Wagstaff Conjecture - Page 14

CRGreathouse · 2010-12-31, 05:24

Quote:

Originally Posted by science_man_88

what says we can't do something like what we've been doing to get the line

I say that. Your project is flawed through lack of understandings of random processes, especially the Poisson and exponential distributions.

You're welcome to spend whatever time you like on this, but it will be a waste. That may not bother you.

science_man_88 · 2010-12-31, 12:28

Quote:

Originally Posted by CRGreathouse

I say that. Your project is flawed through lack of understandings of random processes, especially the Poisson and exponential distributions.

You're welcome to spend whatever time you like on this, but it will be a waste. That may not bother you.

So what articles should I read to get better understanding? How I read Wikipedia a Poisson distribution rely's on a Vec() filled with random data. and i've basically stated this but nobody has pointed me to a good enough article on it so that's that. Exponential distrubutions I'm guessing would rely on a base raised to a power multiplied by the first value making them non random.

science_man_88 · 2011-01-01, 19:26

Code:

(15:20)>solve(x=1,10,(x^6/2^5)-1)
%236 = 1.781797436280678609480452411181025015974425231756320806767513984503861606631524985275051534501114395

I had a dream once involving a power of 5 and power of 6 relationship and Mersenne primes this is devilishly close in my mind to the base I was using in my assessment.

CRGreathouse · 2011-01-02, 08:19

Quote:

Originally Posted by science_man_88

How I read Wikipedia a Poisson distribution rely's on a Vec() filled with random data.

Yes, I have no idea ( not even a little bit of one) as to what that means. It's as though you said "baking is a quickly sleep oven furiously".

CRGreathouse · 2011-01-02, 08:21

Quote:

Originally Posted by science_man_88

Code:

(15:20)>solve(x=1,10,(x^6/2^5)-1)
%236 = 1.781797436280678609480452411181025015974425231756320806767513984503861606631524985275051534501114395

I had a dream once involving a power of 5 and power of 6 relationship and Mersenne primes this is devilishly close in my mind to the base I was using in my assessment.

You understand that solve(x=1,10,(x^6/2^5)-1) gives a solution to x^6/32 - 1 = 0, that is, a sixth root of 32?

science_man_88 · 2011-01-02, 12:14

Quote:

Originally Posted by CRGreathouse

You understand that solve(x=1,10,(x^6/2^5)-1) gives a solution to x^6/32 - 1 = 0, that is, a sixth root of 32?

my point was I've talked of a x^6/y^5 type relationship and it comes close to the 1.78 I tried in the equation.

davar55 · 2011-01-10, 04:56

Getting back to the OP:

Quote:

Wagstaff Conjecture
OK
Not proved.
A few subtleties.
Some folk can't get their head round it.
But is there anything simpler going around to base our assumptions on?

How about this three-parter conjecture:

Let M_n = the nth Mersenne Prime exponent (MPE).

(1) The ratios R_n = M_n+1/M_n are bounded above.

(2) In particular, 1 < R_n < 10 for all integral n >= 1.

(2') Consequently, there is at least one MPE for each number of
decimal digits > 0.

(2'') This implies the Mersenne Prime sequence is infinite.

(2''') This implies the Even Perfect Number sequence is infinite.

(3) The YJ-Conjecture:
lim (n->infinity) R_n= M_n+1/M_n = 3/2 = 1.500.

Take this as:

define function yj(K,M,N) = K * M^N

then there exists a real K in 0.5 < K < 2.0 and an M in 1 < M < 2 s.t.
the values of R_n hover around yj(K,M,N), i.e.
(similarly to the prime distribution function hovering around li(x))
the values of R_n grow like yj(K,M,N) and continue to
exceed it and then be exceeded by it infinitely often (cyclicly, i.e.
repeatedly), at varying intervals which may be estimated based on
the "best" values for K and M and for no other such values

(3') base M = 3/2 = 1.500, with best coefficient K t.b.d.
(possibly K = 1.0 or 2/3 or 4/3 or 3/2 or 2.0).

(Note especially MPE23 = 11213 for my best guess for K).

(3'') There's more, it's conjecturalisimo, and controversial.

CRGreathouse · 2011-01-10, 06:15

My counter-conjectures:

Let q_n be the n-th Mersenne prime exponent, A000043.

