Timeline for Product of exponents of prime factorization
Current License: CC BY-SA 3.0
13 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Oct 7, 2013 at 20:11 | answer | added | Will Jagy | timeline score: 0 | |
| Oct 3, 2013 at 22:15 | answer | added | Will Jagy | timeline score: 3 | |
| Oct 3, 2013 at 12:29 | comment | added | Joseph O'Rourke | So now we have a new sequence, $2, 4, 16, 1296, 9681819840000, \ldots$, but with little hope of extending it. | |
| Oct 3, 2013 at 10:40 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Oct 3, 2013 at 7:35 | answer | added | Will Jagy | timeline score: 3 | |
| Oct 3, 2013 at 0:21 | comment | added | Joseph O'Rourke | Interesting! The next number in the sequence is about the size of Will's number, but $P(1586874322944)=4$. | |
| Oct 3, 2013 at 0:16 | comment | added | Gerry Myerson | There is one occurence of 2, 4, 16, 1296 at the Online Encyclopedia of Integer Sequences, but it's not related (oeis.org/A070283). | |
| Oct 2, 2013 at 17:41 | comment | added | The Masked Avenger | Let r(n)=n/rad(n), which bumps the exponents of n down 1. Your p(n) satisfies p(n)=d(r(n)), the number of divisors of r(n). You might look to literature on iterates of d(n) to get a further sense of p. I would start a search using Guy's book on unsolved problems in number theory. | |
| Oct 2, 2013 at 17:00 | vote | accept | Joseph O'Rourke | ||
| Oct 2, 2013 at 17:00 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Oct 2, 2013 at 0:12 | answer | added | Will Jagy | timeline score: 5 | |
| Oct 1, 2013 at 23:47 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |