Twin Primes (2)

When one accepts that the sum of twin primes (apart from the first pairing 3 and 5) is divisible by 12, this then implies that the product of these twin primes when divided by 12 will always leave a remainder of 1.

This is due tho the fact that (n + 1)(n – 1) = n² – 1. And if the sum of twin primes = 2n is divisible by 12, (i.e. n is divisble by 6), then n² is thereby divisible by 12; so when n² – 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 n² is divisible by 12.

In fact it always appears to be divisible by 36!

When we calculate the result of  n²/36 with the value of  n here relating to  the average of successive twin primes, an interesting pattern results.

For 5 and 7,           n²/36 = 36/36      = 1 

For 11 and 13,       n²/36 = 144/36    = 4

For 17 and 19,       n²/36 = 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,       n²/36 = 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,      n²/36 = 900/36  = 25

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

For 35 and 37,     n²/36 = 1296/36 = 36 (even though 35 is not a prime).

With next pair,  normal service is again resumed.

For 41 and 43,   n²/36 = 1764/36  = 49

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

For 47 and 49,  n²/36 = 2304/36 = 64 (though 49 is not prime),

For 53 and 55,  n²/36 = 2916/36 = 81 (though 55 is not prime).

For 59 and 61  n²/36 =  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
n²/36 = k² (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 k² (where k = 1, 2, 3, .....) = ζ( – 2).

Twin Primes (1)

I have mentioned before how the sum of all twin primes (excepting the first pair of 3 and 5) appears to be divisible by 12.

I then decided to look at the pattern of results obtained when the sum of successive pairs is divided by 12 and found an unexpectedly interesting pattern.

So when we divide the sum of 5 + 7 by 12, we obtain 1.

Then for the next successive pairs,

(11 + 13)/12     =  2
(17 + 19)/12     =  3
(29 + 31)/12     =  5
(41 + 43)/12     =  7

So there is a perfect matching here with the first four prime numbers. Though this pure pattern breaks down with the next few results, there is still a remarkably close relationship with the sequence of primes (with a difference of only 1 from the correct results in the sequence in evidence. So

(59 + 61)/12     = 10 (as opposed to 11)
(71 + 73)/12     = 12 (as opposed to 13)
(101 + 103)/12 = 17
(107 + 109)/12 = 18 (as opposed to 19)
(137 + 139)/12 = 23

Now with the next pair of twin primes the pattern breaks down somewhat with
(149 + 151)/12 = 25. So this value seems to be sticking out on its own (as compared to the next prime 29 . Then this is followed by,

(179 + 181)/12 = 30 (as opposed to 31) and
(191 + 193)/12 = 32
(197 + 199)/12 = 33

So we seem to have two additional results, here not matched by a corresponding prime before the next
(227 + 229)/12 = 38 (as opposed to 37) and
(239 + 241)/12 = 40 (as opposed to 41)

Then the next,

(269 + 271)/12 = 45 (which again appears as an additional results before the next,

(281 + 283)/12 = 47 and
(311 + 313)/12 = 52 (as opposed to 53).

Then we seem to bypass the primes i.e. 59, 61 and 67 before the nest two results

(419 + 421)/12 = 70 (as opposed to 71) and
(431 + 433)/12 = 72 (as opposed to 73)

And the next we have

(461 + 463) 12 = 77 (as opposed to 79)

So here the difference is 2, and the result is indicative of the fact that close relationship with the sequence of primes is slowly breaking down. So the remaining results up to 100 are

(521 + 523)/12 = 87
(569 + 571)/12 = 95

So again we seem to be missing a value for 83 while 87 and 95 differ from the next two primes 89 and 97 by 2 respectively.

However, it still seems remarkable that the results when dividing the sums of successive pairs of twin primes, should match the corresponding sequence of prime numbers so closely.