December 10th, 2007, 09:45 PM  #1 
Senior Member Joined: Nov 2007 Posts: 258 Thanks: 0  "recurrent" mersenne primes
Is there a prime p for which M(p) = 2^p1 is prime, and also M(M(p)), and also M(M(M(p))), etc.? Imagine if there was such a prime for which M(p) was always prime... the days of GIMPS would be well over! Hehe.

December 11th, 2007, 05:04 AM  #2 
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0 
I severely doubt that. The odds of finding a simple formula that turns out nothing but primes doesn't seem to be a likely thing, at least to me. 
December 11th, 2007, 05:25 AM  #3 
Senior Member Joined: Nov 2007 Posts: 258 Thanks: 0 
I was mostly kidding about such a prime. However I'm still curious to see if there are any mersenne primes for which 2^m1 is also prime and so on? For instance 31 is a mersenne prime, as well as 2^311. 127 also. 
December 11th, 2007, 05:42 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
M(M(2)) is a 1digit prime. M(M(3)) is a 3digit prime. M(M(5)) is a 10digit prime. M(M(7)) is a 39digit prime. M(M(11)) is a 617digit composite. M(M(13)) is a 2,466digit composite. M(M(17)) is a 39,457digit composite. M(M(19)) is a 157,827digit composite. M(M(23)) is a 2,525,223digit composite. M(M(29)) is a 161,614,249digit composite. M(M(31)) is a 646,456,993digit composite. M(M(37)) is a 41,373,247,568digit composite. M(M(41)) is a 661,971,961,084digit composite. M(M(43)) is a 2,647,887,844,335digit composite. M(M(47)) is a 42,366,205,509,364digit composite. M(M(53)) is a 2,711,437,152,599,296digit composite. M(M(59)) is a 173,531,977,766,354,911digit composite. M(M(61)) has 694,127,911,065,419,642 digits. The deepest recursion: M(M(M(M(2)))) = 170141183460469231731687303715884105727 is prime. 
December 11th, 2007, 02:37 PM  #5 
Senior Member Joined: Nov 2007 Posts: 258 Thanks: 0 
Nice! you totally rule, Greathouse. I know it's not known whether there are infinitely many mersenne primes, but, if it was proven that there are, my next question would be : are there sequences of such recurring primes of any arbitrary length? 
December 11th, 2007, 09:31 PM  #6 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
I will ponder this question.

December 13th, 2007, 05:26 AM  #7  
Senior Member Joined: Dec 2006 Posts: 1,111 Thanks: 0  Quote:
I guess the big problem is that you would not know exactly where the recurring sequence of arbitrary length was, so you would still have to check all of the primes you got, and you would only know that the sequence existed after you had gone through the usual primality checking of all the possible candidates. Thus, I imagine that the knowledge of the existence of such sequences would not speed up the finding of Mersenne primes.  
December 13th, 2007, 09:28 AM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Good heuristics suggest a chance of about e^gamma (1.8 + lg p)/p that a prime p is a Mersenne prime. Since the sum of the reciprocal of the primes diverges, the expected number of Mersenne primes also diverges since the above expression is greater than 1/p. Iterated Mersenne primes are somewhat rarer than you might expect, since Mersenne primes are always 3 mod 4 and exponents which are 1 mod 4 are more likely to produce Mersenne primes. Taking these things into account, the chance that a Mersenne prime with index p is also the index for the double Mersenne prime 2^(2^p1)1 is roughly e^gamma (p lg (p+1))/2^p Ignoring (!) the condition that M(p) be prime, which condition is assumed in the above formula, the expected number of double Mersenne primes with index > 13 is about 0.0013. Since we know that there are no double Mersenne numbers with index between 13 and 59 inclusive, we can calculate the expected number of double Mersenne primes as 4 + sum(p = 61, infinity, formula above). But the sum over all primes from 61 on is just 2.86 e16, so there are most probably only a finite number of double Mersenne primes, with overwhelming chance of there being either 4 or 5. Of course this is not a proof. Actual results may vary. Talk to your healthcare professional to see if the Wagstaff heuristic is right for you. 
December 13th, 2007, 10:16 AM  #9 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Actually looking back at it I'm not sure about the 1 or 3 mod 4 stuff*, but either way the infinite sum is convergent with an expected value of what I posted (since it's a small term and I didn't post many digits). This does heuristically solve brunojoyal's question: there are expected to be a finite number of prime M(M(p)), and this would imply that there are not iterated Mersenne numbers of arbitrary depth. M(M(M(M(2)))) has four recursions, and it's unlikely anything else will match this. * If I got it wrong, replace the 1 in the second equation with lg 6. 
July 1st, 2008, 07:18 AM  #10 
Member Joined: Jul 2008 Posts: 38 Thanks: 0  Re: "recurrent" mersenne primes
M(M(2)) is a 1digit prime. M(M(3)) is a 3digit prime. M(M(5)) is a 10digit prime. M(M(7)) is a 39digit prime. M(M(11)) is a 617digit composite. M(M(13)) is a 2,466digit composite. M(M(17)) is a 39,457digit composite. M(M(19)) is a 157,827digit composite. M(M(31)) is a 646,456,993digit composite. M(M(61)) is a 694,127,911,065,419,642digit composite. M(M(89)) is a 186,328,542,329,173,367,306,815,834digit composite. M(M(107)) is a 48,844,909,400,338,823,199,277,929,902,126digit composite. M(M(127)) is a 51,217,599,719,369,681,875,006,054,625,051,616,350digit prime. 

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