20060220, 01:50  #1 
Feb 2006
BrasÃlia, Brazil
3×71 Posts 
Any other primes in this sequence?
Hi everyone! What I have here is just a simple and humble curiosity. I'm not a native English speaker nor a mathematician, not even a college level student at that. So please forgive me for any mistakes in spelling, grammar etc. and also in mathematical terms, concepts and symbols. Please do correct me too if I'm wrong, as I really want to learn.
Once again, I don't know if I'm using the correct language and math symbols here. But while I don't know exactly how to call these numbers, I'll call P_p the perfect number you get by multiplying the Mersenne prime M_p (with exponent p) by 2^(p1). Thus P_2 = 6, P_3 = 28, P_4=496 and so on. (the idea here is that the appropriate letters and numbers are subscript, sorry if they aren't) My question is: besides P_2 + 1 = 7, P_3 + 1 = 29, P_13 + 1 = 33550337 and P_19 + 1 = 137438691329, are there any other primes with the form P_p + 1? I mean, any other integers immediately following the (even) perfect numbers in the logical sequence of the integers (1, 2, 3, 4, 5...) which are prime? I've tried to check these numbers here (http://primes.utm.edu/curios/include...primetest.html) and here (http://www.alpertron.com.ar/ECM.HTM). Sorry I still don't know how to input links, but those are Chris Caldwell's Prime Pages and Dario Alpern's ECM factoring applet. I've tried to factor all the even perfect numbers up to P_11213; does anyone know if are there any primes for Mersennes larger that M23? It is also possible I've missed some prime there, I may have mistyped some formula. Another question, whose answer will most probably involve number and group theories I'm completely clueless about, but I'll ask it anyway: I think I've read somewhere in these forums that it's relatively simple to check if a number P is prime if we know all of the factors of P1 (I'm clueless because I have no idea why that is true). But since we know all factors in the numbers I've mentioned, is it simple to check them for primality? Could someone please explain to me if there's some way to prove there can't be any more primes in this form, or something like that? Sorry for the long and repetitive post, I just wanted to be sure I was minimally clear. 
20060220, 10:06  #2 
"Nancy"
Aug 2002
Alexandria
2464_{10} Posts 
This is, in fact, a good question and well posed. Promoted to "Math". Unfortunately, I don't know the answer.
Alex 
20060220, 11:18  #3 
Jul 2005
602_{8} Posts 
[EDIT]  I may have this wrong. I'm using the definition that P_n = ((2^n)1)*(2^(n1))
Hope that is what you meant, I'm slightly hungover today and my brain hasn't started working fully (despite it being 11am and I've been at work for 2 hours). Ok, here's how it goes after P_19: P_45+1 is prime, but 45 is not. P_46+1 is prime, but 46 is not. P_58+1 is prime, but 58 is not. P_141+1 is prime, but 141 is not. P_271+1 is prime AND 271 is prime! (163 digits) P_336+1 is prime, but 336 is not. At this point I've stopped doing primality testing on P_n and do Fermat Pseudoprimes for all prime bases below 100, not much should slip through that but you can't be 100% sure. Following link has more info: http://mathworld.wolfram.com/FermatPseudoprime.html P_562+1 is a strongpseudoprime, but 562 is not prime. P_601+1 is a strongpseudoprime AND 601 is prime! (362 digits) P_1128+1 is a strongpseudoprime, but 1128 is not prime. P_1635+1 is a strongpseudoprime, but 1635 is not prime. P_2718+1 is a strongpseudoprime, but 2718 is not prime. P_2920+1 is a strongpseudoprime, but 2920 is not prime. That's it until n=3000. Last fiddled with by Greenbank on 20060220 at 11:22 
