[QUOTE=DanielBamberger;475364]How long is the longest string of 1's? :innocent:[/QUOTE]If there is only one of something can it be considered the longest? Compared to what?

[size=1]So much impatience in this thread. Don't ya'all have something else more important to worry about?[/size]

So if you wanted to print the full decimal expansion on US letter sheets at 6 point size with ½" margins, it would take approximately 2500 pages.

[QUOTE=Mark Rose;475367]So if you wanted to print the full decimal expansion on US letter sheets at 6 point size with ½" margins, it would take approximately 2500 pages.[/QUOTE]
Sounds about right. [URL]https://www.youtube.com/watch?v=tlpYjrbujG0[/URL]

[QUOTE=DanielBamberger;475368]Sounds about right. [URL]https://www.youtube.com/watch?v=tlpYjrbujG0[/URL][/QUOTE]

I guess if I double-sided it, it would take only 1250 pages. I forgot you could do that with paper haha

[QUOTE=chalsall;475251]Is that anything like "What is the sound of one hand clapping?" :wink:[/QUOTE]
[url]https://www.youtube.com/watch?v=RUzbmIKVAHo[/url]

[QUOTE=DanielBamberger;475368]Sounds about right. [URL]https://www.youtube.com/watch?v=tlpYjrbujG0[/URL][/QUOTE]

Now they can make a sequel.

Yay! 2017 didn't get passed by without a Mersenne Prime being found.

It was a close one though! We need to get more compute!

CUDALucas run is finished at 4320K:

M( 7xxxxxxx )P, n = 4320K, CUDALucas v2.05.1

That "P" is the very subtle sign saying "Prime!".


Also another Prime95 run using 29.4 build 5 at 4096K. The original and Madpoo's run was both using Prime95 28.9:

M7xxxxxxx is prime! Wg4: F9C5D09F,00000000

The Penultimate LL step is a + this time, so we are at 25+ and 24-.
(There are only 49 as M1 = 2^2-1 = 3 does not have a penultimate step)..

Well, well, looks like we´ve done it again, didn´t we? :showoff::toot::beer2::beer2:
What a nice way of wrapping 2017 up!

It will be 16 years next month since I joined the project, and it´s good to see that our little community is still alive and kicking.
Congrats everyone! (and no, I will not disclose the good news before the official announcement...) :smile::smile:

[QUOTE=Mark Rose;475367]So if you wanted to print the full decimal expansion on US letter sheets at 6 point size with ½" margins, it would take approximately 2500 pages.[/QUOTE]
I calculated that this number have approx 23.1 million digits,so this number close to M77000000.

[QUOTE=ATH;475449]The Penultimate LL step is a + this time, so we are at 25+ and 24-.
(There are only 49 as M1 = 2^2-1 = 3 does not have a penultimate step)..[/QUOTE]

More importantly, is the new exponent "Team 1" or "Team 3" ?

[CODE]
       2, 2
       3, 3
       5, 1
       7, 3
      13, 1
      17, 1
      19, 3
      31, 3
      61, 1
      89, 1
     107, 3
     127, 3
     521, 1
     607, 3
    1279, 3
    2203, 3
    2281, 1
    3217, 1
    4253, 1
    4423, 3
    9689, 1
    9941, 1
   11213, 1
   19937, 1
   21701, 1
   23209, 1
   44497, 1
   86243, 3
  110503, 3
  132049, 1
  216091, 3
  756839, 3
  859433, 1
 1257787, 3
 1398269, 1
 2976221, 1
 3021377, 1
 6972593, 1
13466917, 1
20996011, 3
24036583, 3
25964951, 3
30402457, 1
32582657, 1
37156667, 3
42643801, 1
43112609, 1
57885161, 1
74207281, 1
[/CODE]

So far the counts are:

[CODE]
[B]mod 4[/B]
29 1
 1 2
19 3
[/CODE]

The [URL="https://primes.utm.edu/mersenne/heuristic.html"]Wagstaff conjecture[/URL] says that 1 is actually more common than 3, although this effect is more pronounced at lower exponent ranges.


On a related note, if it's 1 or 7 mod 8, then the search for [URL="http://maths-people.anu.edu.au/~brent/trinom.html"]primitive trinomials[/URL] can begin again.

The counts for that are:

[CODE]
       2, 2
       3, 3
       5, 5
       7, 7
      13, 5
      17, 1
      19, 3
      31, 7
      61, 5
      89, 1
     107, 3
     127, 7
     521, 1
     607, 7
    1279, 7
    2203, 3
    2281, 1
    3217, 1
    4253, 5
    4423, 7
    9689, 1
    9941, 5
   11213, 5
   19937, 1
   21701, 5
   23209, 1
   44497, 1
   86243, 3
  110503, 7
  132049, 1
  216091, 3
  756839, 7
  859433, 1
 1257787, 3
 1398269, 5
 2976221, 5
 3021377, 1
 6972593, 1
13466917, 5
20996011, 3
24036583, 7
25964951, 7
30402457, 1
32582657, 1
37156667, 3
42643801, 1
43112609, 1
57885161, 1
74207281, 1
[/CODE]

[CODE]
[B]mod 8[/B]
19 1
1 2
9 3
10 5
10 7
[/CODE]

[QUOTE=ATH;475449]...
The Penultimate LL step is a + this time, so we are at 25+ and 24-.
(There are only 49 as M1 = 2^2-1 = 3 does not have a penultimate step)..[/QUOTE]
I suppose you will get the s[SUB]0[/SUB] = 10 value as well, no doubt.
As for s[SUB]0[/SUB] = 2/3, I am curious to learn if the Penultimate LL value is proven to be a simple function of p ...or only conjectured? Anyone?