[QUOTE=Unregistered;346275]Still, why 43rd number is an exception?at least to this point from all the Mersenne Numbers that we know of. Not trying to make people angry at me but just out of curiosity.[/QUOTE]

I'm guessing it come down to this the odds of a prime < 59 million has digit sum 25 is roughly 1 in 425 if you allow it to be any number it's roughly 1 in 37 there are 34 Mersenne prime exponents above the minimum exponent that has digital sum of 25 ( 799 by the way) so its not far off from a random occurrence at least if you look at this part of the math.

[QUOTE=science_man_88;346313]I'm guessing it come down to this the odds of a prime < 59 million has digit sum 25 is roughly 1 in 425 if you allow it to be any number it's roughly 1 in 37 there are 34 Mersenne prime exponents above the minimum exponent that has digital sum of 25 ( 799 by the way) so its not far off from a random occurrence at least if you look at this part of the math.[/QUOTE]

sorry I realized now that the mistake is I divided by 59 million so that's when over all the numbers. okay maybe I messed more up:

[CODE]? a=0;forstep(x=799,59000000,9,if(sum(y=1,#eval(Vec(Str(x))),eval(Vec(Str(x)))[y])==25,a=a+1));a
%1 = 1606628
? a=0;forstep(x=799,59000000,9,if(isprime(x) && sum(y=1,#eval(Vec(Str(x))),eval(Vec(Str(x)))[y])==25,a=a+1));a
%2 = 139060
? %1/59000000.
%3 = 0.02723098305084745762711864407
? %2/59000000.
%4 = 0.002356949152542372881355932203
? %1/59000000
%5 = 401657/14750000
? 1/%3
%6 = 36.72287548828975967056468579
? 1/%4
%7 = 424.2772903782539910829857615[/CODE]

is my math if anyone wants to check.

[QUOTE=cheesehead;346305]Left as the only fourteen odd possibilities are: 3, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37 and 41. They're all prime except 25 and 35.

So, there are only two non-even, non-prime numbers less than 42 that can be digit-sums of Mersenne prime exponents ... and, as you've seen, we already have one of them occurring among the 49 exponents known so far.

I expect that, probabilistically, it's not very strange that in 49 "tries" of the 34 possible digit-sums less than 42, we've (a) not seen the particular case of 35 yet, and (b) have seen the particular case of 25.[/QUOTE]

Well, as a first approximation let's say that "odd" or "even" have equal probability, and that an odd is 25 or 35 with probability 1/14 each. The expected number of 25s and 35s is 1/2 * 1/14 * 2 * 49 = 3.5. But this ignores the fact that most are too small to generate large sums like 25 and 35.

So let's go through the known Mersenne exponents and choose a number chosen randomly such that (1) both have the same number of digits (2) and the same first digit, and (3) the new number is not divisible by 2, 3, or 5. Then we'll see what the chances are that we get a 25 or a 35. Note that (2) essentially takes care of Benford and (3) takes care of digit sums which are multiples of 3.

We'll start with 13, since for smaller exponents the method does nothing. For 13 there are 4 possibilities: 11, 13, 17, and 19. So this contributes 0 to 25 and 0 to 35. The first number with a positive contribution is 1279 which generates 25 in 6 of the 268 cases (2.2%).

There are 44 exponents to consider. In total, the expected number of 25s in this model is 1.96 and the expected number of 35s is 1.37. A Poisson distribution would say that would be only slightly surprising (alpha = 0.14) to see no 25s but not particularly surprising (alpha = 0.25) to see no 35s.

Cracked it!
 

[QUOTE=CRGreathouse;346325]Well, as a first approximation let's say that "odd" or "even" have equal probability, and that an odd is 25 or 35 with probability 1/14 each. The expected number of 25s and 35s is 1/2 * 1/14 * 2 * 49 = 3.5. But this ignores the fact that most are too small to generate large sums like 25 and 35.

