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 (MWOLF@proton.ift.uni.wroc.pl), 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)
Correction:
Mr. Wolf himself has sent the
following email 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
http://www.lynchburg.edu/public/academic/math/nicely/
....it 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:
http://www.ift.uni.wroc.pl/~mwolf "
***
J. K. Andersen wrote, Jan 2009:
666digit 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.
***
