Predict M49... - Page 5

davar55 · 2013-02-17, 03:31

Upgrading my prediction from the earlier thread,
I predicted a prime exponent near 52,100,000,
one near 77,000,000, and then a large gap to
199,000,000 (slightly adjusted) and another large
gap to 290,000,000. These could be M48, M49,
M50, and M51. Basis; catching up with the true
limiting ratio of consecutive Mersenne prime
exponents, and expectation of a P95 breakthrough.

As for when:
M49 @ 3/5/2015
M50 @ 2017/09/02
M51 @ 2020/201

philmoore · 2013-02-25, 05:52

Quote:

Originally Posted by davieddy

BTW Caldwell seems to think (log2 + log6)/2 is negligible compared to log(exponent). I make it ~7%

I make it ~6% at the level of the most recent discovery, i.e., an exponent congruent to 1 mod 4 is about 6% more likely to yield a prime than one congruent to 3 mod 4, so at this level, we would expect the exponent of any newly discovered prime to have ~51.4% chance of being 1 mod 4 and ~48.6% chance of being 3 mod 4. On the other hand, the 14 discoveries by GIMPS have included 10 primes with exponents congruent to 1 mod 4 and only 4 primes with exponents congruent to 3 mod 4. Luck of the draw? Certainly possible, chances are better than 7% of that particular mix. I'll look up Wagstaff's paper again tomorrow, as R. Gerbicz seemed to have issues with it, and Chris Caldwell's says that his account is a simplification of the original argument:
http://primes.utm.edu/mersenne/heuristic.html

davieddy · 2013-02-25, 19:32

Quote:

Originally Posted by philmoore

I make it ~6% at the level of the most recent discovery, i.e., an exponent congruent to 1 mod 4 is about 6% more likely to yield a prime than one congruent to 3 mod 4, so at this level, we would expect the exponent of any newly discovered prime to have ~51.4% chance of being 1 mod 4 and ~48.6% chance of being 3 mod 4. On the other hand, the 14 discoveries by GIMPS have included 10 primes with exponents congruent to 1 mod 4 and only 4 primes with exponents congruent to 3 mod 4. Luck of the draw? Certainly possible, chances are better than 7% of that particular mix. I'll look up Wagstaff's paper again tomorrow, as R. Gerbicz seemed to have issues with it, and Chris Caldwell's says that his account is a simplification of the original argument:
http://primes.utm.edu/mersenne/heuristic.html

I love you Phil.
But 6% of 1.47 is ~ 0.09

Knowing you like your music:
7 and 7 is

RIP Authur Lee

D

davieddy · 2013-02-26, 01:34

We conventionally say (round here) that the expected number of Mprimes with expos between e1 and e2 is log(e2/e1)/log(1.47).

If we are underestimating the probability by 6%, then the expected number should be 1.06*log(e2/e1)/log(1.47)
~ log (e2/e1)/log(1.44).

Jeez it's tricky getting math right while listening to ]Country Joe and the Fish

davieddy · 2013-02-27, 21:51

From PNT, the probability of an integer x being a prime = 1 mod 4 is
1/(2 lnx).
If this is the case, Caldwell says that Wagstaff says the probability of 2^x-1 being prime is
e^gamma(lnx + ln6)/(x ln2).
Combining these, the probability is A(1 + ln6/lnx)/x
where A = e^gamma/(2 ln2) = 1.285

Integrating from k₁to k₂ gives
A(ln(k₂k₁ + ln6 ln(lnk₂/lnk₁))

For exponents = 1 mod 4 between 600K and 60M (GIMPS era),we expect
A(ln100 + ln6*ln 1.346) = A(4.605 + 0.532)

For exponents = 3 mod 4 between 600K and 60M (GIMPS era),we expect
A(ln100 + ln2*ln 1.346) = A(4.605 + 0.206)

NOTE THAT THE SECOND TERM IS DISTINCTLY NON-NEGLIBLE!

