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## Twin prime curio

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• Let n be (p(p+2)+1)/2 where p and p + 2 are potential twin primes. If p and p+2 are twin primes then the sum of the products of (n+x)(n-x) that divide
Message 1 of 2 , Oct 3, 2004
Let n be (p(p+2)+1)/2 where p and p + 2 are potential twin primes.

If p and p+2 are twin primes then the sum of the products of (n+x)(n-x) that divide integrally by p or p+2, with x starting at 0 to x = n-2, will equal p(p+2)(a) ((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1)) where a = (p+1) / 6. If either of p or p+2 are composite then the sum will be greater than p(p+2)(a)((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1)).

As I'm not that good at expressing myself mathematically as an example

p = 5, p+2 = 7

18 * 18 = 324

17 * 19 = 323

16 * 20 = 320

15 * 21 = 315

..

2 * 34 = 68

Note that we only have x to n-2 = 16 i.e. we don't include 1 and 35.

Of all the generated products 5 will divide 155, 180, 260, 275, 315 and 320. 7 will divide 203, 224, 308 and 315. As 315 divides by both 5 and 7 it is added only once. The sum of all these products is 2240.

So with p = 5 then (p+ 1) / 6 = 1 in this example so

p(p+2)(a)((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1))

(35)(1) ((6(6-1) + (6-1)(6+1))) - (3(1(1-1)+1))

35(30 + 35 - 1) = 2240

The reason this works out this way is due to the fact if p or p+2 is composite then more of the products are divisible than with two primes when one expects only 2p of the products to be divisible.

Anybody fancy putting this into more formal mathematical language?

Cheers,

David

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• Hi all, (Hope this formatting comes out) n-2 ____ / (x+n)(x-n) ... k=0 I think (?) this is the sum of all (x+n)(x-n) for all n starting from 0 to n-2. But
Message 2 of 2 , Oct 4, 2004
Hi all,
(Hope this formatting comes out)

n-2
____
\
/ (x+n)(x-n)
-------
k=0

I think (?) this is the sum of all (x+n)(x-n) for all n starting from 0 to n-2. But how would I modify it to show that only those (x+n)(x-n) that divide by p or p+2 should be summed?
Cheers,
David

David Litchfield <dwlitchfield@...> wrote:

Let n be (p(p+2)+1)/2 where p and p + 2 are potential twin primes.

If p and p+2 are twin primes then the sum of the products of (n+x)(n-x) that divide integrally by p or p+2, with x starting at 0 to x = n-2, will equal p(p+2)(a) ((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1)) where a = (p+1) / 6. If either of p or p+2 are composite then the sum will be greater than p(p+2)(a)((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1)).

As I'm not that good at expressing myself mathematically as an example

p = 5, p+2 = 7

18 * 18 = 324

17 * 19 = 323

16 * 20 = 320

15 * 21 = 315

..

2 * 34 = 68

Note that we only have x to n-2 = 16 i.e. we don't include 1 and 35.

Of all the generated products 5 will divide 155, 180, 260, 275, 315 and 320. 7 will divide 203, 224, 308 and 315. As 315 divides by both 5 and 7 it is added only once. The sum of all these products is 2240.

So with p = 5 then (p+ 1) / 6 = 1 in this example so

p(p+2)(a)((6a(6a-1) + (6a-1)(6a+1))) - (3(a(a-1)+1))

(35)(1) ((6(6-1) + (6-1)(6+1))) - (3(1(1-1)+1))

35(30 + 35 - 1) = 2240

The reason this works out this way is due to the fact if p or p+2 is composite then more of the products are divisible than with two primes when one expects only 2p of the products to be divisible.

Anybody fancy putting this into more formal mathematical language?

Cheers,

David

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