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Old 2012-04-02, 14:06   #1
Stargate38
 
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Default Sequence 4198862272

I noticed this sequence has been stuck at a C110 since Valentines Day:

http://www.factordb.com/sequences.ph...20&fr=0&to=100

C110=
Code:
51709575528507073763595405281798526430796959961820579558786286970151157498771359600065634235127123647562961971
Why isn't it factored by now? It could have been factored at least 40 times over by now. How long would it take (in CPU hours)?

Last fiddled with by Stargate38 on 02 Apr 12 at 07:10

Last fiddled with by schickel on 2012-04-04 at 08:40 Reason: Adding [code] tags
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Old 2012-04-02, 15:34   #2
fivemack
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This is me with my supermoderator hat on.

I would strongly appreciate it if you stopped sending messages complaining that nobody has yet done a straightforward calculation (it will take you about three days counting the time that it takes you to learn how to use gnfs) that you (and, for this sequence, only you) are interested in.

If you continue doing so then I will consider transforming you for some medium-length period into some silent amphibian-like object.
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Old 2012-04-02, 15:56   #3
firejuggler
 
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i will avance it a few iteration in the hope it will satisfy him
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Old 2012-04-02, 16:46   #4
axn
 
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Quote:
Originally Posted by Stargate38 View Post
Why isn't it factored by now? It could have been factored at least 40 times over by now. How long would it take (in CPU hours)?
Composites in factordb don't get magically factored. Somebody has to do the factoring. There are some automated scripts that hits the unreserved sequences < 1M. For obvious reasons, this sequence won't be picked up by them.

If you want to see this sequence extended, you'll have to DIY.

Last fiddled with by axn on 2012-04-02 at 16:47
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Old 2012-04-02, 19:17   #5
Dubslow
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Quote:
Originally Posted by fivemack View Post
it will take you about three days counting the time that it takes you to learn how to use gnfs
Stargate, as axn said, do it yourself. With regards to learning how to do gnfs, you don't have to: all you need is YAFU. You can type "yafu 'factor(<number here>)'" and it will automatically throw some ECM at it (among other methods) for about 10 minutes, and then move on to SIQS or GNFS automatically and complete them without input from the user. PM me if you want more help (but I highly recommend just downloading YAFU and reading README, that should get you all you need to know).

Last fiddled with by Dubslow on 2012-04-02 at 19:17 Reason: s/others/other methods/
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Old 2012-04-02, 20:20   #6
firejuggler
 
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There, i advanced it by 10 line, you have now a C112(300 curves at 1e6), half-hearteadly ecm'ed.. now go play with your rage and your demands somewhere else... it took me 7 hours.
Now i ll go back to test R468 (n @72214) and R207 (n @24820)

Last fiddled with by firejuggler on 2012-04-02 at 20:22
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Old 2012-04-03, 14:30   #7
Stargate38
 
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What are the criteria for losing the 2^3*3*5 driver? Does it have to be 2^3*3^2*5^2 or 2^3*3^3*5^3?

Look how straight the graph has gotten: http://www.factordb.com/aliquot.php?...&aq=4198862272
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Old 2012-04-03, 15:52   #8
fivemack
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Have a play in gp: try running things of the form

Code:
for(u=10^12,10^12+10^5,s=u*2^3*3*5;if(gcd(s,2^4*3^2*5^2)==120,S1=sigma(s)-s;S2=sigma(S1)-S1;S3=sigma(S2)-S2;S4=sigma(S3)-S3;if(gcd(S4,15)==1,print(u,factor(u),factor(S1),factor(S2),factor(S3),factor(S4)))))
(it ought to be clear what I'm doing there, the first gcd test is to ensure that you have exactly 2^3*3*5 rather than higher powers)

You never lose the 2^3 on the first step. If you've lost it on the second step, you've gone via 2^3*3*5^W*X^2*p where W is 2 or more and X can either be 1 or something larger than seven.

In 'actual escape' cases where you lose the 3 and 5, the trajectory seems always to be

* get an extra power of five and have the remainder be a prime (2^3*3*5^2*p)
* lose the extra power of five in exchange for changing the power of two (2^k*3*5)
* have the power of five change
* lose the three

and this happens with probability about 0.0002 for numbers around 15 digits and 0.0001 for numbers around 25 digits. Of course you have higher probability (I think about 0.003) of escape routes that take a little longer, but the first step on an escape route appears always to be 2^3*3*5^2*p.

(try working it through with 120*50221*389318203*51145795021)
(or 120*1000000009991 where you lose the 2^3 only for a short period)
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Old 2012-04-03, 16:02   #9
axn
 
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Quote:
Originally Posted by fivemack View Post
You never lose the 2^3 on the first step.
2^3*3*5*x^2 will lose it on the first step.
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Old 2012-04-03, 16:17   #10
firejuggler
 
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If my memory serve, it require that 2^3*3^2*5^2 and every other prime 1 mod 4.
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Old 2012-04-03, 17:57   #11
Raman
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Quote:
Originally Posted by Stargate38 View Post
I noticed this sequence has been stuck at a C110 since Valentines Day:

http://www.factordb.com/sequences.ph...20&fr=0&to=100

Code:
C110=51709575528507073763595405281798526430796959961820579558786286970151157498771359600065634235127123647562961971
Why isn't it factored by now? It could have been factored at least 40 times over by now. How long would it take (in CPU hours)?
I found out a new confluence for the Aliquot Sequence

Sequence starting value = 3*2[sup]164[/sup]
for merging with the following
Aliquot sequence = 772032

Straight downdriver run quite directly from 95 digits
to 7 digits in the size

Stargate38, why this number 4198862272 = 26*11*5964293
what's special with this number?

The number that I mentioned above, which is as well as,
since it is being struck at C110, at the iteration number 3324
at all, this is again being a C110, I'd like to see it as well
as being factored, again, thereby being gone ahead

Code:
C110 = 22570951157462245129386607801622813637611303162132159250374656707973168765663980862177943354450043517261333717
Why isn't it this being factored by now?
at all yet still

Last fiddled with by wblipp on 2012-04-03 at 18:10 Reason: add code box
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