(1) The ratios q_n+1/q_n can be arbitrarily close to 1 or $+\infty.$

(1') This implies that there are infinitely many Mersenne primes and hence even perfect numbers.

(2) In particular, for any N and ε > 0, there are m,n > N with q_m+1/q_m < 1 + ε and q_n+1/q_n < 1 / ε.

(2') Consequently, there are infinitely many numbers for which no Mersenne prime exponents with that number of decimal digits exists.

(3) $\lim_{n\to\infty}q_{n+1}/q_n$ does not exist.

(3')* $\sqrt[n]{q_n}=2^{e^{-\gamma}}+o(1)$

I reserve the right to modify or retract conjecture 3' if I miscalculated the expected value or error. I'm quite confident about the others, at least as confident as one can be about non-foundational conjectures in mathematics.

In short, I think that your conjectures (1), (2), (2'), (3), and (3') are wrong.

davar55 · 2011-01-10, 14:12

Quote:

Originally Posted by CRGreathouse

My counter-conjectures:

Let q_n be the n-th Mersenne prime exponent, A000043.

(1) The ratios q_n+1/q_n can be arbitrarily close to 1 or $+\infty.$

(1') This implies that there are infinitely many Mersenne primes and hence even perfect numbers.

(2) In particular, for any N and ε > 0, there are m,n > N with q_m+1/q_m < 1 + ε and q_n+1/q_n < 1 / ε.

(2') Consequently, there are infinitely many numbers for which no Mersenne prime exponents with that number of decimal digits exists.

(3) $\lim_{n\to\infty}q_{n+1}/q_n$ does not exist.

(3')* $\sqrt[n]{q_n}=2^{e^{-\gamma}}+o(1)$

I reserve the right to modify or retract conjecture 3' if I miscalculated the expected value or error. I'm quite confident about the others, at least as confident as one can be about non-foundational conjectures in mathematics.

In short, I think that your conjectures (1), (2), (2'), (3), and (3') are wrong.

While you have expressed this better than I could have, using Tex and all,
I and others understood that this is the current conjecture. Our key
difference is my claims based on my conjecture (1), that the ratio is in
fact bounded above, which denies the fundamental assumption of the
current conjecture, namely that the distribution of primes can be "modeled"
as a random (poisson or otherwise) process. They can not, except as an
approximation. The primes, just as the integers, are immutable, not random.

science_man_88 · 2011-01-10, 14:51

Quote:

Originally Posted by CRGreathouse

(3')* $\sqrt[n]{q_n}=2^{e^{-\gamma}}+o(1)$

if I knew what the smal o was for maybe I could calculate how close you were.

CRGreathouse · 2011-01-10, 19:45

Quote:

Originally Posted by science_man_88

if I knew what the smal o was for maybe I could calculate how close you were.

http://en.wikipedia.org/wiki/Big_O_n...tle-o_notation

2011-01-01, 19:26	#146
science_man_88 "Forget I exist" Jul 2009 Dumbassville 8384₁₀ Posts	Code: (15:20)>solve(x=1,10,(x^6/2^5)-1) %236 = 1.781797436280678609480452411181025015974425231756320806767513984503861606631524985275051534501114395 I had a dream once involving a power of 5 and power of 6 relationship and Mersenne primes this is devilishly close in my mind to the base I was using in my assessment.

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2011-01-10, 06:15	#151
CRGreathouse Aug 2006 3·1,993 Posts	My counter-conjectures: Let q_n be the n-th Mersenne prime exponent, A000043. (1) The ratios q_n+1/q_n can be arbitrarily close to 1 or $+\infty.$ (1') This implies that there are infinitely many Mersenne primes and hence even perfect numbers. (2) In particular, for any N and ε > 0, there are m,n > N with q_m+1/q_m < 1 + ε and q_n+1/q_n < 1 / ε. (2') Consequently, there are infinitely many numbers for which no Mersenne prime exponents with that number of decimal digits exists. (3) $\lim_{n\to\infty}q_{n+1}/q_n$ does not exist. (3')* $\sqrt[n]{q_n}=2^{e^{-\gamma}}+o(1)$ I reserve the right to modify or retract conjecture 3' if I miscalculated the expected value or error. I'm quite confident about the others, at least as confident as one can be about non-foundational conjectures in mathematics. In short, I think that your conjectures (1), (2), (2'), (3), and (3') are wrong.