This proof covers both parts of the original assertion, i.e. n = 0 mod 3 implies both that nth and (n-1)th terms of sequence are equal to 0 mod 3.

The assertion is clearly true for n=3.

Now for the case where you're adding n on at the end. This number looks like:

{the (n-3)rd number in the sequence}{digits of n-2}{digits of n-1}{digits of n}

The n-3rd number is divisible by 3 by the induction hypothesis. The digits of n-2 and n-1 are respectively -2 and -1 mod 3, so adding them together is 0 mod 3, and n itself is 0 mod 3, so the nth term of the sequence must also be divisible by 3.

Alternative version : If I get annoyed about their poor grammar, I am likely to become one of them.

Say d(n) = p(n+1) - p(n).

Write down d(n) as a list of numbers.

It'll bounce around all the time (see other posts for proof that you can always find a value as large as you like), but you'll also always be able to find a value that's 600 or less, and at that point you've found a "nearby pair."

The holy grail of the full twin prime conjecture is just saying that there are an infinite number of 2s in the sequence of d(n)s.

It isn't trying to cover every prime, just saying that no matter how far you go along them, you'll always be able to find another "nearby pair" further on.

The holy grail of the exercise is to bring the 600 down to 2, so that we'll be able to say there are an infinite number of so-called twin primes.