Factorization of 6 consecutive 1404-digit integers
m = 67440294559676054016000
a = (y-22^2)*(y-61^2)*(y-86^2)*(y-127^2)*(y-140^2)*(y-151^2)/(6*m)
b = (y-35^2)*(y-47^2)*(y-94^2)*(y-121^2)*(y-146^2)*(y-148^2)/m
Then b = 6*a+1. Moreover, with
y = (m*(10^96+9581328)+22)^2
each element of
[3*a-1, 6*a-1, (3*a+1)/(23*24943*13071241), 2*a+1]
By factorizing all 24 of the 119-digit algebraic cofactors
in a and b, we obtain complete factorizations of the 6
successive 1404-digit integers 6*a+k-2 with k in [0,5].
subjected to 320 ECM curves at B1=250000
20 August 2007
- David Broadhurst wrote:
> By factorizing all 24 of the 119-digit algebraic cofactorsCongratulations!
> in a and b, we obtain complete factorizations of the 6
> successive 1404-digit integers 6*a+k-2 with k in [0,5].
http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm is updated.
The algebraic cofactors have been upgraded from
"relatively easy to factor" to "possible to factor"
since your 1104-digit record for 6 numbers.
Donovan Johnson has set a 2135-digit record for 5 numbers.
It contains a twin prime but it and others were computed to
set this record, so it doesn't count as a record with an "origin".
Jens Kruse Andersen
- --- In email@example.com, "Jens Kruse Andersen"
> http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm is
I liked the peculiarity that 3/4 of the 1400-digit primes
were provable by BLS. Thay was not my intent;
it just happened that way: the 3-constellation
[3*a-1, 6*a-1, 2*a+1] fell out of pfgw -f -d,
In the long term, the k=5 record will need
a 4-constellation, to avoid ECPP. But I leave
that to the master sievers of linear forms.
I prefer shallow sieving of 12th (resp. 8th)
order polynomials for k = 6,7 (resp. 8,9,10),
followed by serious ECM in a minority of
I chose 1400 digits for k=6 so that any
significant later improvement would entail SNFS,
rather than GNFS, which works fine for me
below 120 digits.