## Factorization of 4 consecutive 8193-digit numbers

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• Let n = 378149751*2^27186-2 n = 2*(378149751*2^27185-1) (2*prime) n+1 = 378149751*2^27186-1 (prime) n+2 = 378149751*2^27186 = 2^27186*3^2*7*17*353081 n+3 =
Message 1 of 5 , May 2, 2010
Let n = 378149751*2^27186-2

n = 2*(378149751*2^27185-1) (2*prime)
n+1 = 378149751*2^27186-1 (prime)
n+2 = 378149751*2^27186 = 2^27186*3^2*7*17*353081
n+3 = 378149751*2^27186+1 = 5*1019*24923*43651*989239*P8174

The 8174-digit cofactor of n+3 has been proven prime through a joint effort of Geoffrey Hird and myself, and makes the Top-20 for ECPP proofs:

http://primes.utm.edu/primes/page.php?id=92563

This prime is currently the fifth-largest candidate certified by Primo, with a certificate available at:

http://www.ellipsa.eu/public/primo/files/ecpp8174.zip

The two primes 378149751*2^27185-1 and 378149751*2^27186-1 are a Sophie Germain byproduct of a recent CC3 (1st kind) search.

n-1 = 378149751*2^27186-3 = 3*53*127*7043*626921*24117041920337*C8166
n+4 = 378149751*2^27186+2 = 2*11*561139037952529*C8177

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 with a note that the certificate is being verified.
Message 2 of 5 , May 3, 2010
Tom wrote:
> Let n = 378149751*2^27186-2
>
> n = 2*(378149751*2^27185-1) (2*prime)
> n+1 = 378149751*2^27186-1 (prime)
> n+2 = 378149751*2^27186 = 2^27186*3^2*7*17*353081
> n+3 = 378149751*2^27186+1 = 5*1019*24923*43651*989239*P8174
>
> The 8174-digit cofactor of n+3 has been proven prime through a joint
> effort of Geoffrey Hird and myself, and makes the Top-20 for ECPP proofs

Congratulations!
http://users.cybercity.dk/~dsl522332/math/consecutive_factorizations.htm
is updated with a note that the certificate is being verified. Marcel Martin
lists it on the Primo Top-20 and I assume there will be no problems.
I'm pleasantly surprised that somebody has run Primo at 8174 digits for
my record page.

--
Jens Kruse Andersen
• ... When http://primes.utm.edu/primes/page.php?id=92563 was posted, by Geoffrey, I wondered: what cunning problem might this prime solve? Thanks, Tom, for your
Message 3 of 5 , May 4, 2010
--- In primeform@yahoogroups.com,
"tjw99" <tjw99@...> wrote:
>
> Let n = 378149751*2^27186-2
>
> n = 2*(378149751*2^27185-1) (2*prime)
> n+1 = 378149751*2^27186-1 (prime)
> n+2 = 378149751*2^27186 = 2^27186*3^2*7*17*353081
> n+3 = 378149751*2^27186+1 = 5*1019*24923*43651*989239*P8174

When
http://primes.utm.edu/primes/page.php?id=92563
was posted, by Geoffrey, I wondered:
what cunning problem might this prime solve?

David
• ... When I saw Tom Wu in the prover code I felt certain about the problem it solved and only wondered about a little detail. A few tests confirmed my
Message 4 of 5 , May 4, 2010
David wrote:
> When
> http://primes.utm.edu/primes/page.php?id=92563
> was posted, by Geoffrey, I wondered:
> what cunning problem might this prime solve?

When I saw Tom Wu in the prover code I felt certain about the problem it
solved and only wondered about a little detail.
A few tests confirmed my assumption about the problem and I posted the below
before the record had been announced or submitted.

----- Original Message -----
From: "Jens Kruse Andersen" <jens.k.a@...>
To: "tjw99" <tjw99@...>
Sent: Monday, May 03, 2010 2:27 AM
Subject: Proven factorization of 4 consecutive 8193-digit numbers

> Congratulations on http://primes.utm.edu/primes/page.php?id=92563
> I'm honored that somebody would run Primo at 8174 digits for my record page.
>
> I'm a little curious why it was submitted as
> (378149751*2^27186+1)/(989239*5542921182935)
>
> I would have expected either
> (378149751*2^27186+1)/5483273808085436465 or
> (378149751*2^27186+1)/(5*1019*24923*43651*989239)
> but it's not important for the Prime Pages.
>
> --
> Jens Kruse Andersen

It turned out the latter form was submitted but the Prime Pages
canonicalization converted it.

--
Jens Kruse Andersen
• ... thus confirming that Jens is quicker on the uptake than am I. However, my slower brain recently had a (perhaps) neat idea that may (or may not) result in
Message 5 of 5 , May 4, 2010
--- In primeform@yahoogroups.com,
"Jens Kruse Andersen" <jens.k.a@...> wrote:

> When I saw Tom Wu in the prover code I felt certain about
> the problem it solved

thus confirming that Jens is quicker on the uptake than am I.

However, my slower brain recently had a (perhaps) neat
idea that may (or may not) result in another update for
Jens to perform, perhaps not before too long.

Festina lente!

David
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