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 2016-09-14, 09:25 #2619 rajula     "Tapio Rajala" Feb 2010 Finland 13B16 Posts Let's see what happens this time. Again below 120 digits with the down-driver.
 2016-09-14, 10:58 #2620 LaurV Romulan Interpreter     Jun 2011 Thailand 25·5·59 Posts Well, with the same notation as before, say t=2*F, then its s is 3(1-F*)-2F, therefore this can never be multiple of 3. To "break", it needs to change the power of 2, and that is only possible if F=Qp is a product of an odd perfect square Q and a prime p with p=1 mod 4. It also works if p or q are void. For that to happen below 100 digits the chances (i.e. bad odds) are: at ~50 digits: below 1% at ~40 digits: below 1.6% at ~30 digits: about 2.4% at ~20 digits: about 3.6% at ~15 digits: about 4.8% at ~10 digits: about 7.2% We are "safe" for now (yeah, well, I know, "don't jinx it!" hehe). Last fiddled with by LaurV on 2016-09-14 at 11:08 Reason: some wrong calculus
 2016-09-14, 11:08 #2621 unconnected     May 2009 Russia, Moscow 13×197 Posts < 100 digits again!
 2016-09-14, 11:28 #2622 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
2016-09-14, 11:34   #2623
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

100000110000002 Posts

Quote:
 Originally Posted by Dubslow Booooooooooooooooooo. i8359 = 2 * 7^2 * P91
yeah but luckily since sigma(2^3) = 15 we would need a factor of 3 or 5 to drive it up at last check of what's been said. okay now it's is 2^2 and so would need a 7 I think. and now at 90 digits and a single 2.

Last fiddled with by science_man_88 on 2016-09-14 at 11:49

2016-09-14, 11:51   #2624
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3·29·83 Posts

Quote:
 Originally Posted by science_man_88 and now at 90 digits and a single 2.
Huzzah!
Code:
Checked	8417	93 (show)	2687306884...12<93> = 2^2 · 3^4 · 8294157051...13<90>
$$\tau(P90) = 1$$ and we have recovered the downdriver!

Down below 90! 86 and dropping, now 81. We're at the point where the DB is factoring most of the lines by itself.

Quote:
 Originally Posted by science_man_88 yeah but luckily since sigma(2^3) = 15 we would need a factor of 3 or 5 to drive it up at last check of what's been said. okay now it's is 2^2 and so would need a 7 I think.
Yeah these are/were correct.

Edit: 1205Z and I got a refresh down at 37 digits, subsequent refresh attempts are still pending. Exciting! Termination or more disappointment?

Last fiddled with by Dubslow on 2016-09-14 at 12:08

2016-09-14, 11:53   #2625
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by Dubslow Huzzah! Code: Checked 8417 93 (show) 2687306884...12<93> = 2^2 · 3^4 · 8294157051...13<90> $$\tau(P90) = 1$$ and we have recovered the downdriver!
we are about i8433 right now actually. i8458 is this going to be like yesterday where I couldn't keep up with the edit needed to keep it recent lol. edit: yeah probably ... now at or above i8507. and around 74 digits. might update this edit near my end of edit time. driver:2^4*31 now back at over 90 digits and climbing.

Last fiddled with by science_man_88 on 2016-09-14 at 12:11

 2016-09-14, 12:08 #2626 mshelikoff     "Marc Shelikoff" Dec 2014 Cambridge, MA 3·5 Posts 314718 just passed i15,000. Even though it's a bit (but not extremely) unlikely that the sequence will terminate soon and it's very unlikely that the sequence will catch itself on the way down to create a large cycle, I don't think the probability of those events are in "winning the lottery" territory, and it certainly wouldn't be like the coin flips in Tom Stoppard's "Rosencrantz and Guildenstern Are Dead" at https://youtu.be/KchhSIVwMdY?t=1m30s .
 2016-09-14, 12:09 #2627 Dubslow Basketry That Evening!     "Bunslow the Bold" Jun 2011 40
 2016-09-14, 12:11 #2628 unconnected     May 2009 Russia, Moscow 1010000000012 Posts 10000 lines and counting! 2^4*31 isn't easy to survive.
 2016-09-14, 12:59 #2629 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26×131 Posts 102 digits now i10133 1700 iterations from when I posted in response to dubslow. and 66 minutes roughly. average of roughly 2.33 seconds per iteration roughly. ( to within a minute timing total). Last fiddled with by science_man_88 on 2016-09-14 at 13:03

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