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## Factorization of 4 consecutive 5638-digit numbers

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• Let n = 25390425*2^18703-1 n = 7*15569*150611*1338793*P5622 n+1 = 25390425*2^18703 = 2^18703*3*5^2*43*7873 n+2 = 25390425*2^18703+1 (prime) n+3 =
Message 1 of 2 , Jan 23, 2010
Let n = 25390425*2^18703-1

n = 7*15569*150611*1338793*P5622
n+1 = 25390425*2^18703 = 2^18703*3*5^2*43*7873
n+2 = 25390425*2^18703+1 (prime)
n+3 = 25390425*2^18703+2 = 2*(25390425*2^18702+1) (2*prime)

n-1: 25390425*2^18703-2 = 2*4657349734574297*C5622
n+4: 25390425*2^18703+3 = 3^2*13*47*10528525069*3236067162807210923*C5606

The 5622-digit cofactor of n has been proven prime by Primo, with a certificate at http://xenon.stanford.edu/~tjw/pp/p5622.zip

The two primes 25390425*2^18702+1 and 25390425*2^18703+1 are a CC2 (2nd kind) byproduct of a recent CC3 search.

n-1 and n+4 have been tested with 221 ECM curves at B1=50000.

Tom
• ... Congratulations on improving your month old 5257-digit record in http://tech.groups.yahoo.com/group/primeform/message/9948 That also used a CC2 from a CC3
Message 2 of 2 , Jan 25, 2010
Tom wrote:
> Let n = 25390425*2^18703-1
>
> n = 7*15569*150611*1338793*P5622
> n+1 = 25390425*2^18703 = 2^18703*3*5^2*43*7873
> n+2 = 25390425*2^18703+1 (prime)
> n+3 = 25390425*2^18703+2 = 2*(25390425*2^18702+1) (2*prime)
>
> The 5622-digit cofactor of n has been proven prime by Primo, with a
> certificate at
> http://xenon.stanford.edu/~tjw/pp/p5622.zip
>
> The two primes 25390425*2^18702+1 and 25390425*2^18703+1
> are a CC2 (2nd kind) byproduct of a recent CC3 search.

Congratulations on improving your month old 5257-digit record in
http://tech.groups.yahoo.com/group/primeform/message/9948
That also used a CC2 from a CC3 search.
I have verified the certificate and updated
http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm

--
Jens Kruse Andersen
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