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:

C  O  N  J  E  C  T  U  R  E

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

Questions:

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 k-th sequence is  (2^k)n+1  and we take  n  to be a Sierpinski number, then the n-th 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  b-2n) 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).

***

 Records   |  Conjectures  |  Problems  |  Puzzles