mersenneforum.org  

Go Back   mersenneforum.org > Great Internet Mersenne Prime Search > Math

Reply
 
Thread Tools
Old 2008-10-06, 00:27   #1
AntonVrba
 
AntonVrba's Avatar
 
Jun 2005

2·72 Posts
Default New Class of primes - proving algorithm

Using my new proof for Wagstone numbers I hope to present a proof for a new class of primes.

\blue \text{Let }p \text{ be any odd prime, and }V_{qsp}=\frac{2^q+sp} {(p-2)}\text{ and } s=\pm1

\blue V_{qsp}= \text{ is prime }  \Longleftrightarrow \ S_{n} \equiv S_{r\times q } \pmod{V_{qsp}} \text{ , where: } S_0=c \text{ , and } \ S_{i}=S_i^2-2 \ \pmod{V_{q p}}  \ .

\blue \text{where } n=1 \text{ or }n \text{ is such that } 2^n \text{ divides either } (V_{qsp}+1) \text{ or } (V_{qsp}-1)\text{ and }r \text{ an integer not necessary 1}

\blue \text{( and }c\text{ is such that all }S_x\text { are unique )}

allow a day or two for me to adapt the paper presented in my post of thread

This method will only be of value if one can pick a S_0=c to complete the cycle within q and I am not sure yet if that can be done.

to convince you, you could try:
V_(12+5) = (2^12+5)/3=1367
5 -> 23 -> 527 -> 226 -> 495 -> 330 -> 905 -> 190 -> 556-> 192 -> 1320 -> 840 -> 226
or
19 -> 359 -> 381 -> 257 -> 431 -> 1214 -> 168 -> 882 -> 99 -> 230 -> 952-> 1348 -> 359

now the minus case and that q is prime is coincidental
V_(13-5)= (2^13-5)/3 = 2729
2434 -> 2424 -> 237 -> 1587 -> 2429 -> 2670 -> 750-> 324 -> 1272 -> 2414 -> 979 -> 560 -> 2492 -> 1587
or
2220 -> 2553 -> 955 -> 537 -> 1822 -> 1218 -> 1675 -> 211 -> 855 -> 2380 -> 1723 -> 2304 -> 509-> 2553

or you can try for yourself:

V_(13-23) = (2^13-23)/21
S_0 = 199, then S_2 = 15 = S_39
or
S_0 = 6, then S_1 = 34 = S_13

Last fiddled with by AntonVrba on 2008-10-06 at 00:58
AntonVrba is offline   Reply With Quote
Old 2008-10-06, 00:43   #2
retina
Undefined
 
retina's Avatar
 
"The unspeakable one"
Jun 2006
My evil lair

126428 Posts
Default

Quote:
Originally Posted by AntonVrba View Post
5 -> 23 -> 527 -> 226 -> 495 -> 330 -> 905 -> 190 -> 556-> 192 -> 1320 -> 840 -> 226
Gives a cycle length of 9.

But earlier you say:
Quote:
Originally Posted by AntonVrba View Post
\ S_{q} \equiv S_{r\times q } \pmod{P}
Which suggests a cycle length of 12.

Hey this tex thing is interesting, I didn't know we could do that.
retina is online now   Reply With Quote
Old 2008-10-06, 00:53   #3
AntonVrba
 
AntonVrba's Avatar
 
Jun 2005

2·72 Posts
Default

Quote:
Originally Posted by retina View Post
Gives a cycle length of 9.

But earlier you say:Which suggests a cycle length of 12.

Hey this tex thing is interesting, I didn't know we could do that.
12 includes 5 -> 23 -> 527 ->

Cycle bad terminology and my mistake S_q should have read S_n which has now been corrected in the post.

Last fiddled with by AntonVrba on 2008-10-06 at 00:55
AntonVrba is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Proving, for A > 1, (A+1)^n mod A = 0 MushNine Number Theory Discussion Group 9 2018-01-04 03:29
Mersenne primes and class field theory Nick Math 4 2017-04-01 16:26
Primality proving CRGreathouse Software 13 2011-01-30 14:30
fastest general number primality-proving algorithm? ixfd64 Math 3 2003-12-17 17:06
Countdown to proving M(6972593) is #38 down to less than 10! eepiccolo Lounge 10 2003-02-03 05:15

All times are UTC. The time now is 15:29.

Sat Jul 11 15:29:20 UTC 2020 up 108 days, 13:02, 1 user, load averages: 1.59, 1.49, 1.37

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.