Twin Primes: Occur in Pairs, 90k+11/13/17/19 Proven?

[mathman]

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?

 

[Martin Rattigan]

A less cursory check might throw up 5,7,11,13.

 

[mathman]

Sorry - I meant after the single digits. The case you described is the only one where a number ending in 5 could appear.

 

[rasmhop]

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 $n \equiv 1\pmod 3$, then 3 divides n+17 which is a contradiction.
If $n \equiv 2\pmod 3$, then 3 divides n+11 which is a contradiction.
Thus 3|n.

If $n \equiv 1\pmod 5$, then 5 divides n+19 which is a contradiction.
If $n \equiv 2\pmod 5$, then 5 divides n+13 which is a contradiction.
If $n \equiv 3\pmod 5$, then 5 divides n+17 which is a contradiction.
If $n \equiv 4\pmod 5$, then 5 divides n+11 which is a contradiction.
Thus 5|n.

We now have 2*3*5=30|n.

 

[ramsey2879]

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 $n \equiv 1\pmod 3$, then 3 divides n+17 which is a contradiction.
If $n \equiv 2\pmod 3$, then 3 divides n+11 which is a contradiction.
Thus 3|n.

If $n \equiv 1\pmod 5$, then 5 divides n+19 which is a contradiction.
If $n \equiv 2\pmod 5$, then 5 divides n+13 which is a contradiction.
If $n \equiv 3\pmod 5$, then 5 divides n+17 which is a contradiction.
If $n \equiv 4\pmod 5$, then 5 divides n+11 which is a contradiction.
Thus 5|n.

We now have 2*3*5=30|n.

What about the twin primes 29 and 31?

 

[Martin Rattigan]

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.

 

[Martin Rattigan]

What about the twin primes 29 and 31?

mathman was talking about sequences of 4 primes with consecutive differences of 2,4 and 2.

 

[CRGreathouse]

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.

Agreed. But of the 28387 up to 10^9 only 9339 are of that form, reversing that trend.

 

[mathman]

Agreed. But of the 28387 up to 10^9 only 9339 are of that form, reversing that trend.

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?