## Factorization of 4 consecutive 10673-digit numbers

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• Let n = 245363571*2^35426-3 n = 3*349*111602773267*953666301013*43440278284896679*prp10630 n+1 = 245363571*2^35426-2 = 2*(245363571*2^35425-1) (2*prime) n+2 =
Message 1 of 2 , Apr 27, 2010
Let n = 245363571*2^35426-3

n = 3*349*111602773267*953666301013*43440278284896679*prp10630
n+1 = 245363571*2^35426-2 = 2*(245363571*2^35425-1) (2*prime)
n+2 = 245363571*2^35426-1 (prime)
n+3 = 245363571*2^35426 = 2^35426*3^2*1117*24407

My recent search for a gigantic CC3 yielded about 260 gigantic Sophie Germain prime pairs. For each SG prime pair p, 2p+1, I attempted to factor 2p-1 and 2p+3, first by trial division, and then by ECM. Trial division and ECM up to B1=2000 yielded nothing but composite cofactors, but ECM at B1=11000 finally produced a 10630-digit PrP cofactor.

n-1 = 245363571*2^35426-4 = 2^2*5*720418081*209358848887*C10652
n+4 = 245363571*2^35426+1 = 5*7*17*67*137*C10666

n-1 and n+4 have been tested with 74 ECM curves at B1=11000.

Tom
• ... Congratulations! http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm is updated. -- Jens Kruse Andersen
Message 2 of 2 , Apr 28, 2010
Tom wrote:
> Let n = 245363571*2^35426-3
>
> n = 3*349*111602773267*953666301013*43440278284896679*prp10630
> n+1 = 245363571*2^35426-2 = 2*(245363571*2^35425-1) (2*prime)
> n+2 = 245363571*2^35426-1 (prime)
> n+3 = 245363571*2^35426 = 2^35426*3^2*1117*24407

Congratulations!
http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm
is updated.

--
Jens Kruse Andersen
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