Thread for posting tiny primes - Page 63

kar_bon · 2010-09-24, 20:21

Quote:

Originally Posted by science_man_88

ms = micro seconds ? or do you know a trick you haven't said lol.

ms -> milli seconds = 10^-3
$\mu$ s -> micro seconds = 10^-6

science_man_88 · 2010-09-24, 20:30

Quote:

Originally Posted by kar_bon

ms -> milli seconds = 10^-3
$\mu$ s -> micro seconds = 10^-6

I know I thought that but he said about micro seconds by pari reports ms.

CRGreathouse · 2010-09-24, 20:37

Quote:

Originally Posted by science_man_88

I know I thought that but he said about micro seconds by pari reports ms.

Right. I ran 100,000 tests and divided to get the time per test. Each test took about 0.025 milliseconds.

kar_bon · 2010-09-24, 20:39

Quote:

Originally Posted by science_man_88

I know I thought that but he said about micro seconds by pari reports ms.

Sure you can't output or measure such timings for only one test. But do 10,000,000 tests and divide the total time by the number of tests!

CRG was faster.

3.14159 · 2010-09-24, 20:43

Quote:

Originally Posted by rajula

I have the impression that 3.14 is testing numbers of the form k*2^m+1 with some small range of k's. That would fit the earlier posts and his use of the expression "no factors below x digits". (But of course you noticed this and just want to point out the silly claim

)

Precisely. The strawmen were invalid.

The range I used was k = 1 to 5000.

3.14159 · 2010-09-24, 21:06

So far, the best idea is to manually use Proth's theorem to prove a number such as 3773512084578210152449 prime, with no computer assistance.

19^((3773512084578210152449-1)/2) modulo 3773512084578210152449, anyone?

science_man_88 · 2010-09-24, 21:27

Quote:

Originally Posted by 3.14159

So far, the best idea is to manually use Proth's theorem to prove a number such as 3773512084578210152449 prime, with no computer assistance.

19^((3773512084578210152449-1)/2) modulo 3773512084578210152449, anyone?

well if you look at it the exponent simplifies to 1886756042289105076224 if i did the math correct.

science_man_88 · 2010-09-24, 21:57

I would think the next attempt should use 19^ n = x mod(what number Pi wants to try modulo the whole thing by)

science_man_88 · 2010-09-24, 22:16

because:

1111111111

is what:

1111111111
1111111111
1111111111
1111111111
1111111111
1111111111
1111111111
1111111111
1111111111
1111111111

depends on: so really what's the number mod 19 that will help me a lot lol.

science_man_88 · 2010-09-24, 22:30

if you want to know what I'm talking of I believe it can be stated as:

$x^{y} % z = ((x % z)^{y}) % z$

3.14159 · 2010-09-24, 22:43

Yes, yes:

x^y mod z = (x mod z)^y. Next?

2010-09-24, 21:06	#688
3.14159 May 2010 Prime hunting commission. 2⁴·3·5·7 Posts	So far, the best idea is to manually use Proth's theorem to prove a number such as 3773512084578210152449 prime, with no computer assistance. 19^((3773512084578210152449-1)/2) modulo 3773512084578210152449, anyone? Last fiddled with by 3.14159 on 2010-09-24 at 21:11

2010-09-24, 22:30	#692
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶×131 Posts	if you want to know what I'm talking of I believe it can be stated as: $x^{y} % z = ((x % z)^{y}) % z$ Last fiddled with by science_man_88 on 2010-09-24 at 22:31

2010-09-24, 22:43	#693
3.14159 May 2010 Prime hunting commission. 2⁴·3·5·7 Posts	Yes, yes: x^y mod z = (x mod z)^y. Next? Last fiddled with by 3.14159 on 2010-09-24 at 22:56

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2010-09-24, 21:57	#690
science_man_88 "Forget I exist" Jul 2009 Dumbassville 2⁶·131 Posts	I would think the next attempt should use 19^ n = x mod(what number Pi wants to try modulo the whole thing by)

2010-09-24, 22:16	#691
science_man_88 "Forget I exist" Jul 2009 Dumbassville 20300₈ Posts	because: 1111111111 is what: 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 1111111111 depends on: so really what's the number mod 19 that will help me a lot lol.