twin-prime heuristic proof
- Hello, Group.
I'm optimistic that no one will find a
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.
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)
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???
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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
1) z-2 =
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 =
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 doesIt is undoubtedly nice but I guess it's just a phenomenon
> someone find it interesting???
for small numbers (as the twin primes are quite common in that region).
[Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
- 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