Problems & Puzzles:
Conjectures
Conjecture 102.
Conjecture related to
A359387
Alain Rochelli sent the
following conjecture, on Feb 19, 2023
In Puzzle 1122 OEIS A005385, we have proven that, for any
Safe prime p>7, A(p) = (2^(p-1) - 1) / 3 has as smallest prime factor p.
OEIS A359387 is relative to primes p such that the smallest prime factor
of A(p) is p. Of course, all Safe primes (> 7) are included in A359387.
Let q be the A359387 primes which are not Safe primes. The first
suitable primes q are as follows :
443, 647, 1847, 2243, 2687, 2699, 6263, 6563, 7487, 7583, 8627, 8663,
9419
For any value of q, it is conjectured that q = 2*u*v + 1 where u and v
are distinct primes not in A005384 = non-Sophie Germain primes (cf.
A053176).
For example :
443 = 2*13*17 + 1 (13 and 17 in A053176)
647 = 2*17*19 + 1 (17 and 19 in A053176)
1847 = 2*13*71 + 1 (13 and 71 in A053176)
2243 = 2*19*59 + 1 (19 and 59 in A053176)
2687 = 2*17*79 + 1 (17 and 79 in A053176)
- - -
9419 = 2*17*277 + 1 (17 and 277 in A053176)
Q : Can you get an explanation of this or find
a counterexample ?

During the week from 7-13 Oct 2023,
Contributions came from Oscar Volpatti, Alessandro Casini, Emmanuel
Vantieghem
***
Oscar wrote:
Conjecture 102 is false.
These primes are the first ten counterexamples:
41039 = 2*17^2*71+1
72923 = 2*19^2*101+1
1516583 = 2*31*61*401+1
1902347 = 2*31*61*503+1
2039243 = 2*31^2*1061+1
2235287 = 2*31^2*1163+1
2638907 = 2*31^2*1373+1
2984867 = 2*31^2*1553+1
3817259 = 2*61*67*467+1
4398467 = 2*31*61*1163+1
***
Alessandro wrote:
The conjecture, such as it is
stated, is incorrect. In fact, q = 41039 = 2 * 17^2 * 71 + 1 is in
A359387 but has the 17 squared. Moreover, it isn’t even mandatory to
have only two distinct odd prime factors u and v, although quite
rare. Indeed, there are only four q < 5M with 3 distinct odd primes,
with 1516583 the smallest. However, the fact remains that the prime
factors cannot be Sophie Germain primes. Therefore, the conjecture
can be restated as follows:
If q is a no-safe prime in
A359387, then q = 2*C + 1, where C is a product of non-Sophie
Germain primes.
This statement is actually
always true, here’s the proof.
Since q = 11 (mod 12), q-1
cannot be divisible by 4, so C = (q - 1) / 2 is an odd composite
cofactor. Let’s assume C is divisible by a Sophie Germain prime p.
By Fermat’s theorem, 2 ^ (2p) = 1 (mod 2p + 1) and raising to the
appropriate power we get that 2p + 1 divides 2 ^ (q-1) - 1. But 3 <
2p + 1 < 2C + 1 = q, hence q isn’t in A359387, absurd.
In addition, It isn’t important
that p is a prime number, just that 2p+1 is. Therefore, 2*d+1 must
be composite for every proper divisor d of C.
Lastly, C cannot be a square,
else q = 2C+1 will be a multiple of 3.
I wonder if it is possible to
show further limitations on their shape.
***
Emmanuel wrote:
I think this conjecture has
been settled as "False" in your Conjecture 96.
There I gave many
counterexamples of type 2*p*r*s +1 and 2*(p^2)*r + 1 (p,r s,
primes with p < r < s).
(I conjectured that there
would be no others
but Oscar Volpatti also gave
an example of type 2*p*q^2 + 1 and 2*q^3 + 1.
Recently, I found a
counterexample q = 5165748299123 = 2*(613^3)*11213 + 1.
But, a double check is
wanted here.
***
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