2p+1

clowns789 · 2003-06-07, 14:14

Can they do that instead of 2p-1?

Axel Fox · 2003-06-07, 14:48

Could you please fully state your question, because I don't unerstand what you're trying to ask ?

2003-06-07, 17:59

Cunningham chains of the first and second kind:
2p+1 & 2p-1 respectively

Axel Fox · 2003-06-07, 18:35

Oh, sorry, I don't know anything about that.

cheesehead · 2003-06-08, 00:42

Quote:

Originally Posted by clowns789

Can they do that instead of 2p-1?

Are you referring to 2^p-1, meaning (2 to the power p) minus one, or 2*p-1, meaning (2 multiplied by p) minus one?

2^p -1 is a Mersenne number, but 2*p-1 is usually not.

Sometimes 2^p is written with the p as a superscript, but without a "^". When copied from that form into a text file that doesn't support superscripts, it comes out as "2p", looking like it means "2 times p" rather than "2 to the power p".

clowns789 · 2003-06-08, 02:30

It's 2 superscript p +1 so it would be 2p+1 just with the p superscript and there are no minus signs.

S80780 · 2003-06-08, 05:12

Exept of p=2, 2^p+1=5, all these numbers can be divided by 3 = 2 + 1.
This is just a special case of this formula:

(a + b) * Sum(0 <= k <= 2n)[(-1)^k * a^(2n-k) * b^k] = a^(2n + 1) + b^(2n + 1)

wblipp · 2003-06-08, 12:41

Quote:

Originally Posted by S80780

Exept of p=2, 2^p+1=5, all these numbers can be divided by 3 = 2 + 1.

2^4+1=17
2^6+1=65
2^8+1=257

Only the odd powers are divisible by 3.

Half the evens, those not divisible by 4, are divisible by 5.

S80780 · 2003-06-08, 17:39

True. But as the question was stated 2^p+1, I expect the p to be prime.
Nevertheless, testing 16^n+1 might be interesting.

Benjamin

cheesehead · 2003-06-08, 21:30

Folks,

Whenever someone asks about the primality of 2^n+1, remember the Fermat numbers F(i) = 2^(2^i)+1.

F(0) = 3 is prime, F(1) = 5 is prime, F(2) = 17 is prime, and so are F(3) = 257 and F(4) = 65537, but no others are known to be prime. And you probably know thet for years there have been searches for another Fermat prime, with no success so far.

So, you ask, why should we recall the Fermat primes when we're facing the more general case of 2^n+1, where n is not necessarily a power of 2?

Because it's been proven that 2^n+1 can never be prime unless n is a power of 2, that's why. Any 2^n+1 that is prime must have n eqial to some power of 2.

Links to a couple of proofs:

http://www.bearnol.pwp.blueyonder.co...th/fermatp.htm

http://www.matheprisma.uni-wuppertal...m/FermPri2.htm

clowns789 · 2003-06-09, 01:58

You can just test powers of two, right?

In the first 6 tries with 2^p+1 and 16^p+1 it's 3 composites and 3 primes, while with the usual 2^p-1 all of the first four tries there all primes (I start to the 1st power and go up)

So can you test S80780's and my theory?

Also when you multiply 16 by itself two or more times it seems to always end with 6, is that always true?

2003-06-07, 14:14	#1
clowns789 Jun 2003 The Computer 2³×7² Posts	2p+1 Can they do that instead of 2p-1?

2003-06-07, 14:48	#2
Axel Fox May 2003 2⁵·3 Posts	Could you please fully state your question, because I don't unerstand what you're trying to ask ?

2003-06-07, 17:59	#3
TTn 2²·3·5·7·13 Posts	Cunningham chains of the first and second kind: 2p+1 & 2p-1 respectively

2003-06-07, 18:35	#4
Axel Fox May 2003 2⁵×3 Posts	Oh, sorry, I don't know anything about that.

2003-06-08, 02:30	#6
clowns789 Jun 2003 The Computer 2³×7² Posts	It's 2 superscript p +1 so it would be 2p+1 just with the p superscript and there are no minus signs.

2003-06-08, 05:12	#7
S80780 Jan 2003 far from M40 5³ Posts	Exept of p=2, 2^p+1=5, all these numbers can be divided by 3 = 2 + 1. This is just a special case of this formula: (a + b) * Sum(0 <= k <= 2n)[(-1)^k * a^(2n-k) * b^k] = a^(2n + 1) + b^(2n + 1)

2003-06-08, 17:39	#9
S80780 Jan 2003 far from M40 125₁₀ Posts	True. But as the question was stated 2^p+1, I expect the p to be prime. Nevertheless, testing 16^n+1 might be interesting. Benjamin

2003-06-08, 21:30	#10
cheesehead "Richard B. Woods" Aug 2002 Wisconsin USA 2²·3·641 Posts	Folks, Whenever someone asks about the primality of 2^n+1, remember the Fermat numbers F(i) = 2^(2^i)+1. F(0) = 3 is prime, F(1) = 5 is prime, F(2) = 17 is prime, and so are F(3) = 257 and F(4) = 65537, but no others are known to be prime. And you probably know thet for years there have been searches for another Fermat prime, with no success so far. So, you ask, why should we recall the Fermat primes when we're facing the more general case of 2^n+1, where n is not necessarily a power of 2? Because it's been proven that 2^n+1 can never be prime unless n is a power of 2, that's why. Any 2^n+1 that is prime must have n eqial to some power of 2. Links to a couple of proofs: http://www.bearnol.pwp.blueyonder.co...th/fermatp.htm http://www.matheprisma.uni-wuppertal...m/FermPri2.htm

2003-06-09, 01:58	#11
clowns789 Jun 2003 The Computer 110001000₂ Posts	You can just test powers of two, right? In the first 6 tries with 2^p+1 and 16^p+1 it's 3 composites and 3 primes, while with the usual 2^p-1 all of the first four tries there all primes (I start to the 1st power and go up) So can you test S80780's and my theory? Also when you multiply 16 by itself two or more times it seems to always end with 6, is that always true?