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Re:k difference primes theorems

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• hi, Consider 1. If p is prime and p+k is prime then p+k divides p^(p+k) + k. 2. the converse If p is prime and p+k is composite and p+k divides p^(p+k) +k,
Message 1 of 1 , Jan 9, 2005
hi,

Consider

1. If p is prime and p+k is prime then p+k divides p^(p+k) + k.
2. the converse

If p is prime and p+k is composite and p+k divides p^(p+k) +k, then p+k is a
pseudokprime.

For k=2 we get the case for twin primes. There are 33 pseudotwinprimes for
1517 twin
primes less than 130000. In other words of the 12159 primes p less than
130000 the divide
by p+2 test failed 33 times to detect if p and p+2 were or were not twin
primes.

for k=4 or quad primes, there are 11 pseudoquadprimes for 344 quadprimes
less than 20000.

Thus it appears, at least statistically, that it is highly probable that
p+k is prime if p+k divides
p^(p+k) + k.

It is interesting to note that Charmichael numbers pop up in these
calculations.

Maybe someone can prove 1. and elaborate. I have included a pari script
below.

ktokpk(n,n2,k) =
{
local(x,y,x2,c);
c=0;
forprime(x=n,n2,
x2=x+k;
y=x^x2+k;
if(y%x2==0&!isprime(x2),c++;
print1(x2",");
);
);
print();
print(c","pikprimes(n2,k))
}

pikprimes(n,k) = \\The number of k difference prime pairs less than n.
{
local(c,x);
c=0;
forprime(x=3,n,
if(isprime(x+k),c++)
);
return(c)
}

Cino
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