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Old 2017-08-14, 14:54   #12
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Quote:
Originally Posted by ericw View Post
https://oeis.org/A100289
Numbers n such that (1!)^2 + (2!)^2 + (3!)^2 +...+ (n!)^2 is prime.
a(19) = 32841 from Serge Batalov, Jul 29 2017
I'll probably leave that one alone. I haven't working on it.
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Old 2017-08-15, 14:59   #13
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Sieving completed a couple of days ago. I've tested to n=30000 and have verified known results. I should reach Serge's find by the end of the week.
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Old 2017-08-18, 15:02   #14
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I've tested to n=40000 and have verified known results.
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Old 2017-08-23, 12:50   #15
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I've tested to n=50000 and have verified known results, including Serge's find. Continuing.
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Old 2017-08-31, 16:37   #16
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I've tested to n=60000. No new PRPs.
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Old 2017-09-14, 02:07   #17
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I've tested to n=70000. No new PRPs.
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Old 2017-09-18, 15:22   #18
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I don't know how I missed this one. The 25th term is for n=59961 which is 260447 digits (if I calculated that correctly).

The search is around n=74000. I'm not certain how much further I'm going to search, but I have sieved to n=100000.
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Old 2017-10-23, 13:30   #19
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After a hiatus, I've tested to n=80000. No new PRPs.
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Old 2017-11-15, 15:05   #20
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Completed to 90000. No new PRPs. Continuing.
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Old 2017-11-15, 19:53   #21
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Quote:
Originally Posted by rogue View Post
I don't know how I missed this one. The 25th term is for n=59961 which is 260447 digits (if I calculated that correctly).
Great! (I verified that it's 3-prp.) I see that you've added it to A001272.
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Old 2017-11-20, 14:02   #22
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Quote:
Originally Posted by rogue View Post
I don't know how I missed this one. The 25th term is for n=59961 which is 260447 digits (if I calculated that correctly).
(Actually 260448 decimal digits.)

Congratulations again :)
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