20210224, 14:07  #1 
Aug 2020
2^{4}·7 Posts 
MersenneWoodall Primes
I saw that is a Mersenne prime and also the Woodall prime .
A Mersenne number is a Woodall number if the exponent p is of the form . So obviously we need p to be prime for the MersenneWoodall number to be prime. Just for fun I looked up which n < 101 result in a prime exponent: Code:
n p 1 3 3 11 5 37 9 521 15 32783 39 549755813927 75 37778931862957161709643 81 2417851639229258349412433 89 618970019642690137449562201 
20210224, 18:07  #2 
Nov 2016
2,819 Posts 
This is doubleexponent sequence, thus it is conjectured that the number of such primes is finite, and 2^5211 may be the largest such prime, like 65537 may be the largest Fermat prime, and 2^1271 may be the largest doubleMersenne prime.

20210224, 18:38  #3 
Aug 2020
2^{4}×7 Posts 
I didn't know about this conjecture. That is also why generalized Fermat numbers are supposed to only have a finite number of primes per base?
Does that conjecture have a name and is there a simple explanation? I could only find some papers about it, but with my very limited math knowledge I didn't grasp much. And it probably doesn't say anything about the expected number of primes, or does it? 
20210224, 18:41  #4 
"Daniel Jackson"
May 2011
14285714285714285714
2^{3}×3^{4} Posts 
M549755813927 is divisible by 43399809724955419039.
M37778931862957161709643 has no factors <2^110. M2417851639229258349412433 is divisible by 14507109835375550096474599. M618970019642690137449562201 is divisible by 302234395011250596454696928877487. Just helping out so that no one claims them to be prime (there is still that possibility for M37778931862957161709643, but it's extremely unlikely). Last fiddled with by Stargate38 on 20210224 at 18:47 Reason: forgot "." 
20210224, 18:52  #5  
Nov 2016
2,819 Posts 
Quote:
The conjecture in post https://mersenneforum.org/showpost.p...&postcount=675 may be true, but may be false when c = 1, b = q^m, k = q^r, and the exponent of highest power of 2 dividing r >= the exponent of highest power of 2 dividing m (where (k*b^n+c)/gcd(k+c, b1) corresponding to GFN's to even bases or half GFN's to odd bases, for the reference of half GFN's to odd bases, see http://www.fermatquotient.com/PrimSerien/GenFermOdd.txt) Last fiddled with by sweety439 on 20210224 at 19:17 

20210224, 18:53  #6  
Nov 2016
5403_{8} Posts 
Quote:
Last fiddled with by sweety439 on 20210224 at 19:17 

20210226, 08:55  #7 
Aug 2020
2^{4}·7 Posts 
Thanks for the links, sweety439. To be honest, understanding that Boklan and Conway paper is beyond me. So the conjecture says the number of primes is not only finite, but that there are only a few, i.e. < 10 or so?
Stargate38, I'm impressed and disappointed. ;) Is that the result of TF? But how can you get so quickly to 2^110, or is TF that fast on a single candidate? So, for n < 1300: Code:
n p 317 266998379490113760299377713271194014325338065294581596243380200977777465722580068752870260867389 701 10520271803096747014481979765760257331100679605646347718996561806137464308594161644227333072555176902453965937712356435426038864500367607726255629541303761699910447342256889196383327515768645434542586503471563453 735 180736893357325919804742965901096183254486650358500961579737723575212405143116703993975930943694719806137463391890175780999890999416217020648099397663164550811570949854893831716648452639533025774320471006645409943407035103 Last fiddled with by bur on 20210226 at 09:00 
20210226, 19:49  #8  
Nov 2016
2,819 Posts 
Quote:
Since such numbers are too large to test the primality (even the probableprimality, such as a Miller–Rabin primality test and Baillie–PSW primality test, in fact, since the N1 or N+1 for these numbers are immediately smooth, there are no pseudoprimes), thus we cannot know whether these numbers are primes or not, unless a divisor of the number can be found (we can only use trial division to disprove the primality of these numbers). Note: Since these numbers are of the form Phi(n,2)/gcd(Phi(n,2),n) (where Phi is the cyclotomic polynomial, and gcd is greatest common divisor), and all numbers of the form Phi(n,2)/gcd(Phi(n,2),n) are strongprobableprimes to base 2, so don't test with that base. In generally, all numbers of the form Phi(n,b)/gcd(Phi(n,b),n) (b>=2) are strongprobableprimes to base b, so don't test with that base. Last fiddled with by sweety439 on 20210226 at 20:01 

20210226, 23:54  #9 
"Alexander"
Nov 2008
The Alamo City
1001000100_{2} Posts 
Remember that all factors of Mersenne numbers are of the form 2kp+1, so larger exponents have fewer possible factors to check for a given bit range. A backoftheenvelope calculation says it's actually significantly less than an equivalent GIMPS TF amount of work.

20210227, 01:23  #10  
Nov 2016
2,819 Posts 
Quote:


20210227, 02:08  #11  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{3}×5^{2}×47 Posts 
Quote:
Only prime n are of interest for this thread, and for them gcd() is "1", and Phi() is 2^n1. So you typed 159 characters, that added 0 substance to what Happy already said. ___ * and in general, every statement that start with a guarding qualifier "generally,..." is good for social science arguments and is simply nonsense for math. When you look at a statement, "Generally, <blah>" with a trained eye, you parse out the word "generally" and then skip the rest as irrelevant until next dot. 

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