How unlucky?

Prime95 · 2012-10-12, 03:35

I know we are dealing with large k values, but have we been inordinately unlucky in not finding a single factor yet?

Batalov · 2012-10-12, 03:46

Unlucky?! I found 20!

The source works!

But seriously unlucky for Fermats, I agree.

Xyzzy · 2012-10-12, 03:47

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

LaurV · 2012-10-12, 04:31

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...

aketilander · 2012-10-12, 06:00

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?

Prime95 · 2012-10-12, 13:45

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.

ATH · 2012-10-13, 14:45

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

Batalov · 2012-10-13, 19:01

It should be the probability of f=k*2^N+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:

ixfd64 · 2012-10-13, 23:37

I think it's more unlucky that we haven't found a new Mersenne prime since April 2009.

aketilander · 2012-10-14, 11:15

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.

ATH · 2012-10-14, 11:38

In your screenshot they have the number of primes k*2ⁿ+1 for k<K is G(K)=K/(ln(K*2ⁿ)-1).

So trying this for example on n=25 and 500T<k<2000T

G(2000T)-G(500T) = 2.88*10¹³ primes

Now each of these you say has a 1/k chance of dividing a fermat number, so if we divide the number of primes with the average k in the interval which is 1250T it should be an ok estimate?

2.88*10¹³/1250*10¹² = 0.0231, so 2.31% chance of finding a fermat factor in that interval.

Doing this for all the ranges gives expected fermat factors in the ranges done so far at 0.95, unless my calculations are all wrong?

Code:

Fermat:
n	k
25	500T-2000T	0.02305923714
28	550T-1000T	0.01081348535
29	550T-1000T	0.01067572583
33	700T-1000T	0.006163534161
34	700T-1000T	0.006089842254
37	4500T-5000T	0.001704400774
40	600T-1000T	0.008058379498
41	600T-1000T	0.00796937526
42	600T-1000T	0.007882315112
43	600T-1000T	0.007797136063
44	400T-700T	0.008464224672
45	500T-1000T	0.0101876489
50	300T-1000T	0.01566992576
51	350T-1000T	0.01386193619
52	300T-1000T	0.01536014979
53	200T-1000T	0.01886175148
54	200T-1000T	0.01867863009
55	200T-1000T	0.01849902961
56	200T-1000T	0.01832284948
57	280T-1000T	0.01529457624
58	280T-1000T	0.01515181805
60	200T-1500T	0.02016286432
61	200T-1000T	0.01748999027
62	200T-1000T	0.01733242097
63	200T-1000T	0.01717766503
71	300T-1000T	0.01293148238
72	300T-1000T	0.01282475399
73	300T-1000T	0.01271977269
74	300T-1000T	0.01261649593
83	35T-281T	0.01727941477
84	35T-281T	0.01714751685
85	35T-281T	0.01701761711
86	25T-281T	0.01816263451
87	25T-281T	0.01802700773
88	25T-281T	0.01789339132
89	25T-281T	0.01776174093
90	100T-1000T	0.0170126916
100	4T-100T		0.01833508944
101	16T-100T	0.01425457045
102	16T-100T	0.01415798841
103	16T-100T	0.01406270624
104	16T-100T	0.01396869789
105	16T-100T	0.01387593798
106	16T-100T	0.0137844018
107	16T-100T	0.01369406531
108	16T-100T	0.01360490506
109	16T-100T	0.01351689825
110	30T-100T	0.009970885797
111	30T-100T	0.009907309386
112	30T-100T	0.009844538526
113	30T-100T	0.009782558006
114	30T-100T	0.009721352994
115	30T-100T	0.009660909025
116	30T-100T	0.009601211995
117	30T-100T	0.009542248144
118	30T-100T	0.009484004048
119	30T-100T	0.009426466609
120	30T-100T	0.009369623046
121	30T-100T	0.009313460883
122	30T-100T	0.009257967941
123	30T-100T	0.009203132331
124	30T-100T	0.009148942443
125	30T-100T	0.009095386938
126	30T-100T	0.009042454743
127	30T-100T	0.00899013504
128	30T-100T	0.008938417258
129	30T-100T	0.008887291072
130	12T-50T		0.01012143297
131	12T-50T		0.01006383816
132	12T-50T		0.01000689509
133	12T-50T		0.009950592732
134	12T-50T		0.009894920352
135	12T-50T		0.00983986743
136	12T-50T		0.009785423686
137	12T-50T		0.009731579065
138	12T-50T		0.009678323732
139	12T-50T		0.009625648066
--------------------------------------
Total			0.9522655104

2012-10-12, 03:35	#1
Prime95 P90 years forever! Aug 2002 Yeehaw, FL 7911₁₀ 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, 04:31	#4
LaurV Romulan Interpreter "name field" Jun 2011 Thailand 9,973 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

2012-10-13, 19:01	#8
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·4,931 Posts	It should be the probability of f=k*2^N+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

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2012-10-12, 03:46	#2
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2×4,931 Posts	Unlucky?! I found 20! The source works! But seriously unlucky for Fermats, I agree.

2012-10-12, 03:47	#3
Xyzzy Aug 2002 8477₁₀ 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

2012-10-13, 23:37	#9
ixfd64 Bemusing Prompter "Danny" Dec 2002 California 2,467 Posts	I think it's more unlucky that we haven't found a new Mersenne prime since April 2009.