Assorted formulas for exponents of Mersenne primes

[Uncwilly]

Originally Posted by Lee Yiyuan

The recursive function f(x) = 2^f(x-1) -1 where f(1) = 3
f(x) will always be a Mersenne prime for all positive integers x > 1

Originally Posted by ewmayer

Others have conjectured that this, the 5th (and smallest of unknown status) Catalan-Mersenne Number, a.k.a. MM127, is prime. It's reminiscent of Fermat's conjecture about primality of the number sequence that now bears his name.

AFAIK no one has ever produced any number-theoretic evidence that MM127 is more likely to be prime than one would expect for any other M-number with a prime exponent in that size range.

Originally Posted by CRGreathouse

What makes you think it's prime? It's a huge number, 170141183460469231731687303715884105727 bits long, and the chance that a random number that size is prime is tiny. This one is better than most -- it has no prime factors under 2^128 -- but if you pick a random number that size with no prime factors under 2^200 it has only about a 0.0000000000000000000000000000000002% chance to be prime. (I don't know exactly how far the number has been checked for small factors, but probably not that far.)

Lee Yiyuan, please respond to these posts. You are only conjecturing. You have no proof.

[Lee Yiyuan]

Originally Posted by Uncwilly

Lee Yiyuan, please respond to these posts. You are only conjecturing. You have no proof.

I didn't proof the prime, that's why i came looking for a test to proof it.

[Lee Yiyuan]

• What hardware did you use to run your test? - I am looking for a test.
• Did you write your own software? - I'm not a programmer
• Did you check your number by trial division up to at least 85 bits? - ???
• Did you run P-1 factoring or ECM on the number? (If so at what bounds?) - I do not know how to do this.
• How does your program perform the L-L test? (FFT's and such) - No program yet
• If your test program does not perform a Lucas test, why not? - the bignum library in java only allows input lesser than 2^31 -1.

[Lee Yiyuan]

In fact in the function f(x) = 2^f(x-1) -1, f(1) might be any Mersenne Prime.

f(2) where f(1) = 5
= 31
f(3) where f(1) = 5
= 2147483647

A Mersenne Prime might be produced if the exponent is a Mersenne Prime. But this is unproven.

[10metreh]

Originally Posted by Lee Yiyuan

An exponent that ends with a 7 or 1 makes it more probable for the calculated value of 2^170141183460469231731687303715884105727 - 1 to be prime. The chances of a randomly picked number lesser than 2^200 being a prime is 1/ln(2^200) = 0.007213475 ( 7 significant figures)

Firstly, there is no way to test this number for primality with current hardware and algorithms. It is just too big. We could, however, find a factor, and that would prove that it is composite.
Secondly, why do you seem so sure it is prime? The Catalan-Mersenne sequence is just a sequence - it's like saying that the first 5 Fermat numbers are prime, so the 6th should be as well. Read this.
If you think that the exponent ending with a 7 or 1 makes it more likely to be prime, whatever the reason, shouldn't that work for exponents 170141183460469231731687303715884105757, 170141183460469231731687303715884105851 etc. as well? Why is MM127 so special?

[Lee Yiyuan]

Originally Posted by 10metreh

Firstly, there is no way to test this number for primality with current hardware and algorithms. It is just too big. We could, however, find a factor, and that would prove that it is composite.
Secondly, why do you seem so sure it is prime? The Catalan-Mersenne sequence is just a sequence - it's like saying that the first 5 Fermat numbers are prime, so the 6th should be as well. Read this.
If you think that the exponent ending with a 7 or 1 makes it more likely to be prime, whatever the reason, shouldn't that work for exponents 170141183460469231731687303715884105757, 170141183460469231731687303715884105851 etc. as well? Why is MM127 so special?

Thank you all for your kind help. as a grade 10 student outside the US, i have yet to master the art of proving. I will stop here and maybe continue next time when i become more knowledgable in the field of mathematics.

Thank you all so much. Have a great day.

[akruppa]

Originally Posted by ewmayer

I notice that 2^127-1 is 39 digits in length.

... and we have a winner!

[Lee Yiyuan]

Originally Posted by akruppa

... and we have a winner!

What winner?

[mart_r]

Originally Posted by Lee Yiyuan

(P.S. This is the formula for Catalan-Mersenne Numbers, just realized it after reading ewmayer's post)



The recursive function f(x) = 2^f(x-1) -1 where f(1) = 3
f(x) will always be a Mersenne prime for all positive integers x > 1


E.G. f(3) = 2^f(2) - 1
= 2^(2^f(1) - 1) -1
= 2^(2^3 - 1) - 1
= 2^(7) - 1
= 128 - 1
= 127 (Prime)


f(1) = M3,
f(2) = M7,
f(3) = M127,
f(4) = M170141183460469231731687303715884105727,
f(5) = M(2^170141183460469231731687303715884105727 -1)

See, some years ago I found a formula that seemed to produce ever new primes:
f(n)=41+n*(n-1)
f(1)=41 is prime
f(2)=43 is prime
f(3)=47 is prime
f(4)=53 is prime
...
f(40)=1601 is prime

but suddenly...
f(41)=1681=41*41

so, what happened?

To prove your f(5) prime is not feasible today and will not be within at least the next 30 or 50 years. (Just a guess...)

[ewmayer]

Lastly, note that the fact that the first 4 terms of the sequence in question are prime is not as "special" as it might seem. Owing to the special form of factors of M(p) (every factor must be of form q = 2kp+1), the first 3 terms of the sequence *have* to be prime, since e.g. for f(3) = 2^7-1 = 127, the smallest possible factor of the required form is 2*7+1 = 15, which is larger than sqrt(f(3)), ergo f(3) is prime. Thus the first term of the sequence which is not prime-by-construction is f(4), and in essence claiming f(5) to be prime amounts to extrapolating from a single data point.

Lee Yiyuan, since based on your youth it's understandable why you thought you found something very special by way of this sequence, and it would be a shame for one swing-and-a-miss to diminish your interest for the subject, here is a little study assignment for you:

The Fermat numbers have a similar special-form-of-factors as the Mersennes. Use that to examine how "special" it is that the first 5 F_n (F₀ through F₄) are all prime.

[Uncwilly]

Originally Posted by mart_r

To prove your f(5) prime is not feasible today and will not be within at least the next 30 or 50 years. (Just a guess...)

Doing some rough calculations, here is what I came up with and put in Mersennewiki.org

It would take about 4,700,000,000,000,000,000,000,000,000,000,000 GHz-days to run the Lucas-Lehmer test on that number. Having the top 250 supercomputers working together at full power on this, it would take 161,600,000 times as long as the universe has existed.