David

philmoore · 2013-02-28, 06:33

It has occurred to me, David, that we may both be barking up the wrong tree. The classical "probability" that a number m is prime is exp(gamma)*ln(factoring limit)/ln(m). For a Mersenne number m = 2ⁿ-1, the a priori factoring limit is 2n or 6n, depending on whether n is 3 mod 4 or 1 mod 4, but by the time this number is trial factored to, say, the 73-bit level, isn't this relative difference erased? Sure, we would expect that a higher proportion of mod 1 exponents than mod 3 exponents would survive to this level, but once we get to this factoring level, we would expect that the remaining probability that the number is prime to be the same in both cases, right? At any rate, we should be able to test the survivability of mod 1 exponents versus mod 3 exponents at any given factoring level. (My guess is that there are about 6% more mod 1 exponents than mod 3 exponents to a given factoring level. Any data relevant to this prediction would be most gratefully appreciated!) My point is that it would not confer any advantage to test only mod 1 exponents because the increased a priori probability that they are prime would be exactly outweighed by the increased number of test candidates.

davieddy · 2013-02-28, 10:06

I agree entirely (except you've just suggested a different tree up which to bark!)

What we are expecting is that in a large batch of candidates for LL testing (past present or future), 51.4% will have a remainder 1 when divided by 4. You only have to look at the last two bits or the last two decimal digits.

In 1,000,000 candidates, we expect 514,000 "hits".
The standard deviation would be 500.

Any takers?

science_man_88 · 2013-02-28, 13:58

Quote:

Originally Posted by davieddy

I agree entirely (except you've just suggested a different tree up which to bark!)

What we are expecting is that in a large batch of candidates for LL testing (past present or future), 51.4% will have a remainder 1 when divided by 4. You only have to look at the last two bits or the last two decimal digits.

In 1,000,000 candidates, we expect 514,000 "hits".
The standard deviation would be 500.

Any takers?

wouldn't it more depend on the fact that all Mersenne numbers>7 are 6n+1 for some n, so really the odds that a Mersenne number is prime kind of depend on the odds that 6n+1 is prime no ?

davieddy · 2013-02-28, 16:55

Quote:

Originally Posted by science_man_88

wouldn't it more depend on the fact that all Mersenne numbers>7 are 6n+1 for some n, so really the odds that a Mersenne number is prime kind of depend on the odds that 6n+1 is prime no ?

For that matter, 2⁽²ⁿ⁺¹⁾ - 1 = 2*4ⁿ - 1
but we are talking about the prime exponents, half of which are 4n + 1
and the rest 4n + 3.
2m(4n+1) + 1 = -/+ 1 mod 8 when m = 3 or 4 mod 4
2m(4n+3) + 1 = -/+ 1 mod 8 when m = 1 or 4 mod 4

Smaller candidate factors are more likely to divide the Mersenne than larger ones, so Mersennes with (4n+3) expos are slightly more likely to get factored and less likely to be prime than those with (4n+1) expos.

Since the sample of known Mprimes is far too small to infer anything from,
we want to look at how many of each type get factored.

D

davieddy · 2013-03-03, 21:36

As Phil pointed out, the distinction between 4n+1 and 4n+3 exponents lies in the number of small factors we expect to find.

Time LMH put their efforts to some use:

Factors of 2^p-1 are of the form 2mp+1

For p<1 Billion, could you tell me the number of factors with m<32,
broken down according to m and p mod 4 please?

David

davieddy · 2013-03-14, 10:43

2013-01-18 expected new primes <79.3M is 0.948
Expected time to next prime = 71/0.042 days = 1690 days.

2013-03-14 expected new primes <79.3M is 0.912
Expected time to next prime = 55/0.036 days = 1528 days.

