Problems & Puzzles:
Conjectures
Conjecture 69. A Conjecture
& proposed Axiom
Luis Rodríguez sent the following
conjecture, that at the same time he proposes as a candidate to axiom:
1. If k Arithmetic
Progressions:
a.n + b (n = 0, 1, 2,
3....inf.) a, b coprimes (a even)
are arranged in a matrix of k
rows and there are two columns with de same permissible* pattern of
primes, then there exists infinitely many columns with that pattern.
If the number of rows are
infinite then all the columns will contains infinitely many primes.
This part is equivalent to
Dickson Conjecture (1904)
2. If k = 2 with the first
progression = 2n + 1 (from n = 1) and the second a decreasing
positive: b  2n (from n =1), b odd >=5, then there will be
at least one match of two primes.
* "Permisible in the sense that
there are not congruence relations that prohibits it.
Luis suggests also that if this
Conjecture is not disconfirmed and is added as an axiom in the field
of number theory, the following conjectures will be proved on basis of
the axiom:
a. Polignac and twin prime
conjectures
b. k tuples conjecture
c. Infinitude of Sophie Germain's primes
d. Infinitude of Cunningham chains
e Goldbach's conjecture
Q1. Can you
prove it of find some counterexample, as isolated Conjecture?
Q2. Can you discuss the proposal to add it as an axiom?
Q3. Can you propose an application to another conjecture?
Contribution came from Emmanuel
Vantieghem.
***
Emmanuel wrote:
As to my opinion, the
part of the conjecture about infinitely many arithmetic progressions is
not trrue. If the kth sequence is (2^k)n+1 and we take n to be a
Sierpinski number, then the nth column contains no prime at all.
For the finite part,
the notion of 'inadmissibility' is a littlebit obscure for me. For
instance, if we take the sequences 2n+1, 2n+3, 2n+5, we have the
configuration (prime,prime,prime) just once (for n = 1). But is
this configuration 'inadmissible' ?
Part 2 of the
conjecture (about 2n+1 and b2n) is obviously equivalent to the
Golbach conjecture.
***
Luis replied Emmanuel this:
The Emmanuel's
objection is interesting. But it is resolved by the expressed
condition that the sequences that are impossible by reasons of
congruences are not admissible. The existence of sequences of
Sierpinsky: S(n) = k.2^n 1 that not have primes, means that all its
terms are divisible by some other primes p(i), that is S(n) is
congruent 0 mod p(i).
***
