mersenneforum.org > Math New test
 Register FAQ Search Today's Posts Mark Forums Read

 2010-08-02, 13:43 #1 allasc   Aug 2010 SPb 2·17 Posts New test Estimate please the test for simplicity of number http://oeis.org/classic/A175625 a(i)=2*i+7; ((2*i+7) mod 3)> 0; (a(i) does not share on 3); 4^(i+3) == 1 (mod (2*i+7)); 4^(i+2) == 1 (mod (i+3)) The first compound number a(i)=536870911 It is found by the user "venco" (from http://dxdy.ru) It is interesting to notice that i+3=268435455=2^28-1 The second compound number a(i)=46912496118443 It is found by the user "venco" It is interesting to notice that i+3=268435455=(2^46-1)/3 Exceptions are defined if to prove that they share on numbers (2^x-1) or (2^x+1) Excuse for my English
2010-08-03, 06:13   #2
allasc

Aug 2010
SPb

2·17 Posts

Quote:
 Originally Posted by allasc Estimate please the test for simplicity of number http://oeis.org/classic/A175625 a(i)=2*i+7; ((2*i+7) mod 3)> 0; (a(i) does not share on 3); 4^(i+3) == 1 (mod (2*i+7)); 4^(i+2) == 1 (mod (i+3)) The first compound number a(i)=536870911 It is found by the user "venco" It is interesting to notice that i+3=268435455=2^28-1 The second compound number a(i)=46912496118443 It is found by the user "venco" It is interesting to notice that i+3=268435455=(2^46-1)/3 Exceptions are defined if to prove that they share on numbers (2^x-1) or (2^x+1) Excuse for my English
sorry :)

i+3=23456248059221=(2^46-1)/3

Last fiddled with by wblipp on 2010-08-03 at 19:52

 2010-08-03, 17:34 #3 CRGreathouse     Aug 2006 5,987 Posts I don't understand the sequence. Can you explain how you get 11, 23, 31, ...?
2010-08-03, 19:31   #4
allasc

Aug 2010
SPb

2×17 Posts

Quote:
 Originally Posted by CRGreathouse I don't understand the sequence. Can you explain how you get 11, 23, 31, ...?
For any whole (i) condition performance
(2*i+7) mod 3)> 0; (a(i) does not share on 3); 4^(i+3) == 1 (mod (2*i+7)); 4^(i+2) == 1 (mod (i+3)

Then the number 2*k+7 will be simple

2010-08-03, 19:54   #5
CRGreathouse

Aug 2006

5,987 Posts

Quote:
 Originally Posted by allasc For any whole (i) condition performance (2*i+7) mod 3)> 0; (a(i) does not share on 3); 4^(i+3) == 1 (mod (2*i+7)); 4^(i+2) == 1 (mod (i+3) Then the number 2*k+7 will be simple
I take it from your .ru email address that by "simple" you mean prime and by "compound" you mean composite?

"(2*i+7) mod 3)> 0", I guess, means that 2i + 7 is not divisible by 3.

My guess is that your sequence is odd numbers a = 2i+7 such that
• 2i + 7 is not divisible by 3;
• $4^{i+3}\equiv1\pmod{2i+7}$
• $4^{i+2}\equiv1\pmod{i+3}$
I haven't checked this yet.

2010-08-03, 20:03   #6
CRGreathouse

Aug 2006

5,987 Posts

Quote:
 Originally Posted by CRGreathouse My guess is that your sequence is odd numbers a = 2i+7 such that 2i + 7 is not divisible by 3; $4^{i+3}\equiv1\pmod{2i+7}$ $4^{i+2}\equiv1\pmod{i+3}$ I haven't checked this yet.
It looks like the guess works out for the terms shown. I then take your statements to mean that an odd number a such that
• a is not divisible by 3;
• $2^{a-1}\equiv1\pmod{a}$
• $2^{a-3}\equiv1\pmod{(a-1)/2}$
is some kind of probable-prime.

 2010-08-03, 20:15 #7 allasc   Aug 2010 SPb 2216 Posts to 10^15 There are only two exceptions Both exceptions divisible by expression 2^x-1 or 2^x+1 (If it was possible to prove it (What is the exception has such dividers).... It would be good) The first compound number a(i)=536870911 It is found by the user "venco" (from http://dxdy.ru) It is interesting to notice that i+3=268435455=2^28-1 The second compound number a(i)=46912496118443 It is found by the user "venco" It is interesting to notice that i+3=23456248059221=(2^46-1)/3 PS Such coincidence happens to mine only in the Indian cinema Last fiddled with by allasc on 2010-08-03 at 20:23
 2010-08-03, 20:25 #8 allasc   Aug 2010 SPb 2×17 Posts Excuse for my English :)
2010-08-03, 20:46   #9
CRGreathouse

Aug 2006

176316 Posts

Quote:
 Originally Posted by allasc to 10^15 There are only two exceptions
I trust you're just checking lists of 2-pseudoprimes?

 2010-08-03, 21:22 #10 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 270216 Posts Аскар, опишите по-русски, и я или кто-нибудь другой переведет, ok? Есть много "ложных друзей переводчика", например "простое" число это вовсе не "simple", a "prime". (We will translate from Russian to English.)
 2010-08-03, 21:30 #11 CRGreathouse     Aug 2006 5,987 Posts Batalov, would you comment on the simple/prime and compound/composite issue? This came up on the seqfans list recently (with another native Russian speaker) and I was reminded of it here. I asked a Ukrainian colleague about it, but she doesn't know the math terms in Russian... Last fiddled with by CRGreathouse on 2010-08-03 at 21:30

 Similar Threads Thread Thread Starter Forum Replies Last Post Trilo Miscellaneous Math 25 2018-03-11 23:20 dh1 Information & Answers 8 2015-12-11 11:50 lidocorc Software 3 2008-12-03 15:12 swinster Software 2 2007-12-01 17:54 T.Rex Math 0 2004-10-26 21:37

All times are UTC. The time now is 19:37.

Mon Nov 28 19:37:42 UTC 2022 up 102 days, 17:06, 0 users, load averages: 1.57, 1.56, 1.44