[QUOTE=science_man_88;231311]ms = micro seconds ? or do you know a trick you haven't said lol.[/QUOTE]

ms -> milli seconds = 10^-3
[TEX]\mu[/TEX]s -> micro seconds = 10^-6

[QUOTE=kar_bon;231312]ms -> milli seconds = 10^-3
[TEX]\mu[/TEX]s -> micro seconds = 10^-6[/QUOTE]

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

[QUOTE=science_man_88;231313]I know I thought that but he said about micro seconds by pari reports ms.[/QUOTE]

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

[QUOTE=science_man_88;231313]I know I thought that but he said about micro seconds by pari reports ms.[/QUOTE]

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.

[QUOTE=rajula;231287]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 :smile:)[/QUOTE]

Precisely. The strawmen were invalid.

The range I used was k = 1 to 5000.

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?

[QUOTE=3.14159;231323]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?[/QUOTE]

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

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

because:

[COLOR="Red"]111111[/COLOR]1111


is what:

[COLOR="Red"]1111111111[/COLOR]
[COLOR="Red"]1111111111[/COLOR]
[COLOR="Red"]1111111111[/COLOR]
[COLOR="Red"]1111111111[/COLOR]
[COLOR="Red"]1111111111[/COLOR]
[COLOR="Red"]1111111111[/COLOR]
[COLOR="Red"]111111[/COLOR][COLOR="Blue"]1111[/COLOR]
[COLOR="Red"]111111[/COLOR][COLOR="Blue"]11[/COLOR][COLOR="Lime"]11[/COLOR]
[COLOR="Red"]111111[/COLOR][COLOR="lime"]1111[/COLOR]
[COLOR="Red"]111111[/COLOR]1111

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

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

[TEX]x^{y} % z = ((x % z)^{y}) % z[/TEX]

Yes, yes:

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