20100803, 21:56  #12 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×3^{2}×283 Posts 
Indeed, "simple" is what bing or babelfish would make of "prime". It is the same word "простое".
"Simplicity" should mean "primality". "Compound" means "composite". "Share" means "divide". (this all makes sense, kind of. Many lay terms are reused in mathematics, but the translation engines use the most straightforward commonusage words. Specialized translation services a.f.a.i.k are not free if at all available. Try to use synecdoche when you read such text, I think...) This not a case of homonyms, rather metonyms..? Computerassisted catachresis? (I am not a linguist, though :) Last fiddled with by Batalov on 20100803 at 22:08 Reason: "сimple" (with a cyrillic "s") ...switching registers is a PITA 
20100803, 22:04  #13  
Aug 2006
2^{2}·3·499 Posts 
Quote:
I just didn't want to think that I understood when I didn't  maybe I was right about compound but wrong about simple, for example... 

20100804, 00:22  #14 
Jun 2003
2^{2}·3·5·7·13 Posts 

20100804, 00:35  #15 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·3^{2}·283 Posts 
Yep, google translate is better than even "should have been good, because it is native" http://translate.ru

20100804, 04:38  #16 
Aug 2010
SPb
2·17 Posts 
All numbers are checked up
проверены все числа до 10^15 Last fiddled with by allasc on 20100804 at 04:39 
20100804, 04:41  #17 
Aug 2006
2^{2}·3·499 Posts 
All composites in this sequence are 2pseudoprimes, so save yourself some effort and use that structure in your testing.

20100804, 05:01  #18 
Aug 2006
13544_{8} Posts 
I edited the sequence.
%I A175625 %S A175625 7,11,23,31,47,59,83,107,167,179,227,263,347,359,383,467,479,503,563, %T A175625 587,683,719,839,863,887,983,1019,1123,1187,1283,1291,1307,1319,1367, %U A175625 1439,1487,1523,1619,1823,1907,2027,2039,2063,2099,2207,2447,2459,2543 %N A175625 Numbers n such that gcd(n, 6) = 1, 2^(n1) = 1 (mod n), and 2^(n3) = 1 (mod (n1)/2). %C A175625 All composites in this sequence are 2pseudoprimes, A001567. That subsequence begins 536870911, 46912496118443, ...; these were found by 'venco' from the dxdy.ru forums. %C A175625 Intended as a pseudoprimality test, but note that most primes are also missing as a result of the third test. %H A175625 Alzhekeyev Ascar M, <a href="b175625.txt">Table of n, a(n) for n = 2..11941610</a> %o A175625 (PARI) isA175625(n)=gcd(n,6)==1&Mod(2,n)^(n1)==1&Mod(2,n\2)^(n3)==1 %K A175625 nonn %O A175625 1,1 %A A175625 Alzhekeyev Ascar M (allasc(AT)mail.ru), Jul 28 2010, Jul 30 2010 
20100804, 05:02  #19  
Aug 2010
SPb
2×17 Posts 
Quote:
это только предположение It is necessary to check all or to prove the statement надо проверять все или доказать утверждение 

20100804, 05:09  #20 
Aug 2006
2^{2}×3×499 Posts 
Really? It's not obvious from post #6?
Proof: 1. By definition, the value of the sequence is 2*i+7 for some i such that ((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). 2. In particular, the value of the sequence is 2*i+7 only for values of i such that 4^(i+3) == 1 (mod (2*i+7)). 3. Let n = 2*i+7, so that i = (n7)/2. Then n is in the sequence only if 4^((n7)/2+3) == 1 (mod (2*(n7)/2+7)). 4. Thus n is in the sequence only if 4^((n1)/2) == 1 (mod n). 5. Thus n is in the sequence only if 2^(n1) == 1 (mod n). 6. By definition, such n are 2probable primes. 7. Any composite 2probable prime is, by definition, a 2pseudoprime. Last fiddled with by CRGreathouse on 20100804 at 05:10 
20100804, 05:43  #21  
Aug 2010
SPb
2·17 Posts 
Quote:
Yes certainly found numbers are 2pseudoprime. It is interesting to me to prove other statement. Whether there will be found (i+3) for Members a subsequence divide on expressions 2^x1 or 2^x+1 2*i+7=536870911 => i+3=268435455=2^281 2*i+7=46912496118443 => i+3=23456248059221=(2^231)*2796203=(2^461)/3 next????  извините. это все мой английский. да конечно наиденные числа являются 2pseudoprime. мне интересно доказать другое утверждение. будут ли найденные значения (i+3) для исключений делится на выражения 2^x1 или 2^x+1 Last fiddled with by allasc on 20100804 at 06:11 

20100804, 06:27  #22  
Aug 2006
2^{2}·3·499 Posts 
Quote:
Code:
isA175625(n)=gcd(n,6)==1&Mod(2,n)^(n1)==1&Mod(2,n\2)^(n3)==1 isA175625Mersenne(n)=Mod(2,n\2)^(n3)==1 nonMersenne(ff)=for(i=2,#A000043,forprime(p=A000043[i1]+1,A000043[i]1,ff(p))) A000043=[2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917]; nonMersenne(p>if(isA175625Mersenne(2^p1),print("2^"p"1"))) The code can be modified to check for multiples of Mersenne numbers instead, if desired. 

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