## Factorization of 9 consecutive 641-digit numbers

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• 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 =
Message 1 of 178 , Nov 23, 2009
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.

PS: I expect this record to be short lived:
Joe Crump has a 804-digit target awaiting SNFS.
• ... Yes. Moreover the proof is rather easy. Let C be a Carmicahel number and let p be its largest prime factor. Then p-1|C-1. and hence p-1|(C/p)-1. Thus p
Message 178 of 178 , Feb 25, 2010
--- In primeform@yahoogroups.com,