- I have found the largest known semi-prime "twins" - not that hard considering
I may be the first to search for it. If you don't have enough skill, luck or
computing power for coveted records then make up your own records :-)
I got a little unknowing help from contributors to an old top 5000. Also
thanks to David Broadhurst who noted there were enough factors for an easy
proof in a similar problem with triplets.
p=1171440861*2^80025+1 was found by David Underbakke and Yves Gallot.
q=2704029*2^98305+1 was found by Phil Carmody.
Both were in 2000 so the exponents may not seem familiar to them.
r=(p*q+1)/2 was found by me using PrimeForm/GW, after testing many
combinations of known primes. I trial factored each combination with my own
p, q and r are primes, so p*q and 2*r are semi-prime twins, i.e. two
consecutive integers which are both the product of two primes.
The twins have 53699 digits where the twin prime record is 51090 digits, but
the semi-prime problem is far easier with the right algorithm. I set my target
to "beat" the twin prime record anyway.
The proof of r uses the factor 2^80024 of r-1:
> Running N-1 test using base 13p and q are factors of 2r-1, not r-1, and did not have to be primes. Therefore
> Calling Brillhart-Lehmer-Selfridge with factored part 44.86%
> ((1171440861*2^80025+1)*(2704029*2^98305+1)+1)/2 is prime!
Underbakke, Gallot and Carmody are not credited in the top 5000 submission of
r - not that they would care, considering the number of submissions they have.
> 1051a ((1171440861*2^80025+1)*(2704029*2^98305+1)+1)/2--
> 53698 p97 2003
Jens Kruse Andersen
- Neat work, Jens. The complexity is a bit like that of finding
an AP3 by picking over enough pairs of old prime carcasses.
But at 53k-digit size it's still impressive!