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Old 2010-08-03, 21:56   #12
Batalov
 
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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 common-usage 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..? Computer-assisted catachresis? (I am not a linguist, though :-)

Last fiddled with by Batalov on 2010-08-03 at 22:08 Reason: "сimple" (with a cyrillic "s") ...switching registers is a PITA
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Old 2010-08-03, 22:04   #13
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Quote:
Originally Posted by Batalov View Post
this all makes sense, kind of. Many lay terms are reused in mathematics, but the translation engines use the most straightforward common-usage words.
Excellent, thanks!

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...

Quote:
Originally Posted by Batalov View Post
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".
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Old 2010-08-04, 00:22   #14
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Quote:
Originally Posted by Batalov View Post
Indeed, "simple" is what bing or babelfish would make of "prime". It is the same word "простое".
"Simplicity" should mean "primality".
"Compound" means "composite".
google translate handles "prime number" and "composite number" correctly.
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Old 2010-08-04, 00:35   #15
Batalov
 
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Yep, google translate is better than even "should have been good, because it is native" http://translate.ru
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Old 2010-08-04, 04:38   #16
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Quote:
Originally Posted by CRGreathouse View Post
I trust you're just checking lists of 2-pseudoprimes?
All numbers are checked up
проверены все числа до 10^15

Last fiddled with by allasc on 2010-08-04 at 04:39
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Old 2010-08-04, 04:41   #17
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All composites in this sequence are 2-pseudoprimes, so save yourself some effort and use that structure in your testing.
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Old 2010-08-04, 05:01   #18
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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^(n-1) = 1 (mod n), and 2^(n-3) = 1 (mod (n-1)/2).
%C A175625 All composites in this sequence are 2-pseudoprimes, 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)^(n-1)==1&Mod(2,n\2)^(n-3)==1
%K A175625 nonn
%O A175625 1,1
%A A175625 Alzhekeyev Ascar M (allasc(AT)mail.ru), Jul 28 2010, Jul 30 2010
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Old 2010-08-04, 05:02   #19
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Quote:
Originally Posted by CRGreathouse View Post
All composites in this sequence are 2-pseudoprimes, so save yourself some effort and use that structure in your testing.
It only the assumption
это только предположение

It is necessary to check all or to prove the statement
надо проверять все или доказать утверждение
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Old 2010-08-04, 05:09   #20
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Quote:
Originally Posted by allasc View Post
It is necessary to check all or to prove the statement
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 = (n-7)/2. Then n is in the sequence only if 4^((n-7)/2+3) == 1 (mod (2*(n-7)/2+7)).
4. Thus n is in the sequence only if 4^((n-1)/2) == 1 (mod n).
5. Thus n is in the sequence only if 2^(n-1) == 1 (mod n).
6. By definition, such n are 2-probable primes.
7. Any composite 2-probable prime is, by definition, a 2-pseudoprime.

Last fiddled with by CRGreathouse on 2010-08-04 at 05:10
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Old 2010-08-04, 05:43   #21
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Quote:
Originally Posted by CRGreathouse View Post
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 = (n-7)/2. Then n is in the sequence only if 4^((n-7)/2+3) == 1 (mod (2*(n-7)/2+7)).
4. Thus n is in the sequence only if 4^((n-1)/2) == 1 (mod n).
5. Thus n is in the sequence only if 2^(n-1) == 1 (mod n).
6. By definition, such n are 2-probable primes.
7. Any composite 2-probable prime is, by definition, a 2-pseudoprime.
Excuse. It is all my English.
Yes certainly found numbers are 2-pseudoprime.
It is interesting to me to prove other statement.
Whether there will be found (i+3) for Members a subsequence divide on expressions 2^x-1 or 2^x+1

2*i+7=536870911 => i+3=268435455=2^28-1
2*i+7=46912496118443 => i+3=23456248059221=(2^23-1)*2796203=(2^46-1)/3
next????
----

извините. это все мой английский.
да конечно наиденные числа являются 2-pseudoprime.
мне интересно доказать другое утверждение.
будут ли найденные значения (i+3) для исключений делится на выражения 2^x-1 или 2^x+1

Last fiddled with by allasc on 2010-08-04 at 06:11
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Old 2010-08-04, 06:27   #22
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Quote:
Originally Posted by allasc View Post
It is interesting to me to prove other statement.
Whether there will be found (i+3) for Members a subsequence divide on expressions 2^x-1 or 2^x+1
Here's some quick Pari code that shows that there are no other terms (beside 2^29-1) of your sequence in A054723, at least as far as you have patience to check.
Code:
isA175625(n)=gcd(n,6)==1&Mod(2,n)^(n-1)==1&Mod(2,n\2)^(n-3)==1
isA175625Mersenne(n)=Mod(2,n\2)^(n-3)==1
nonMersenne(ff)=for(i=2,#A000043,forprime(p=A000043[i-1]+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^p-1),print("2^"p"-1")))
This code assumes that numbers of the form 2^p-1 (p > 2 prime) are 2-pseudoprimes that are not divisible by 3; if you like, you can change the word "isA175625Mersenne" to "isA175625" from the last line to remove that assumption if you like.

The code can be modified to check for multiples of Mersenne numbers instead, if desired.
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