Call me lazy but...

[Orgasmic Troll]

Originally Posted by davieddy

We are extracting more mileage out of this thread than I anticipated.
Does n=r?

More to come.

David

yes, that "n" is supposed to be an "r"

[davieddy]

Originally Posted by Orgasmic Troll

I think r = 1.21495... is what you're looking for.

This is assuming that the expected number of mersenne primes in the interval (x,rx) is log₂(2ⁿ)*1.78 (i.e. in the interval (x,4x), we expect to find 3.56 mersenne primes)

We seem to have frightened off the riff raff, judging by the
deafening response

First let me say that "ugly" would refer to your avatar (if anything)
When I say "shober", that is comparatively speaking.

I think you meant log₂r*1.78.

My old tutor Joe Hatton's first tutorial told me about
the "uniqueness theorem": if you find an answer that fits
by whatever means, then it iis the only one.

David
x

[davieddy]

Originally Posted by Orgasmic Troll

I think there are large assumptions (on top of the ones you've already made) that are necessary before this is solvable. I'm pretty sure that once those assumptions are made, the result will have nothing to do with Mersenne primes

The assumption I had made was that the expected primes
was a function of the ratio r of the exponents E(r).

From the additive definition of "expected primes",
we can quickly infer that E(r) + E(s) = E(rs) so
E is the logarithm function.

I stated E(2) = 1.78
So E(r) = 1.78 * log₂r

Now the r you quoted gives 0.5 expected prrmes
whereas I asked that the probability of no primes be 0.5.

The probabity of no primes is e^(-E(r))



David

[Orgasmic Troll]

Originally Posted by davieddy

The assumption I had made was that the expected primes
was a function of the ratio r of the exponents E(r).

From the additive definition of "expected primes",
we can quickly infer that E(r) + E(s) = E(rs) so
E is the logarithm function.

I stated E(2) = 1.78
So E(r) = 1.78 * log₂r

Now the r you quoted gives 0.5 expected prrmes
whereas I asked that the probability of no primes be 0.5.

The probabity of no primes is e^(-E(r))



David

Where do you get that the probability of no primes is e^(-E(r))? I don't think that knowing the expected values recovers the probability distribution. If I'm wrong, feel free to convince me.

I should have gone with my gut, instead, I convinced myself that looking for expected numbers of primes in the interval is what you were really looking for.

[wblipp]

Originally Posted by Orgasmic Troll

Where do you get that the probability of no primes is e^(-E(r))? I don't think that knowing the expected values recovers the probability distribution. If I'm wrong, feel free to convince me.

Poisson Distribution.

It's the result of assuming the probability of a prime in an infintesimal interval is proportional to the measure of the interval.

[lfm]

Originally Posted by davieddy

Sybil is wizard on "Mastermind" - speciality topic:
The Bleedin' Obvious.

Can you get that pigeon out of the water tank
please. The hotel inspectors will be round in a minute.
Manuel: "Que?"
This isn't a proposition from Wittgenstein

Are you still drunk?

[davieddy]

Originally Posted by lfm

Are you still drunk?

No
That post was a week ago.

[davieddy]

Originally Posted by Orgasmic Troll

Where do you get that the probability of no primes is e^(-E(r))? I don't think that knowing the expected values recovers the probability distribution. If I'm wrong, feel free to convince me.

Originally Posted by wblipp

Poisson Distribution.

It's the result of assuming the probability of a prime in an infintesimal interval is proportional to the measure of the interval.

I refrained from mentioning Poisson because I am
so accustomed to deriving "exponential decay" directly from the
assumption you mentioned (inferring it from E(t)= 1.78t
where t = log₂r is the "measure".)

The reason the Poisson distribution applies well to the
"expected new primes" as shown in the Classic Status page,
is that George sums the very small probabilities of a large
number of exponents (taking into account trial factoring).


David

[davieddy]

Originally Posted by davieddy

1)E(r) = 1.78 * log₂r

2)The probabity of no primes is e^(-E(r))

David

I wanted the probability of no primes to be 0.5 (as in "halflife")

From 2) we get E(r) = ln(2)
From 1) we get E(r) = 1.78*ln(r)/ln(2)

Giving r = 1.31

I was trying to give as sensible answer as possible to the
question "When will the next Mersenne prime be discovered?"

If the centre of the region of completed first time LLtests is 45M,
then I guess there is a 50% chance of none being found
before the "wavefront" has advanced by 0.31 * 45M.
If the the rate of advance is 4M per year, we get a halflife
of 3.5 years.