Problems & Puzzles: Problems

Problem 13.- The Gap of the Beast

"A few moments ago I was wondering what would be the smallest two
consecutive primes that differ exactly by 666, our beast of course !"
(Problem sent by Patrick De Geest, 25/oct/98)


18691113008663 + 666 = 18691113009329

Author: Mr. Marek Wolf
(, from Poland, who has calculated all the gaps between consecutive primes from 2 to 916.

(this information was sent by Mr. Wolf to Jud McCranie who sent it to Patrick De Geest who finally sent it to me yesterday, November 2, 1998)


Mr. Wolf himself has sent the following e-mail comments:

" ...The gap 666 really appears for the first time after prime  18691113008663. It can be found in J. Young and A. Pottler, "First occurrence prime gaps",  Math. Comp.  52 (1989), p. 221.  I have send to Jud the whole table  of first occurences which I have at present at my disposal  and I asked for credits when somebody will use my large data. I enclose this huge table below. It mainly comes from my own search up to 2^44 and remaining entries I have compiled from different sources, I should mention mainly Thomas Nicely, see is not true, that I have calculated all gaps up to 916, I have stopped after 716 when I have learned that Nicely is doing it also and that he reached over 10^15... Please notice my approximate formula for p: the first occurence of a gap d after a prime p:

p ~ sqrt(d)*exp(sqrt(d)).

The paper can be found on my www page:         "


J. K. Andersen wrote, Jan 2009:

666-digit primes are called apocalypse primes.
The first apocalypse case is (p, p+666) for p = 10^665+1245313.

The first apocalypse case with two consecutive gaps of 666 is
(p, p+666, p+666+666) for p = 10^665+90344421.

The first apocalypse case with three consecutive gaps of 666 is
(p, p+666, p+666+666, p+666+666+666) for p = 10^665+5259048033991.

PrimeForm/GW made prp tests and Primo proved primes.
Four consecutive gaps of 666 is not possible, because 5 would divide one of the 5 gap ends.


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