So let's go through the known Mersenne exponents and choose a number chosen randomly such that (1) both have the same number of digits (2) and the same first digit, and (3) the new number is not divisible by 2, 3, or 5. Then we'll see what the chances are that we get a 25 or a 35. Note that (2) essentially takes care of Benford and (3) takes care of digit sums which are multiples of 3.

We'll start with 13, since for smaller exponents the method does nothing. For 13 there are 4 possibilities: 11, 13, 17, and 19. So this contributes 0 to 25 and 0 to 35. The first number with a positive contribution is 1279 which generates 25 in 6 of the 268 cases (2.2%).

There are 44 exponents to consider. In total, the expected number of 25s in this model is 1.96 and the expected number of 35s is 1.37. A Poisson distribution would say that would be only slightly surprising (alpha = 0.14) to see no 25s but not particularly surprising (alpha = 0.25) to see no 35s.[/QUOTE]
Sounds like we are on the same track.

Take a number N, 100 or larger say, not divisible by 3.
If we expect 6 Mprimes with exponents between x and 10x (as we do),
show that we expect 2 Mprimes to have N as the sum of the digits of their exponents.

David

[QUOTE=xilman;346299]But give him high marks for effort. Hardly anyone, myself included, would have taken that much trouble to try to educate someone who is presently clueless but may yet be cured of that condition.[/QUOTE]I'll give him an extramark for spelling his name correctly.
In the midst of his verbiage, I might of missed some inference that he was talking about odd sums of digits.
But I never was that enthusiastic about marking that kind of answer, and even less so now.

:davieddy:

[QUOTE=BudgieJane;346292]25 is not prime, put it is a prime power. Can you make your theory fit that?[/QUOTE]

With enough data you can make any assumptions, and with large amount of data I'm pretty sure I can predict the number quite faster than going one by one as you guys do. Going as far as the probability go, you still don't give a realistic approach when 35(sum of Mersenne prime) will appear, as you only give probabilities rather than a realistic view. As one said as we take smaller (decrease in size) of Mersenne prime numbers we MUST expect the sum to be even or prime. However you don't extrapolate your analysis and tell what WILL HAPPEN as we INCREASE Mersenne numbers, not what happens as we decrease.

To the all " experts" with all the knowledge that you possess ( LL test and etc) it still takes quite a long time to discover new Mersenne primes?

PS. Please Moderator and Admin to close this thread. Thank you.

With Regards,

M.

[QUOTE=Unregistered;346389]With enough data you can make any assumptions, and with large amount of data I'm pretty sure I can predict the number quite faster than going one by one as you guys do.[/QUOTE]

If you believe this, you have entirely missed the points that have been made. You may not realize that the quality of the response you have received is substantially higher than most internet sites - several of the responses you have received have been from real mathematicians - people who really make their living doing mathematics and have publications in the field to back it up. While I'm not a believer in "credentials must make it right," I'm also not a believer in dismissing real experts just because I don't like their opinions.

Here is the message they are trying to get across to you - What you have found has no predictive power. It is an unsurprising result for any set of comparably sized numbers, so it says nothing about the likelihood of Mersenne Primes. Here is a simple exercise to test this out. I haven't yet run this experiment - so it's an honest chance for you to prove me wrong.

Take all the Mersenne Exponents, and find the Next Prime after them. You can easily do this using[URL="http://www.alpertron.com.ar/ECM.HTM"] Dario Alpern's Java Applet[/URL] with N(x). The resulting set of numbers should be a lot like the Mersenne exponents - similar size and prime. Now apply your analysis to THIS set of numbers - I'm betting you will find the same "amazing" result. You can easily try this a few more times, stepping a bit further - N(N(N(x))) is the third prime beyond the Mersenne Exponent.

I propose that you actually do this three times and report back. The experts here strongly expect you to find the same results for all three sets, and therefor that your discovery has no predictive value. If you find something different, then you have found a reason for the experts to reconsider you discovery.

