## Factorization of 9 consecutive 552-digit numbers

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• Let x = (1320*(10^20 + 13065906))^3 y = (x*(5*x + 9)/2 - 31)^2 N = (y - 23^2)*(y - 24^2)/55440 - 7 then p544 = N/(2^2*85580203) p552a = (N + 1)/3 p546 = (N
Message 1 of 178 , Nov 11, 2009
Let

x = (1320*(10^20 + 13065906))^3
y = (x*(5*x + 9)/2 - 31)^2
N = (y - 23^2)*(y - 24^2)/55440 - 7

then

p544 = N/(2^2*85580203)
p552a = (N + 1)/3
p546 = (N + 2)/(2*7^2*16127)
p540 = (N + 5)/(223*9887*252181)
p552b = (N + 6)/2
p533 = (N + 8)/(2^2*9479*90863*15503173813)

are prime. Moreover N + 3, N + 4 and N + 7 are completely
factorized, with no prime factor having more than 128 digits.
These 9 consecutive factorizations at 552 digits are in

Notes: I used OpenPFGW, GMP-ECM, Msieve, ggnfs, Pari-GP.
All primality proofs were performed by Pari-GP.
The primality proof for p552b is easy, since
we know the factorization of p552b - 1 = (N + 4)/2.
The composite numbers
c542 = (N - 1)/(5*53*151*259499)
c524 = (N + 9)/(5*7*31*37*288232349*593280102498871)
are unlikely to have any prime divisor less than 10^30.
This particular constellation was studied because
p539 = (N + 10)/(2*3*60647*24846803)
is prime. Thus a complete factorization of N + 9
would have yielded 11 consecutive factorizations.

• ... 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,