And proving the last two PRPs will also enable a proof that [url]http://factordb.com/index.php?id=1100000000213107698[/url], 2^67016+1-2^1496, is prime. Both large primes have 2^65520-1 as a factor of N-1.

Chris

[QUOTE=chris2be8;401924]Proving [url]http://factordb.com/index.php?id=1100000000777707215[/url] (1035 digits) and [url]http://factordb.com/index.php?id=1100000000777707223[/url] (687 digits) should enable a N+1 proof that [url]http://factordb.com/index.php?id=1100000000213107758[/url], 2^66878-2^1358-1, (20133 digits) is prime. I've checked both PRPs with bases 3, 5, 7, 11 and 13 so they should not turn out to be composites this time.

Chris[/QUOTE]

done, done, added to queue

Primitive part of (6^3453-1) = (6^3453-1)/(6^1151-1)/43 ([url]http://www.factordb.com/index.php?id=1100000000776538724[/url]) proved by N-1

Here's a nice chain, using algebraic factors of an algebraic factor

Pa (1780 digits):
[URL="http://www.factordb.com/index.php?id=1100000000779266164"](314829443463887129828729347511255641510518601^41-1)/314829443463887129828729347511255641510518600[/URL]

The largest algebraic factor of Pa-1 is Pb (712 digits):
[URL="http://www.factordb.com/index.php?id=1100000000779268025"](314829443463887129828729347511255641510518601^20+1)/(314829443463887129828729347511255641510518601^4+1)[/URL]

Entering algebraic factors of Pb-1 was sufficient to prove both numbers prime by N-1

100^189-((100^189+1)/101-1)*8/99-1 ([url]http://factordb.com/index.php?id=1100000000784871986[/url]) can be proved prime by proving either [url]http://factordb.com/index.php?id=1100000000784915360[/url] ((100^189-((100^189+1)/101-1)*8/99-2)/14943055223338988475827489718) or [url]http://factordb.com/index.php?id=1100000000784915365[/url] ((100^189-((100^189+1)/101-1)*8/99)/10178531516294521783668245600). Both are 350 digits.

The first enables a N-1 proof, the second a N+1 proof. It's the first time I've seen a number with both N-1 and N+1 having a large PRP.

Chris

Proving [url]http://factorization.ath.cx/index.php?id=1100000000789416221[/url] prime will enable a N-1 proof that [url]http://factorization.ath.cx/index.php?id=1100000000789118791[/url] ((314^499+1)/315) is prime.

Chris

Adding algebraic factors to (254^883-1)/253 [url]http://factorization.ath.cx/index.php?id=1100000000801406148[/url] I found that proving [url]http://factorization.ath.cx/index.php?id=1100000000801436073[/url] prime will enable a N-1 proof for (254^883-1)/253.

Chris

Proving [url]http://factorization.ath.cx/index.php?id=1100000000802860234[/url] prime should enable a N-1 proof for [url]http://factorization.ath.cx/index.php?id=1100000000802860221[/url], 1+439!-85!.

Chris

10^3612*2-3 ([url]http://www.factordb.com/index.php?id=1000000000024453636[/url]) proved using combined proof (small help from N+1)

I've spotted another pair where proving the smaller will enable a N+1 proof for the larger: [code]
1100000000805313831 (((10^332-1)/9+332)/15832838441-1)/4223758
1100000000805289544 ((10^332-1)/9+332)/15832838441
[/code]
Chris

Another pair, proving the second will enable a N+1 proof the first is prime: [code]
1100000000820318551 (13007^179-1)/13006
1100000000820316172 ((13007^179-1)/13006+1)/2
[/code]
Chris