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

program.

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 13

> Calling Brillhart-Lehmer-Selfridge with factored part 44.86%

> ((1171440861*2^80025+1)*(2704029*2^98305+1)+1)/2 is prime!

p and q are factors of 2r-1, not r-1, and did not have to be primes. Therefore

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

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Jens Kruse Andersen