I make the following statement based only on a limited number of

calculations. I was unable to find any web references. Has anyone come

across anything like this?

For any two consecutive positive even integers A and B, there exists at

least one set of three consecutive primes C<D<E such that A equals (D-C)

and B equals (E-D) OR that A equals (E-D) and B equals (D-C).

For example for the 2 consecutive even integers A=2 and B=4, the 3

consecutive primes are C=5, D=7 and E=11.

For the 2 consecutive even integers A=10 and B=12, the 3 consecutive

primes are C=619, D=631 and E=641.

For the 2 consecutive even integers A=94 and B=96, the 3 consecutive

primes are C=327418141, D=327418237 and E=327418331.

I tested all pairs of consecutive even integers (2, 4) to (98, 100). Each

pair had a matching prime triad. As expected, the larger the pair of

consecutive even integers, the longer it takes to find the triad. I have

only limited access to a computer so I stopped checking.

My reason for posting this is to ask whether there is a neat way of

selecting ranges of primes among which one would be most likely to find a

record triad example, assuming the statement is true. Also I thought this

idea of "twin gaps" might interest someone in the group. Thanks folks for

your time.

Bill Sindelar