Thanks to whoever did the last one, I've just found another case:

Proving [url]http://factorization.ath.cx/index.php?id=1100000000748602274[/url] (((10^274+10^137+1)*(2^452-3)/3+10^411*1972-1)/53366409470) will enable a N-1 proof that [url]http://factorization.ath.cx/index.php?id=1100000000748438030[/url] ((10^274+10^137+1)*(2^452-3)/3+10^411*1972) is prime.

Chris

After adding algebraic factors to [url]http://factorization.ath.cx/index.php?id=1100000000504241133[/url] ((6607^853-1)/6606) I found that proving [url]http://factorization.ath.cx/index.php?id=1100000000755048276[/url] (a 1067 digit PRP) will probably be enough to prove it prime.

Chris

Thanks for proving that.

Here's another:
Proving [url]http://factorization.ath.cx/index.php?id=1100000000755346596[/url] is prime should enable a N-1 proof for [url]http://factorization.ath.cx/index.php?id=1100000000439186935[/url] ((4122^937-1)/4121).

Chris

And another:

Proving [url]http://factorization.ath.cx/index.php?id=1100000000755505371[/url] should be enough to prove [url]http://factorization.ath.cx/index.php?id=1100000000512390486[/url] ((3581^967-1)/3580) is prime.

Chris

And ...

Proving [url]http://factorization.ath.cx/index.php?id=1100000000756068010[/url] will enable a N-1 proof for [url]http://factorization.ath.cx/index.php?id=1100000000439186901[/url] ((4854^1009-1)/4853).

Chris

And proving [url]http://factorization.ath.cx/index.php?id=1100000000759078678[/url] should be enough to prove [url]http://factorization.ath.cx/index.php?id=1100000000439186888[/url] ((4735^1153-1)/4734) is prime. The former is 452 digits and the latter is 4234 digits so it's a bigger ration than usual.

Chris

Another:

Proving [url]http://factorization.ath.cx/index.php?id=1100000000759995166[/url] is probably enough to prove [url]http://factorization.ath.cx/index.php?id=1100000000696698412[/url] ((116^2089+1)/117) is prime.

Chris

And another:

Proving [url]http://factorization.ath.cx/index.php?id=1100000000759997087[/url] will enable a N-1 proof that [url]http://factorization.ath.cx/index.php?id=1100000000296002061[/url] (22^3217+21) is prime (at least this one is certainly large enough for the proof).

Chris

[QUOTE=chris2be8;396571]Another:

Proving [url]http://factorization.ath.cx/index.php?id=1100000000759995166[/url] is probably enough to prove [url]http://factorization.ath.cx/index.php?id=1100000000696698412[/url] ((116^2089+1)/117) is prime.

Chris[/QUOTE]

It wasn't. However, adding some more missing algebraic factors did the trick.

[QUOTE=chris2be8;396571]Another:

Proving [URL]http://factorization.ath.cx/index.php?id=1100000000759995166[/URL] is probably enough to prove [URL]http://factorization.ath.cx/index.php?id=1100000000696698412[/URL] ((116^2089+1)/117) is prime.

Chris[/QUOTE]

That wasn't quite enough. So I started work on the smallest factor of N-1, (116^116+1)/5549418630282309787479951401753137, and found: [code]
********** Factor found in step 2: 1563721650332372051297386247663898037313
Found probable prime factor of 40 digits: 1563721650332372051297386247663898037313
Probable prime cofactor 34571626573845918564783429269016198081709009231700089277835355381228515794200673865236406976328168287033463361692761363974121183975856856744607152825112929989944807057 has 167 digits
[/code] Which was enough to prove (116^2089+1)/117 is prime.

Chris

Edit. I didn't notice AXN's post before I made this one. But I added the factors above to factordb a couple of hours ago but factordb took a while to run the N-1 test.

@AXN, what algebraic factors did you add? I thought I'd added all possible ones before I asked for the smal PRP to be proved.

Edit 2: Is anyone working on the PRP in post 204?

[QUOTE=chris2be8;396649]@AXN, what algebraic factors did you add? I thought I'd added all possible ones before I asked for the smal PRP to be proved.[/QUOTE]

Hmmm... I was just working thru the algebraic factors from bottom up and supplying factors and all of a sudden one of the top level composites got factored. So I took credit for it. Must've been the factor you submitted. :redface: