Twin primes may occur in pairs - i.e. 11, 13, 17, 19. A cursory check seems to indicate that they have to be of the form 90k + 11, 13, 17, 19. Has this ever been proven? If so has it ever been proven that the set of k's is infinite or is it finite?
Sorry - I meant after the single digits. The case you described is the only one where a number ending in 5 could appear.
1481, 1483, 1487, 1489 is the first counterexample. (1491 = 41 + 90 * 16) However what is true is that they are all of the form: 30k + 11, 13, 17, 19. This can easily be proven by supposing we have primes n+11,n+13,n+17,n+19 (with n non-negative). n must be even because otherwise n+11 is even and therefore not prime. So 2|n. If [itex]n \equiv 1\pmod 3[/itex], then 3 divides n+17 which is a contradiction. If [itex]n \equiv 2\pmod 3[/itex], then 3 divides n+11 which is a contradiction. Thus 3|n. If [itex]n \equiv 1\pmod 5[/itex], then 5 divides n+19 which is a contradiction. If [itex]n \equiv 2\pmod 5[/itex], then 5 divides n+13 which is a contradiction. If [itex]n \equiv 3\pmod 5[/itex], then 5 divides n+17 which is a contradiction. If [itex]n \equiv 4\pmod 5[/itex], then 5 divides n+11 which is a contradiction. Thus 5|n. We now have 2*3*5=30|n.
Of 165 occurences of twin prime pairs taken from primes in the range 10-1000000 there are 60 of the form 90k+11,13,17,19. That's slightly more than you would predict from the 30k+11,13,17,19 constraint mentioned in rasmhop's post, but not surprisingly so.
This seems to imply that the 30k + 11, 13, 17, 19 prime sets fall into 3 classes depending on congruence of k mod 3. Have they been shown to be asymptotically equal in size?