Mersenne decay curve
 

Assume that the probability of 2^k-1 being prime is
2.57/k for all integers k>k0.
Let P(k) be the probability of there being no more Mersenne primes
before k. Given P(k0), derive an expression for P(k) for k>k0.

David

[quote=davieddy;136037]Let P(k) be the probability of there being no more Mersenne primes before k.[/quote]Do you mean that P(k) = probability that, for all k0 =< j < k, 2[sup]j[/sup] - 1 is composite? If not, what do "no more" and "before k" mean?

[quote]Given P(k0)[/quote] ... which would be 1 - (2.57/k0) [B]if[/B] your first statement were true for k = k0 -- right?

It is probably easiest to think of the concrete example:
Say k0 is 45M, so no testing has been done above it and
my probability assumption holds.
P(45M) is < 1 because not all exponents below 45M have been
tested. "No more primes" means "no more than the 44 found to date".
Big hint: P(j+1)=P(j)(1-2.57/j).

[QUOTE=davieddy;136037]Assume that the probability of 2^k-1 being prime is
2.57/k for all integers k>k0.
Let P(k) be the probability of there being no more Mersenne primes
before k. Given P(k0), derive an expression for P(k) for k>k0.

David[/QUOTE]

Obviously it's a totally false conjecture. If it would be good, then the probability that 2^(2*n)-1 is prime is c/n, where c>0 is a constant, but sum(n=1,infinity,c/n)=infinity and it means that there are infinte many such primes (by p=1 probability), however all of them are composite, expect for n=1.
Visit: [URL="http://primes.utm.edu/notes/faq/NextMersenne.html"]http://primes.utm.edu/notes/faq/NextMersenne.html[/URL]
for a probably valid conjecture.

It is a model which reflects the density of Mersenne primes,
just as 1/ln(n) gives the probability of n being prime.

It helps (me at least) to regard the exponent j as a continuous real variable.
The probability density is 2.57/j , meaning that the probability
of an exponent between j and (j+dj) yielding a Mersenne prime
is 2.57dj/j for small enough dj.

It's no more "outrageous" than applying the wave equation to
a solid even though we know it is made of atoms.

[QUOTE=davieddy;136088]It's no more "outrageous" than applying the wave equation to a solid even though we know it is made of atoms.[/QUOTE]Eh?

Atoms are well described by a sum of wave functions for their constituent particles (including virtual particles in the full QED treatment); an ensemble of atoms is well described by a sum of wave functions.

What's outrageous about it?


Paul

I was referring to the classical wave equation, and there is
nothing outrageous about it (to a physicist) as long as the
wavelength >> separation of atoms.
In this case the "half life" of the "decay curve" I seek is
of the order of k0 (i.e. millions >>1)

Enough pedantry: someone integrate P(j+dj)= P(j)*(1 - 2.57dj/j)
between j=k0 and j=k.

[QUOTE=davieddy;136097]I was referring to the classical wave equation, and there is
nothing outrageous about it (to a physicist) as long as the
wavelength >> separation of atoms.[/QUOTE]Aah. It appears that you're primarily a classical physicist whereas I tend to think of quantum mechanics first.

Paul

[quote=davieddy;136097]
Enough pedantry: someone integrate P(j+dj)= P(j)*(1 - 2.57dj/j)
between j=k0 and j=k.[/quote]
OK I'll do it myself:

dP/P=-2.57dj/j
integrating from j=k0 to k:
ln(P(k)/P(k0)) = -2.57ln(k/k0)

P(k)/P(k0) = (k0/k)^2.57

Note that this formula can also be arrived at by
observing that the expected number of Mersenne
primes between exponents k0 and k is 2.57ln(k/k0)
and using the Poisson distribution to give
Probability of no primes = 1/e^(expected number of primes).

Surely my decay curve illuminates discussion of "where is M45"
in the same way that "exponential decay" helps predict when
an unstable nucleus will decay?

More anon,
David

[quote=xilman;136101]Aah. It appears that you're primarily a classical physicist whereas I tend to think of quantum mechanics first.

Paul[/quote]
I made a bad mistake when I turned down the offer of
doing a DPhil in solid state physics under the supervision
of Rudolf Peierls. (I thought I'd discover the theory of Everything instead)
Re quantum mechanics, I found that the "explanation" of atomic
structure made me (almost) glad I did Chemistry instead of double
Maths at school.

David