This is due tho the fact that (n + 1)(n – 1) = n

^{2 }– 1. And if the sum of twin primes = 2n is divisible by 12, (i.e. n is divisble by 6), then n

^{2 }is thereby divisible by 12; so when n

^{2 }– 1 is divided by 12, it leaves a remainder of 1.

Expressed in an alternative fashion the result = k – 1/12 (where k is an integer).

This suggests a possible connection with the result of the Riemann zeta function when s = – 1,

i.e. ζ( – 1) = – 1/12.

In fact, in this context it is interesting to note that the denominators of the Riemann zeta function for all negative odd integers (– 1, – 3, – 5,......) - as with the sum of twin primes - appear to be always divisible by 12!

^{ }

Thus if n – 1 and n + 1 represent twin primes (other than 3 and 5) then n

^{2 }is divisible by 12.

In fact it always appears to be divisible by 36!

When we calculate the result of n

^{2}/36 with the value of n here relating to the average of successive twin primes, an interesting pattern results.

For 5 and 7, n

^{2}/36 = 36/36 = 1

For 11 and 13, n

^{2}/36 = 144/36 = 4

For 17 and 19, n

^{2}/36 = 324/36 = 9

Now if we watch the series on the right the terms seem to be increasing by 1, 3, 5,... Alternatively these results represent the successive squares of the natural numbers.

For 23 and 25, n

^{2}/36 = 576/36 = 16 (though of course 25 is not prime). Interestingly 25 also showed up as a "bogus prime" in yesterday's exercise. Then with 29 and 31 normal service is resumed.

For 29 and 31, n

^{2}/36 = 900/36 = 25

Again we would expect that the next result on the right should be 36 and this occurs

For 35 and 37, n

^{2}/36 = 1296/36 = 36 (even though 35 is not a prime).

With next pair, normal service is again resumed.

For 41 and 43, n

^{2}/36 = 1764/36 = 49

Now the next 2 squares of natural numbers are 64, 81 And these occur:

For 47 and 49, n

^{2}/36 = 2304/36 = 64 (though 49 is not prime),

For 53 and 55, n

^{2}/36 = 2916/36 = 81 (though 55 is not prime).

For 59 and 61 n

^{2}/36 = 3600/36 = 100.

What is remarkable here, just as in yesterday's exercise that the twin primes concur with the squares of 2, 3, 5 and 7 (for the first four pairs (after 5 and 7).

And even in all the other cases where the result is a square of a natural number that is not prime, one of the pairings is always prime (certainly up to 100).

This would suggest that a great way of searching for twin prime pairings would be to keep testing for

n

^{2}/36 = k

^{2 }(where k = 1, 2, 3, ....).

Incidentally, it is again very interesting in view of the earlier observation that ζ( – 1) = – 1/12, that the sum of all our results for k

^{2 }(where k = 1, 2, 3, .....) = ζ( – 2).