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twin-prime heuristic proof

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• Hello, Group. I m optimistic that no one will find a counter-example. This method is similar to Euclid s proof for infinite-ly many primes. Let k = 2, 3, 4, 5,
Message 1 of 3 , Jul 8 12:24 PM
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Hello, Group.

I'm optimistic that no one will find a
counter-example.

This method is similar to Euclid's proof for
infinite-ly many primes.

Let k = 2, 3, 4, 5, etc. and then...
compute z = [(p(sub k) primorial) / 2] + 2^k.

Take a look at the results...

z -2 = 7 - 2 = 5 (twin-prime number)
z = 3 + 2^2 = 7 (also twin-prime)
z +2 = 7 + 2 = 9 (analysis not necessary)

Clearly, the twin-prime pair is 5,7 and 5 > 3.

Continuing...

z -2 = 21 = 3 * 7 (again the pair is 5,7 and 7 > 5)
z = 3*5 + 2^3 = 23 (prime but not important)
z +2 = 23 + 2 = 25 (factorization not needed)

Once again...

z -2 = 119 (analysis not necessary)
z = 3*5*7 + 2^4 = 121 = 11 * 11; (the twin-prime
pair is 11,13 and 11 > 7)
z +2 = 123 (ignore this number too)

Either the pair z-2,z or z,z+2 will be a twin-prime
pair or one of the three will contain a factor from a
twin-prime pair that is > p(sub k).

I checked the results up to p(sub 10) = 29.

Also, I found a variation of the formula for primes
that are just four units apart.

Has this relationship already been found? ... or does
someone find it interesting???

Bill

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• Hello Bill, ... Unfortunately, your conjecture seems to fail for k=29 prime(29) = 109 primorial(109)=279734996817854936178276161872067809674997230
Message 2 of 3 , Jul 8 3:45 PM
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Hello Bill,

> I'm optimistic that no one will find a counter-example.

Unfortunately, your conjecture seems to fail for k=29

prime(29) = 109
primorial(109)=279734996817854936178276161872067809674997230
corresponding z=139867498408927468089138080936033905374369527

1) z-2 =
3*5^2*29*43*1495509205120849698894820432355347825441

The small factors do not qualify as they're smaller than 109 and the big
one (t=1495509205120849698894820432355347825441) does not belong to a twin
prime pair (t-2 is divisible by 29, t+2 by 3).

2) z =
373*16159597*3217431548819*7212206181309313175293 = P1*P2*P3*P4

No twin primes here, as
P1+2, P2+2, P3-2, P4+2 are divisible by 3,
P2-2 is divisible by 5,
P1-2 and P3+2 are divisible by 7,
and P4-2 by 29.

3) z+2 =
17*307*1559*10259*393083*4262791622718255614129772517=P1*P2*P3*P4*P5*P6

P1 does not qualify, the others aren't members of twin prime pairs, as

P2+2, P3-2, P4-2, P5-2, P6+2 are divisible by 3,
P2-2, P5+2, P6-2 are divisible by 5,
P3+2 is divisible by 7,
P4+2 is divisible by 31.

Did I do anything wrong?

> Has this relationship already been found? ... or does
> someone find it interesting???

It is undoubtedly nice but I guess it's just a phenomenon
for small numbers (as the twin primes are quite common in that region).

Peter

--
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
• ... No. PARI/GP says the first counter examples are for k = 29, 33, 38, 39. ... That s an awfully short check, especially for a form allowing so many twin
Message 3 of 3 , Jul 8 4:34 PM
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Peter Kosinar wrote:
> Unfortunately, your conjecture seems to fail for k=29
...
> Did I do anything wrong?

No. PARI/GP says the first counter examples are for k = 29, 33, 38, 39.

Bill Bouris wrote:
> I checked the results up to p(sub 10) = 29.

That's an awfully short check, especially for a form allowing
so many twin chances.
There is no justification for the words "heuristic" or "proof".
It was just a poor guess.

--
Jens Kruse Andersen
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