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Problems & Puzzles:
Conjectures
Conjecture 96.
OEIS A360827
Alain Rochelli sent the
following conjecture:
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
The corresponding sequence
is now published as OEIS
A360827,
equals
A005385
subtracted from
A359387.
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 18-24
March, 2023 contribution came from Emmanuel Vantieghemm Oscar
Volpatti
***
Emmanuel wrote:
The conjecture is not true.
I found a lot of 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
4657007 = 2*(31^2)*2423 + 1
5488139 = 2*(101^2)*269 + 1
5533439 = 2*(31^2)*2879 + 1
6821303 = 2*31*269*409 + 1
7367027 = 2*(31^2)*3833 + 1
7484579 = 2*31*61*1979 + 1
7684283 = 2*101*109*349 + 1
8151203 = 2*(31^2)*4241 + 1
8181203 = 2*(101^2)*401 + 1
8655047 = 2*(61^2)*1163 + 1
8693207 = 2*(31^2)*4523 + 1
8799227 = 2*31*347*409 + 1
9223679 = 2*(31^2)*4799 + 1
9815999 = 2*(61^2)*1319 + 1
10284623 = 2*(31^2)*5351 + 1
11102087 = 2*109*127*401 + 1
11703059 = 2*(31^2)*6089 + 1
12996539 = 2*61*307*347 + 1
13663499 = 2*(31^2)*7109 + 1
15769763 = 2*31*347*733 + 1
16684883 = 2*(31^2)*8681 + 1
17030843 = 2*(31^2)*8861 + 1
17105987 = 2*31*61*4523 + 1
17332907 = 2*31*61*4583 + 1
18149819 = 2*31*61*4799 + 1
18841367 = 2*(31^2)*9803 + 1
19417967 = 2*(31^2)*10103 + 1
21274619 = 2*(31^2)*11069 + 1
23358047 = 2*(109^2)*983 + 1
23530763 = 2*(103^2)*1109 + 1
23604083 = 2*(31^2)*12281 + 1
23684807 = 2*(31^2)*12323 + 1
24740363 = 2*127*257*379 + 1
25343183 = 2*31*61*6701 + 1
26037119 = 2*61*457*467 + 1
26543003 = 2*(101^2)*1301 + 1
27696083 = 2*31*103*4337 + 1
27905519 = 2*(31^2)*14519 + 1
28355267 = 2*(31^2)*14753 + 1
29267039 = 2*31*103*4583 + 1
29797403 = 2*61*467*523 + 1
30445223 = 2*(61^2)*4091 + 1
30508883 = 2*109*349*401 + 1
30673199 = 2*(31^2)*15959 + 1
31030823 = 2*61*347*733 + 1
31999379 = 2*(31^2)*16649 + 1
32103167 = 2*(31^2)*16703 + 1
32388479 = 2*101*109*1471 + 1
32483723 = 2*(31^2)*16901 + 1
32910407 = 2*(31^2)*17123 + 1
33030839 = 2*(101^2)*1619 + 1
33466919 = 2*31*61*8849 + 1
35228339 = 2*(31^2)*18329 + 1
36035579 = 2*(31^2)*18749 + 1
36865883 = 2*(31^2)*19181 + 1
(the next one is > 37000000)
I printed all of them because they seem to be
either of type 2*p^2 r or 2*p*r*t (p, r, t primes). This could be a conjecture.
However, it is perfectly provable that every odd
prime divisor p of q - 1 is a non-Sophie Germain prime..
There is an even stronger result :
When q is a prime such that every prime divisor
of (2^(q-1) - 1)/(3q) is > q, then every proper divisor d
of q-1 has the property that 2d + 1 is NOT a prime number.
Proof :
Let d be a proper divisor of q - 1.
Then 2^d - 1 divides 2^(q-1) - 1.
So, if 2d + 1 is prime, then, by Fermat's
little theorem, it is a divisor of 2^(2d) - 1 and thus also of
2^(q-1) - 1 which has no divisors (except 3 and q) of that
kind.
This proves what we wanted.
***
Oscar wrote:
Conjecture 96 is false.
I checked all primes q < 6*10^9, finding 2049 counterexamples q
= 2*t+1 such that t has three odd prime factors (counted with
multeplicity); I'm sending the first counterexample for each of
the four possible forms.
small prime repeated:
q_1 = 41039 = 2*17^2*71+1;
three distinct primes:
q_3 = 1516583 = 2*31*61*401+1;
large prime repeated:
q_657 = 549397727 = 2*503*739^2+1;
prime cube:
q_1615 = 3991233959 = 2*1259^3+1.
***
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