[QUOTE=Unregistered;346389]With enough data you can make any assumptions, and with large amount of data I'm pretty sure I can predict the number quite faster than going one by one as you guys do.[/QUOTE]So? I can predict the winner of the next horse race at Kentucky Downs.

Merely predicting some outcome means nothing unless it has a rational basis and has been demonstrated to be correct. You haven't presented the latter two. We have.

[quote]Going as far as the probability go, you still don't give a realistic approach when 35(sum of Mersenne prime) will appear, as you only give probabilities rather than a realistic view.[/quote]That just demonstrates your ignorance. Probabilities _are_ realistic.

[quote]As one said as we take smaller (decrease in size) of Mersenne prime numbers we MUST expect the sum to be even or prime.[/quote]... or the product of primes greater than three -- referring to the digit sums of the exponents, that is.

[quote]However you don't extrapolate your analysis and tell what WILL HAPPEN as we INCREASE Mersenne numbers,[/quote]Yes, we did! You weren't paying attention!

[U]As the exponents increase their digit sums will continue to be even or prime or the product of primes greater than three.[/U]

Once you actually understand what we've explained, you'll see why the "even or prime or the product of primes greater than three" rule will continue to be true for larger and larger exponents' digit sums.

- - -

There's a more compact way to say "even or prime or the product of primes greater than three". Once you understand, you'll figure that out, too.

[QUOTE]:
Originally Posted by Unregistered
With enough data you can make any assumptions, and with large amount of data I'm pretty sure I can predict the number quite faster than going one by one as you guys do.
[/QUOTE]

Where did you get your math degree? How many papers have you published?
Where do you get the arrogance to say that you can perform mathematics that noone
else (including math PhD's and people with 30 years of studying this subject)
can?


[QUOTE]
Going as far as the probability go, you still don't give a realistic approach when 35(sum of Mersenne prime) will appear, as you only give probabilities rather than a realistic view. As one said as we take smaller (decrease in size) of Mersenne prime numbers we MUST expect the sum to be even or prime. However you don't extrapolate your analysis and tell what WILL HAPPEN as we INCREASE Mersenne numbers, not what happens as we decrease.
[/QUOTE]

Firstly, there does not exist a proof that arbitarily large Mersenne primes exist.
There are strong heuristics that suggest they do, but no proof.

Secondly, it is trivial to see what happens as the Mersenne primes get arbitrarily large.
The probability that their digit sum is prime goes to 0 as M_p --> oo.
Showing how fast the probability goes to 0 as a function of p is a simple
exercize for any first year student in probability and statistics.

This is *trivial*


[QUOTE]
To the all " experts" with all the knowledge that you possess ( LL test and etc) it still takes quite a long time to discover new Mersenne primes?
[/QUOTE]

Now, you are just being an asshole; deprecating those who have much more knowledge
and intelligence and experience than you do.

<plonk>

[QUOTE=R.D. Silverman;346479]Where did you get your math degree? How many papers have you published?
Where do you get the arrogance to say that you can perform mathematics that noone
else (including math PhD's and people with 30 years of studying this subject)
can?




Firstly, there does not exist a proof that arbitarily large Mersenne primes exist.
There are strong heuristics that suggest they do, but no proof.

Secondly, it is trivial to see what happens as the Mersenne primes get arbitrarily large.
The probability that their digit sum is prime goes to 0 as M_p --> oo.
Showing how fast the probability goes to 0 as a function of p is a simple
exercize for any first year student in probability and statistics.

This is *trivial*




Now, you are just being an asshole; deprecating those who have much more knowledge
and intelligence and experience than you do.

<plonk>[/QUOTE]

Let me add: the OP is a classic instance of Dunning & Kruger in action.

WoW, you think it take a long time to discover new Mersenne prime?
Really? we found 13 of them in less than 20 years, and 35 since the beginning of the computer era.
I think that's can be called quick. Sure it's long since you have to wait for the next one. but, remember, compared to the history of mathematic, 60 years is nothing.
Or , if you can prove *us* wrong, please do. Nothing will make mathematician more joyous than disproved long-term-believed true theory.