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 2012-10-12, 03:35 #1 Prime95 P90 years forever!     Aug 2002 Yeehaw, FL 11111100101002 Posts How unlucky? I know we are dealing with large k values, but have we been inordinately unlucky in not finding a single factor yet?
 2012-10-12, 03:46 #2 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 100111000000012 Posts Unlucky?! I found 20! The source works! But seriously unlucky for Fermats, I agree.
 2012-10-12, 03:47 #3 Xyzzy     Aug 2002 7×23×53 Posts We don't see how luck plays in at all. It is what it is. (Right?) Definition of LUCK a : a force that brings good fortune or adversity b : the events or circumstances that operate for or against an individual
 2012-10-12, 04:31 #4 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 3×5×683 Posts I also say it works. It finds all (findable) known factors, why it should be missing unknown factors? (if this was the question, but I smell here the question is more about math, if you are in fact asking about expectancy to find any factor, like "were we unlucky, or the expectancy is really so low?", well, it is quite low too, for the ranges we are testing, only very few factors should be expected). I just think there are pure and simple no factors in the ranges we tested, except one factor which managed to stay hidden, but hiding is bad because now he is cornered somewhere and can't get out and I will put my paw on it soon... Last fiddled with by LaurV on 2012-10-12 at 04:35
2012-10-12, 06:00   #5
aketilander

"Åke Tilander"
Apr 2011
Sandviken, Sweden

2×283 Posts

Quote:
 Originally Posted by Prime95 I know we are dealing with large k values, but have we been inordinately unlucky in not finding a single factor yet?
Well it would of course always be a good idea to try mmff against some known factors (if any) just to make sure everything works, if not done already. MM31 have composits within reach I think.

I did quite some work for OBD and I thought for awhile that I was unfortunate, but then I found 2 factors so now I am more fortunate then I should be according to statistics I think.

Someone could maybe do a little statistics here?

Last fiddled with by aketilander on 2012-10-12 at 06:02

2012-10-12, 13:45   #6
Prime95
P90 years forever!

Aug 2002
Yeehaw, FL

22·43·47 Posts

Quote:
 Originally Posted by LaurV if you are in fact asking about expectancy to find any factor, like "were we unlucky, or the expectancy is really so low?",
Yes, it was a math question. With frmky churning out tons of work and several others contributing, I was wondering if the expected number of factors found was less than 1? 1 to 2? above 2? I guess I'm too lazy to go back through the posted results to come up with an exact figure.

2012-10-13, 14:45   #7
ATH
Einyen

Dec 2003
Denmark

1101010101012 Posts

Quote:
 Originally Posted by ET_ In one month of work with mmff, the 22% of the equivalent work done since 2001 has been completed.
Sounds unlucky to me.

I don't know how to calculate the odds of factors within a certain k range, but here is the completed ranges from results thread up to Batalov "N=25 to 2e15" Oct 13th (excluding the few fermat results with version 0.20).
If someone knows the formula for the odds, I'll be happy to try to calculate it.

Code:
Fermat:
n	k
25	500T-2000T
28	550T-1000T
29	550T-1000T
33	700T-1000T
34	700T-1000T
37	4500T-5000T
40	600T-1000T
41	600T-1000T
42	600T-1000T
43	600T-1000T
44	400T-700T
45	500T-1000T
50	300T-1000T
51	350T-1000T
52	300T-1000T
53	200T-1000T
54	200T-1000T
55	200T-1000T
56	200T-1000T
57	280T-1000T
58	280T-1000T
60	200T-1500T
61	200T-1000T
62	200T-1000T
63	200T-1000T
71	300T-1000T
72	300T-1000T
73	300T-1000T
74	300T-1000T
83	35T-281T
84	35T-281T
85	35T-281T
86	25T-281T
87	25T-281T
88	25T-281T
89	25T-281T
90	100T-1000T
100	4T-100T
101	16T-100T
102	16T-100T
103	16T-100T
104	16T-100T
105	16T-100T
106	16T-100T
107	16T-100T
108	16T-100T
109	16T-100T
110	30T-100T
111	30T-100T
112	30T-100T
113	30T-100T
114	30T-100T
115	30T-100T
116	30T-100T
117	30T-100T
118	30T-100T
119	30T-100T
120	30T-100T
121	30T-100T
122	30T-100T
123	30T-100T
124	30T-100T
125	30T-100T
126	30T-100T
127	30T-100T
128	30T-100T
129	30T-100T
130	12T-50T
131	12T-50T
132	12T-50T
133	12T-50T
134	12T-50T
135	12T-50T
136	12T-50T
137	12T-50T
138	12T-50T
139	12T-50T

k
MM31	18000T-25000T
MM61	3573T-10000T
MM89	7T-1300T
MM107	4T-1000T
MM127	563T-3700T and 4000T-4600T and 5500T-5800T

Last fiddled with by ATH on 2012-10-13 at 14:58

 2012-10-13, 19:01 #8 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 5·1,997 Posts It should be the probability of f=k*2N+1 being prime (which is C/ln f) times probability of dividing a Fermat number which is 1/k, then sum (intergrate is fine) over the range. There's Bjorn/Riesel (1998) with a detailed treatment: Attached Thumbnails
 2012-10-13, 23:37 #9 ixfd64 Bemusing Prompter     "Danny" Dec 2002 California 3×829 Posts I think it's more unlucky that we haven't found a new Mersenne prime since April 2009.
2012-10-14, 11:15   #10
aketilander

"Åke Tilander"
Apr 2011
Sandviken, Sweden

2·283 Posts

Quote:
 Originally Posted by ixfd64 I think it's more unlucky that we haven't found a new Mersenne prime since April 2009.
Well, between M[10,000,000 digits] and M[100,000,000 digits] we are supposed to find 6 Mersenne primes according to the theory (about 6 between M[10n digits] and M[10n+1 digits]) so I would say we have been extremely lucky that have found 3 already.

 2012-10-14, 11:38 #11 ATH Einyen     Dec 2003 Denmark 3,413 Posts In your screenshot they have the number of primes k*2n+1 for k

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