"Brian J. Beesley" <[EMAIL PROTECTED]> wrote,
in response to my earlier message
> The point of course is that there is a formal proof that, if a prime
> p is congruent to 3 modulo 4 and 2p+1 is also prime, then 2^p-1 is
> divisible by 2p+1 - which makes searching for a factor of M(p) by
> trying to factor M(M(p)) somewhat pointless!
> >
> > With c = 2^727 - 1, 2^751 - 1, 2^809 - 1, 2^997 - 1, 2^1051 - 1, I
> > looked for factors 2*k*c + 1 of 2^c - 1, but found none with k <=
> > 20000. I invite those with a special search program for M_(M127)
> > factors to search these further.
>
> None of 727, 751, 809, 997 & 1051 are "3 mod 4" S-G primes (else they
> wouldn't be on the "factors wanted" lists!)
Then I challenge readers to show how to discover the factor
11447 of 2^97 - 1 given the factor 1800*(2^97 - 1) + 1 of M(M97)
97 is not 3 mod 4.
Let c be a large positive integer.
Given k with 1 <= k <= 20000, the chance that 2*k*c + 1
is a prime divisor of 2^c - 1 is about
(1/(2*k)) * (2/ln(2*k*c)) = 1/(k * ln(2*k*c))
Integrate this from k = 1 to k = 20000.
An antiderivative is ln(ln(2*k*c)). The integral is
ln(ln(40000*c)) - ln(ln(2*c))
= ln( ln(40000*c) / ln(2*c) )
= ln(1 + ln(20000) / ln(2*c) )
For c as large as 10^8, the quotient ln(20000) / ln(2*c) exceeds 0.5,
giving an estimated ln(1.5) ~= 0.4 factors with 1 <= k <= 20000.
For c = 2^e - 1 with large e, we estimate 14.2/(e + 1) factors
with k in this range. This is only 0.02 factors when e = 727,
so it should not be surprising that my search was unsuccessful.
Extending the search to k < 10^8 will approximately double
our chance of success.
Peter Montgomery
[EMAIL PROTECTED]
--JAA22954.988529496/hera.cwi.nl--
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