Let

x = (1320*(5*10^23 + 7574922))^3

y = (x*(5*x + 9)/2 - 31)^2

N = (y - 3^4)*(y - 2^10)/55440

then

p628 = (N + 2)/(7621*10607*83911)

p641 = (N + 3)/2

p637 = (N + 5)/(2^2*17*281)

p631 = (N + 6)/(5*7*71*3027473)

p498 = (N + 7)/(2*3*773*1699*10103853977418233687

*41681046033582287059951

*36621959455077044549010329

*4211861143455373066167093525382847

*86528790939635630219352122503699213)

p615 = (N + 8)/(5960137*45290243150024180117)

are prime. Moreover N, N + 1 and N + 4 are completely

factorized, with no prime factor having more than 160 digits.

These 9 consecutive factorizations at 641 digits

have been mailed to Jens Kruse Andersen.

Notes: I used OpenPFGW, GMP-ECM, Msieve, ggnfs, Pari-GP.

All primality proofs were performed by Pari-GP.

The primality proof for p641 is easy, since

we know the factorization of p641 - 1 = (N + 1)/2.

The composite numbers

c610 = (N - 1)/(2*7*88309036301*7938470054221217579)

c578 = (N + 9)/(2^3*1138534864709243017

*2353150702179440759951

*28486552286765416493593)

are unlikely to have any prime divisor less than 10^30.

David Broadhurst, 23 November, 2009

PS: I expect this record to be short lived:

Joe Crump has a 804-digit target awaiting SNFS.