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Old 2016-09-14, 09:25   #2619
rajula
 
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Let's see what happens this time. Again below 120 digits with the down-driver.
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Old 2016-09-14, 10:58   #2620
LaurV
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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
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Old 2016-09-14, 11:08   #2621
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< 100 digits again!
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Old 2016-09-14, 11:28   #2622
Dubslow
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Booooooooooooooooooo. i8359 = 2 * 7^2 * P91
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Old 2016-09-14, 11:34   #2623
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Quote:
Originally Posted by Dubslow View Post
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
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Old 2016-09-14, 11:51   #2624
Dubslow
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Quote:
Originally Posted by science_man_88 View Post
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 View Post
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
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Old 2016-09-14, 11:53   #2625
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Quote:
Originally Posted by Dubslow View Post
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
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Old 2016-09-14, 12:08   #2626
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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 .
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Old 2016-09-14, 12:09   #2627
Dubslow
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Nope, 1208Z and the downdriver went below 20, cycled up to 50 then below 20 again and now back into the 80s with the D4 = 2^4 * 31. Close to 1500 lines added in the last 30 minutes. Edit: 1210Z and we have passed 10,000 total lines! Congrats Ryan, better than nothing

Last fiddled with by Dubslow on 2016-09-14 at 12:10
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Old 2016-09-14, 12:11   #2628
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10000 lines and counting! 2^4*31 isn't easy to survive.
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Old 2016-09-14, 12:59   #2629
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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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