Maybe someone can do this better than I, but....

"Submitted for your approval...."

7235733 digits would be 2^24036582 or the last number could be 3 or 4.

[QUOTE=Uncwilly]Maybe someone can do this better than I, but....

"Submitted for your approval...."[/QUOTE]

I might do something in Photoshop if/when I feel like it :razz:

There are a lot of questions re to have a guessing game about the new prime or not. I submit that half the fun is anticipation and trying to guess the correct number. Based on information in the available files and some simple extrapolation, I had the correct number less than 4 hours after noticing that a new prime had been found. I did not post anything about it until there was no doubt of the actual number and most likely several others did the same. I suggest that a guessing game has more positives than negatives so long as actual confirmation is delayed until multiple validation runs have completed.

As for methods to prevent guessing the next Mp, you might consider a wild goose chase. A few carefully chosen but erroneous hints would send the hue and cry off into the wild blue yonder.

Fusion - congratulations to Jfindley* :flex:

[QUOTE=clowns789]7235733 digits would be 2^24036582 or the last number could be 3 or 4.[/QUOTE]

It would have to be 2^24036583-1 because it's the only one ending in an odd number.

The last digit of (2^24036583) - 1 is 7.

"It looks like Richard Brent can fire up his primitive polynomial machine again soon!". What does that mean?

[QUOTE=Fusion_power]I had the correct number less than 4 hours after noticing that a new prime had been found. I did not post anything about it until there was no doubt of the actual number and most likely several others did the same.[/QUOTE]

I suppose it was the C80125 string that gave it away to you, right?

[QUOTE=ewmayer]There were 2 problems in this instance:

1) The fake residue output by the server contained a highly nonrandom substring;

2) One of the DCers inadvertently included a percent-complete datum in a "latest update" post (easy mistake to make - I almost did similar myself at one point).

Both of these are highly preventable. The solution to (1) is obviously to do a better job randomizing the Res64. To prevent something like (2) from occurring in future, I suggest the DCers all send their day-to-day updates to George, and George posts a daily summary to the board. Moderators should be vigilant in looking for leaks and squashing them ASAP. Simply deleting the % figure from the post yesterday would have sufficed, though a handful of users would've probably still seen it before there was time to edit it.

I also suggested to George that no DC status updates aside from "completed and verified (or not)" be given after (say) 20M, but by that time the cat had already run out the door.

But none of this should detract from the joy of the new discovery - my own DC is still poking along at ~14M iterations right now, but I'd say it's not premature to say, "congratulations!"[/QUOTE]

Even without those two problems, I still would have been able to make a good guess at the exponent if I had been able to go through with my plan :grin:. I'll delay giving out what that plan is until next time.

[QUOTE=jinydu]The last digit of (2^24036583) - 1 is 7.

"It looks like Richard Brent can fire up his primitive polynomial machine again soon!". What does that mean?[/QUOTE]

I believe the poster meant primitive [i]trinomials[/i] - here is the URL for Richard Brent's webpage on that project:

[url]http://web.comlab.ox.ac.uk/oucl/work/richard.brent/trinom.html[/url]

From the introductory paragraph:

[i]"By a theorem of Swan, it is easy to rule out Mersenne exponents r if r = 3 or 5 (mod 8)."[/i]

The present case has r = 24036583 == 7 modulo 8, so admits of a primitive trinomial over GF(2) of form x^r + x^s + 1, 0 < s < r. Actually [i]finding[/i] the magic value of s for the middle term will require a lot of compute time - I'll ask Richard B. what his estimate is.

[QUOTE=jinydu]The last digit of (2^24036583) - 1 is 7.

"It looks like Richard Brent can fire up his primitive polynomial machine again soon!". What does that mean?[/QUOTE]

Well, the last digit is an easy thing to find. But what if I tell you the first many digits and the last many digits?

299410429404157172089048926340446938257367722975418473547677348600097640221100741026265865109912320858493344156415212635335...(sorry, over 6 million digits have been removed to simplify your reading because the total result has 7 984 781 digits)...5207718464378183256423142526858687039800556031269118412915067436921882733969407

It took some time to calculate it, my office is now full of papers...

[QUOTE=Lumo]299410429404157172089048926340446938257367722975418473547677348600097640221100741026265865109912320858493344156415212635335...(sorry, over 6 million digits have been removed to simplify your reading because the total result has 7 984 781 digits)...5207718464378183256423142526858687039800556031269118412915067436921882733969407[/QUOTE]

Look at that. THAT! 407 is the final three digits? Well, well, well. Those 47 conspirators are at it AGAIN! M40 end with ...047, and M41 ends in ...407! It's a CONSPIRACY! A CONSPIRACY I TELL YOU!

Sorry, I of course agree that the number of digits is 7235733. It was a typo.