taylor | ||
| S Jul 27, 2013 at 22:42 | history | suggested | flornquake | CC BY-SA 3.0 |
correction (see 1st line)
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| Jul 27, 2013 at 18:55 | review | Suggested edits | |||
| S Jul 27, 2013 at 22:42 | |||||
| S Jul 27, 2013 at 15:09 | history | bounty started | Peter Taylor | ||
| S Jul 27, 2013 at 15:09 | history | notice added | Peter Taylor | Reward existing answer | |
| Jul 23, 2013 at 21:29 | answer | added | flornquake | timeline score: 9 | |
| May 24, 2013 at 11:14 | history | edited | Peter Taylor |
edited tags
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| May 17, 2013 at 21:26 | history | edited | Vladimir Reshetnikov | CC BY-SA 3.0 |
fixed off-by-one error
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| May 17, 2013 at 21:18 | comment | added | Vladimir Reshetnikov | @moose Thanks, you are right! | |
| May 15, 2013 at 5:37 | comment | added | Martin Thoma |
@Vladimir Reshetnikov: I think there is an off-by-one error in your formula. When you have n twos and C_n=(2n)!/(n+1)!/n! should be the number of parenthesizations, then for n=3 it should be 5, correct? I see (2^2)^2 and 2^(2^2), but what are the other three combinations? I think C_n gives you the number of parenthesizations for n+1 twos.
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| May 9, 2013 at 18:09 | history | tweeted | twitter.com/#!/StackCodeGolf/status/332558049438990336 | ||
| May 9, 2013 at 18:00 | comment | added | Dr. belisarius | Related oeis.org/A003018/a003018.pdf | |
| May 9, 2013 at 17:46 | comment | added | John Dvorak | @PeterTaylor oh... at least we have a test suite | |
| May 9, 2013 at 17:44 | comment | added | Peter Taylor | @JanDvorak, A002845 (no closed form given) | |
| May 9, 2013 at 16:27 | comment | added | John Dvorak | "Search: seq:1,1,1,2,4,8 Displaying 1-10 of 108 results found." -- I think I'm on the right track. Let's see what are the other entries... | |
| May 9, 2013 at 16:24 | comment | added | John Dvorak |
@Fors I guess n is still way too big to compute. Still, well noted. Maybe a recursive representation in the form "1 or 2^(...) or (...)+(...)"; but you still have the problem of how to normalize such representation of a number (or compare two representations for value equality).
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| May 9, 2013 at 15:36 | comment | added | Fors |
I'm just sharing an idea here, but it seems that it should be possible to exclusively use addition and multiplication, as the answer will always be of the form 2^n, and it therefore would be unnecessary to keep track of anything except n. I.e., just using the rules of exponentiation seems wise. However, there surely is a smarter and completely algebraic way to do this.
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| May 9, 2013 at 15:33 | review | First posts | |||
| May 12, 2013 at 9:51 | |||||
| May 9, 2013 at 15:16 | history | asked | Vladimir Reshetnikov | CC BY-SA 3.0 |