20060220, 11:28  #4 
Jul 2005
2×193 Posts 
OK, I've reread it and I understand it now. You only care about P_n where n is the exponent of a known Mersenne prime.
I'll have another look at this... 
20060220, 12:44  #5 
Jul 2005
2·193 Posts 
Mersenne exponents from: http://mathworld.wolfram.com/MersennePrime.html
P_2 + 1 = 7 = prime P_3 + 1 = 29 = prime P_5 + 1 = 497 = 7 * 71 P_7 + 1 = 8129 = 11 * 739 P_13 + 1 = 33550337 = prime P_17 + 1 = 8589869057 = 7 * 11 * 111556741 P_19 + 1 = 137438691329 = prime P_31 + 1 = 2305843008139952129 = 29 * 71 * 137 * 1621 * 5042777503 P_61 + 1 = 2432582681 * 1092853292237112554142488617 P_89 + 1 = 7 * 132599200423201647070231067 * 206381273143696885332153493 P_107 + 1 = 7 * 11 * 11 * 67 * prp60. P_127 + 1 = 11 * 107 * 261697 * 70333627629913 * prp54. P_521 + 1 = 7 * 71 * 1050252439763 * c299 :run 77 ECM curves at 11000:1600000) :running 206 ECM curves at 50000:14000000 (25 digits) P_607 + 1 = 11 * prp365 P_1279 + 1 = ???? :run 77 ECM curves at 11000:1600000) :running 206 ECM curves at 50000:14000000 (25 digits) P_2203 + 1 = 60449 * 1498429 * ???? P_2281 + 1 = 197 * 557 * 1999 * 92033 * ???? P_3217 + 1 = 11 * ???? P_4253 + 1 = 7 * 53 * 8731 * 2353129 * ???? P_4423 + 1 = 2163571 * ???? P_9689 + 1 = 7 * 211 * ???? P_9941 + 1 = 7 * 67 * 1605697 * ???? P_11213 + 1 = 7 * ???? P_19937 + 1 = 7 * 11 * 1129 * 168457 * ???? P_21701 + 1 = 7 * ???? P_23209 + 1 = 35603 * 620377 * ???? P_44497 + 1 = 11 * 13259 * ???? P_86243 + 1 = 7 * 29 * 301123 * ???? P_110503 + 1 = 491 * 1493 * 1529761 * ???? P_132049 + 1 = ???? P_216091 + 1 = 4673 * 6920341 * ???? P_756839 + 1 = 7 * ???? P_859433 + 1 = 7 * ???? P_1257787 + 1 = 11 * ???? P_1398269 + 1 = 7 * 53 * 12713 * ???? P_2976221 + 1 = 7 * 71 * ???? P_3021377 + 1 = 7 * 11 * 49603 * ???? P_6972593 + 1 = 7 * 6007 * ???? P_13466917 + 1 = 11 * 45007 * ???? P_20996011 + 1 = ???? P_24036583 + 1 = 149 * ???? P_25964951 + 1 = 7 * ???? P_30402457 + 1 = 11 * 11 * ???? So, you can see that I have factors for all numbers except P_1279+1, P_132049+1 and P_20996011+1. Looking for these with ECM (15 digit, 20 digit, 25 digit) although P_20996011 is a big number (12640858 digits!) As for knowing the factors of p1 helping find p, no it doesn't. Two ways of thinking about it: 1) If the factorisation of p1 helped to factor p then we could simply find the factors of p1 by looking at the factors of p2. The factors of p2 could be found with the help of the factors of p3...etc...etc...This doesn't even touch on the fact that every time you have an even number you can remove all powers of 2 from it to get a smaller odd number. 2) Consider factoring p+1 where p is prime. How does one big prime factor (p1) help at all? You may be thinking of p1 factoring. The p1 bit refers to factors, not the number you are trying to factor. See: http://mathworld.wolfram.com/Pollard...ionMethod.html and: http://en.wikipedia.org/wiki/Pollard%27s_p1_algorithm 
20060220, 14:21  #6 
Jul 2005
182_{16} Posts 
I've done ECM on 1279 up to, and including, the 25 digit level. Going for the 30 digit level (401 curves at B1=25e4 B2=1.7e8).

20060220, 14:49  #7  
Nov 2005
2^{4}·3 Posts 
Quote:
Quote:
John 

20060220, 16:29  #8 
Jul 2005
386_{10} Posts 
Twice in one thread I manage to misread it. More coffee required. Apologies...

20060221, 16:52  #9  
Feb 2005
2^{2}×3^{2}×7 Posts 
Quote:


20060221, 19:47  #10 
"Nancy"
Aug 2002
Alexandria
100110100000_{2} Posts 
I think 1552147  P_20996011 + 1. Can you confirm?
Alex Last fiddled with by akruppa on 20060221 at 20:03 Reason: +1 missing 
20060221, 20:00  #11  
Feb 2005
2^{2}×3^{2}×7 Posts 
Quote:


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