New test

allasc · 2010-08-02, 13:43

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

allasc · 2010-08-03, 06:13

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

CRGreathouse · 2010-08-03, 17:34

I don't understand the sequence. Can you explain how you get 11, 23, 31, ...?

allasc · 2010-08-03, 19:31

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

CRGreathouse · 2010-08-03, 19:54

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.

CRGreathouse · 2010-08-03, 20:03

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.

allasc · 2010-08-03, 20:15

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

allasc · 2010-08-03, 20:25

Excuse for my English :)

CRGreathouse · 2010-08-03, 20:46

Quote:

Originally Posted by allasc

to 10^15 There are only two exceptions

I trust you're just checking lists of 2-pseudoprimes?

Batalov · 2010-08-03, 21:22

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.)

CRGreathouse · 2010-08-03, 21:30

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...

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)=2i+7; ((2i+7) mod 3)> 0; (a(i) does not share on 3); 4^(i+3) == 1 (mod (2i+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, 20:15	#7
allasc Aug 2010 SPb 2×17 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, 21:30	#11
CRGreathouse Aug 2006 2²·1,493 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

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2010-08-03, 17:34	#3
CRGreathouse Aug 2006 2²·1,493 Posts	I don't understand the sequence. Can you explain how you get 11, 23, 31, ...?

2010-08-03, 20:25	#8
allasc Aug 2010 SPb 2×17 Posts	Excuse for my English :)

2010-08-03, 21:22	#10
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·4,663 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.)