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

http://physics.open.ac.uk/~dbroadhu/cert/ifacgg1.zip
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.

David Broadhurst, Armistice Day, 2009