Predict number of digits in factor of (3^607-1)/2 - Page 2

Raman · 2010-10-23, 19:23

Thus, I would leave it up at your wish to make that poll more appealing.

Well, let them be proper bounds: for example
60 to 69 digits, 70 to 79 digits, 80 to 89 digits, 90 to 99 digits,
100 to 119, etc.
that I had rather mistyped as
60 to 70, 70 to 80, 80 to 90, 90 to 100, 100 to 120, etc.
if in any case that this discussion goes on seriously and then becoming interesting...

(Below 60 digits, and then that next option is 61 to 69 digits, or rather,
that first option is less than 59 digits, and then that 60 to 69 digits, that way?)

firejuggler · 2010-10-23, 19:35

i'm all for the 3 way split : 100, 100 /90

Phil MjX · 2010-10-23, 23:46

And so do I, but rather with 70/90/130...

Batalov · 2010-10-24, 00:20

Finally, we have at least one vote for p60-p69, which is fairly possible.
A quick look at past results >=c260 :

Code:

5501 2,1039- c307 p80 . p227 Kleinjung et al snfs
5959 5,409-  c282 p96 . p186 NFS@Home snfs
5811 2,941-  c280 p69 . p211 mersenneforum+NFS@Home+Dodson+Womack snfs
5874 2,1127- c279 p92 . p187 Kleinjung snfs
5787 5,398+  c274 p62 . p100 . p113 Childers+Dodson snfs
5331 6,353-  c274 p120 . p155 Aoki+Kida+Shimoyama+Ueda snfs
5897 3,568+  c268 p64 . p68 . p136 Batalov+Dodson snfs
5743 2,2086M c268 p97 . p171 Batalov+Dodson snfs
5713 2,908+  c268 p95 . p174 Childers+Dodson snfs
5714 5,383+  c267 p128 . p140 Childers+Dodson snfs
5676 5,389+  c265 p112 . p154 Childers/Dodson snfs
5654 6,392+  c262 p127 . p136 Childers/Dodson snfs
5951 2,1043+ c261 p63 . p198 Batalov+Dodson snfs
5946 2,919-  c261 p126 . p135 Batalov+Dodson snfs
5613 12,241- c260 p66 . p194 Childers/Dodson/Wackerbarth snfs

wreck · 2010-10-24, 01:02

Here is the state of the 68 Cunningham numbers' smallest factor that factored by NFS@Home until now.

[50,59]:2
[60,69]:18
[70,79]:10
[80,89]:16
[90,99]:13
[100,109]:7
[110,119]:1
[120,129]:1
[130,139]:0

Raman · 2010-10-24, 17:34

Now that the voting options seems to approach up with model of Gaussian Distribution.
According to that Central Limit Theorem, which states that if number of random variables is being sufficiently large enough, then it would become closer to follow up with that normal distribution curve:
probability density function, that is

$f(x) = \frac{1}{\sqrt{2 \pi }\sigma} e^{-\frac{(x- \mu)^2}{2 \sigma ^2}}$
for that values of mean = $\mu$ , variance = $\sigma^2$ , standard deviation = $\sigma$
that would become to be centered up, at that value of
x = 0, assume that mean value is 0; where 68% of that options lie up within 1 standard deviation from mean; 95% within 2 standard deviation values from that mean, like that is this case
rather within that way

frmky · 2010-10-24, 18:06

Quote:

Originally Posted by Raman

Now that the voting options seems to approach up with model of Gaussian Distribution.

Not so sure... Looks more log normal to me.

10metreh · 2010-10-24, 20:04

Quote:

Originally Posted by Batalov

Code:

5874 2,1127- c279 p92 . p187 Kleinjung snfs

This one was SNFS difficulty 290, so why isn't it on the champions page?

Batalov · 2010-10-24, 20:40

And that was back on page 116!

If this was not missed, this would render both 5,409- and 3,607- non-champions. That's a tough situation.

frmky · 2010-10-28, 07:32

Recent Lonestar issues have delayed the LA a bit. The new ETA is mid to late next week.