2013-02-26, 01:34	#48
davieddy "Lucan" Dec 2006 England 6474₁₀ Posts	Correction We conventionally say (round here) that the expected number of Mprimes with expos between e1 and e2 is log(e2/e1)/log(1.47). If we are underestimating the probability by 6%, then the expected number should be 1.06*log(e2/e1)/log(1.47) ~ log (e2/e1)/log(1.44). Jeez it's tricky getting math right while listening to ]Country Joe and the Fish

2013-02-27, 21:51	#49
davieddy "Lucan" Dec 2006 England 2×3×13×83 Posts	Expected Mprimes with exponents < 60M From PNT, the probability of an integer x being a prime = 1 mod 4 is 1/(2 lnx). If this is the case, Caldwell says that Wagstaff says the probability of 2^x-1 being prime is e^gamma(lnx + ln6)/(x ln2). Combining these, the probability is A(1 + ln6/lnx)/x where A = e^gamma/(2 ln2) = 1.285 Integrating from k₁to k₂ gives A(ln(k₂k₁ + ln6 ln(lnk₂/lnk₁)) For exponents = 1 mod 4 between 600K and 60M (GIMPS era),we expect A(ln100 + ln6ln 1.346) = A(4.605 + 0.532) For exponents = 3 mod 4 between 600K and 60M (GIMPS era),we expect A(ln100 + ln2ln 1.346) = A(4.605 + 0.206) NOTE THAT THE SECOND TERM IS DISTINCTLY NON-NEGLIBLE! David

2013-03-03, 21:36	#54
davieddy "Lucan" Dec 2006 England 194A₁₆ Posts	Plea for data As Phil pointed out, the distinction between 4n+1 and 4n+3 exponents lies in the number of small factors we expect to find. Time LMH put their efforts to some use: Factors of 2^p-1 are of the form 2mp+1 For p<1 Billion, could you tell me the number of factors with m<32, broken down according to m and p mod 4 please? David

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2013-02-17, 03:31	#45
davar55 May 2004 New York City 108A₁₆ Posts	Upgrading my prediction from the earlier thread, I predicted a prime exponent near 52,100,000, one near 77,000,000, and then a large gap to 199,000,000 (slightly adjusted) and another large gap to 290,000,000. These could be M48, M49, M50, and M51. Basis; catching up with the true limiting ratio of consecutive Mersenne prime exponents, and expectation of a P95 breakthrough. As for when: M49 @ 3/5/2015 M50 @ 2017/09/02 M51 @ 2020/201

2013-02-28, 06:33	#50
philmoore "Phil" Sep 2002 Tracktown, U.S.A. 45F₁₆ Posts	It has occurred to me, David, that we may both be barking up the wrong tree. The classical "probability" that a number m is prime is exp(gamma)ln(factoring limit)/ln(m). For a Mersenne number m = 2ⁿ-1, the a priori* factoring limit is 2n or 6n, depending on whether n is 3 mod 4 or 1 mod 4, but by the time this number is trial factored to, say, the 73-bit level, isn't this relative difference erased? Sure, we would expect that a higher proportion of mod 1 exponents than mod 3 exponents would survive to this level, but once we get to this factoring level, we would expect that the remaining probability that the number is prime to be the same in both cases, right? At any rate, we should be able to test the survivability of mod 1 exponents versus mod 3 exponents at any given factoring level. (My guess is that there are about 6% more mod 1 exponents than mod 3 exponents to a given factoring level. Any data relevant to this prediction would be most gratefully appreciated!) My point is that it would not confer any advantage to test only mod 1 exponents because the increased a priori probability that they are prime would be exactly outweighed by the increased number of test candidates.

2013-02-28, 10:06	#51
davieddy "Lucan" Dec 2006 England 2×3×13×83 Posts	I agree entirely (except you've just suggested a different tree up which to bark!) What we are expecting is that in a large batch of candidates for LL testing (past present or future), 51.4% will have a remainder 1 when divided by 4. You only have to look at the last two bits or the last two decimal digits. In 1,000,000 candidates, we expect 514,000 "hits". The standard deviation would be 500. Any takers?

2013-03-14, 10:43	#55
davieddy "Lucan" Dec 2006 England 2×3×13×83 Posts	2013-01-18 expected new primes <79.3M is 0.948 Expected time to next prime = 71/0.042 days = 1690 days. 2013-03-14 expected new primes <79.3M is 0.912 Expected time to next prime = 55/0.036 days = 1528 days.