Batalov · 2010-10-28, 07:44

I know exactly what's happening:

Quote:

Originally Posted by A. & B. Strugatsky

...some powerful, mysterious, and very selective force impedes their (scientific) work in fields ranging from biology to mathematical linguistics.
The mysterious force is in fact the Universe's reaction to the mankind's scientific pursuit which threatens to harm the very essence of the Universe. This reaction prevents development of "super-civilizations", ones that would be able to counteract the Second law of thermodynamics on a cosmic scale.

I've read about it a long time ago. Still true.

View Poll Results: Predict number of digits in factor of (3^607-1)/2
Less than 60 digits	1	3.57%
60 to 69 digits	2	7.14%
70 to 79 digits	3	10.71%
80 to 89 digits	4	14.29%
90 to 99 digits	5	17.86%
100 to 119 digits	3	10.71%
More than 120 digits	6	21.43%
Number splits up into three parts, rather	2	7.14%
Number splits up into FOUR parts, rather	2	7.14%
Number splits up into FIVE parts, rather	0	0%
Voters: 28. You may not vote on this poll

2010-10-23, 19:23	#12
Raman Noodles "Mr. Tuch" Dec 2007 Chennai, India 3×419 Posts	Thus, I would leave it up at your wish to make that poll more appealing. Well, let them be proper bounds: for example 60 to 69 digits, 70 to 79 digits, 80 to 89 digits, 90 to 99 digits, 100 to 119, etc. that I had rather mistyped as 60 to 70, 70 to 80, 80 to 90, 90 to 100, 100 to 120, etc. if in any case that this discussion goes on seriously and then becoming interesting... (Below 60 digits, and then that next option is 61 to 69 digits, or rather, that first option is less than 59 digits, and then that 60 to 69 digits, that way?) Last fiddled with by Raman on 2010-10-23 at 19:24

2010-10-23, 23:46	#14
Phil MjX Sep 2004 5×37 Posts	And so do I, but rather with 70/90/130... Last fiddled with by Phil MjX on 2010-10-23 at 23:48 Reason: do the sum be equal to 290 ?

2010-10-24, 17:34	#17
Raman Noodles "Mr. Tuch" Dec 2007 Chennai, India 2351₈ Posts	Now that the voting options seems to approach up with model of Gaussian Distribution. According to that Central Limit Theorem, which states that if number of random variables is being sufficiently large enough, then it would become closer to follow up with that normal distribution curve: probability density function, that is $f(x) = \frac{1}{\sqrt{2 \pi }\sigma} e^{-\frac{(x- \mu)^2}{2 \sigma ^2}}$ for that values of mean = $\mu$ , variance = $\sigma^2$ , standard deviation = $\sigma$ that would become to be centered up, at that value of x = 0, assume that mean value is 0; where 68% of that options lie up within 1 standard deviation from mean; 95% within 2 standard deviation values from that mean, like that is this case rather within that way Last fiddled with by Raman on 2010-10-24 at 17:37 Reason: for that purpose of formula error correction only, actually

2010-10-24, 20:40	#20
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2⁴·593 Posts	And that was back on page 116! If this was not missed, this would render both 5,409- and 3,607- non-champions. That's a tough situation. Last fiddled with by Batalov on 2010-10-24 at 20:44 Reason: (in difficulty. They are still champions in size!)

2010-10-23, 19:35	#13
firejuggler Apr 2010 Over the rainbow 19×137 Posts	i'm all for the 3 way split : 100, 100 /90

2010-10-24, 01:02	#16
wreck "Bo Chen" Oct 2005 Wuhan,China 2³·3·7 Posts	Here is the state of the 68 Cunningham numbers' smallest factor that factored by NFS@Home until now. [50,59]:2 [60,69]:18 [70,79]:10 [80,89]:16 [90,99]:13 [100,109]:7 [110,119]:1 [120,129]:1 [130,139]:0

2010-10-28, 07:32	#21
frmky Jul 2003 So Cal 83F₁₆ Posts	Recent Lonestar issues have delayed the LA a bit. The new ETA is mid to late next week.

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