Revisions to Generalising Dirichlet's theorem in arithmetic progressions-prime combinatorics

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Let $M$ be a natural number $M>1$. For every prime $p_i$ not dividing $M$ take an arithmetic progression $A_i=k_i+np_i$ , $n\geq 0$ such that $k_i>p_i^2/M$. Is there any $M$ and some choice of the $k_i's$ such that $\cup A_i= \mathbb{N}$ ?

CONJECTURE: There is not such choice for any $M$

MOTIVATION: It is not hard to see that if there is not such a choice for any $M$ Dirichlet's theorem in arithmetic progressions is an easy consequence. Of course this is much more general. We can modify many well known problems in this way.

For example ,to the previous question, if we take $2$ arithmetic progressions for each $p_i>M$, instead of $1$ with the same condition then we have something stronger than the twin prime conjecture and polignac's conjecture in general, If we take $3$ arithmetic progressions we have something stronger than the problem that infinitely many times exists every logical form with $3$ primes ( for example p, p+2,p+6), etc. To these problems it is enough to take infinitely many $M$, not all...

Secondly it is one of the easiest questions to this spirit and i want to know if we can say something about this kind of questions with known techniques. So

What kind of mathematical techniques we could use to reach this kind of problems?

Can we compute the function of the length of the intervals that we can cover using the primes not dividing $M$ below some $x$ and compare it to the growth function of the number of the primes below some $x$ ?

Notice that even for the primes below $M$ the arithmetic progressions can start from $1$ , the first term of every $A_i$ for $p_i < M$ can be less than $p_i$ this changes and the first term of the $A_i's$ becomes larger and larger comparing with the $p_i's$

I am waiting for any help to this direction, thank you.

Let $M$ be a natural number $M>1$. For every prime $p_i$ not dividing $M$ take an arithmetic progression $A_i=k_i+np_i$ , $n\geq 0$ such that $k_i>p_i^2/M$. Is there any $M$ and some choice of the $k_i's$ such that $\cup A_i= \mathbb{N}$ ?

CONJECTURE: There is not such choice for any $M$

MOTIVATION: It is not hard to see that if there is not such a choice for any $M$ Dirichlet's theorem in arithmetic progressions is an easy consequence. Of course this is much more general. We can modify many well known problems in this way.

For example ,to the previous question, if we take $2$ arithmetic progressions for each $p_i>M$, instead of $1$ with the same condition then we have something stronger than the twin prime conjecture and polignac's conjecture in general, If we take $3$ arithmetic progressions we have something stronger than the problem that infinitely many times exists every logical form with $3$ primes ( for example p, p+2,p+6), etc. To these problems it is enough to take infinitely many $M$, not all...

Secondly it is one of the easiest questions to this spirit and i want to know if we can say something about this kind of questions with known techniques. So

What kind of mathematical techniques we could use to reach this kind of problems?

Can we compute the function of the length of the intervals that we can cover using the primes not dividing $M$ below some $x$ and compare it to the growth function of the number of the primes below some $x$ ?

Notice that even for the primes below $M$ the arithmetic progressions can start from $1$ , the first term of every $A_i$ for $p_i < M$ can be less than $p_i$ this changes and the first term of the $A_i's$ becomes larger and larger comparing with the $p_i's$

I am waiting for any help to this direction, thank you.

Let $M$ be a natural number $M>1$. For every prime $p_i$ not dividing $M$ take an arithmetic progression $A_i=k_i+np_i$ , $n\geq 0$ such that $k_i>p_i^2/M$. Is there any $M$ and some choice of the $k_i's$ such that $\cup A_i= \mathbb{N}$ ?

CONJECTURE: There is not such choice for any $M$

MOTIVATION: It is not hard to see that if there is not such a choice for any $M$ Dirichlet's theorem in arithmetic progressions is an easy consequence. Of course this is much more general. We can modify many well known problems in this way.

For example ,to the previous question, if we take $2$ arithmetic progressions for each $p_i>M$, instead of $1$ with the same condition then we have something stronger than the twin prime conjecture and polignac's conjecture in general, If we take $3$ arithmetic progressions we have something stronger than the problem that infinitely many times exists every logical form with $3$ primes ( for example p, p+2,p+6), etc. To these problems it is enough to take infinitely many $M$, not all...

Secondly it is one of the easiest questions to this spirit and i want to know if we can say something about this kind of questions with known techniques. So

What kind of mathematical techniques we could use to reach this kind of problems?

Can we compute the function of the length of the intervals that we can cover using the primes not dividing $M$ below some $x$ and compare it to the growth function of the number of the primes below some $x$ ?

Notice that even for the primes below $M$ the arithmetic progressions can start from $1$ , the first term of every $A_i$ for $p_i < M$ can be less than $p_i$ this changes and the first term of the $A_i's$ becomes larger and larger comparing with the $p_i's$

I am waiting for any help to this direction, thank you.

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edited Apr 13, 2017 at 12:58

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Let $M$ be a natural number $M>1$. For every prime $p_i$ not dividing $M$ take an arithmetic progression $A_i=k_i+np_i$ , $n\geq 0$ such that $k_i>p_i^2/M$. Is there any $M$ and some choice of the $k_i's$ such that $\cup A_i= \mathbb{N}$ ?

CONJECTURE: There is not such choice for any $M$

MOTIVATION: It is not hard to see that if there is not such a choice for any $M$ Dirichlet's theorem in arithmetic progressions is an easy consequence. Of course this is much more general. We can modify many well known problems in this way.

For example ,to the previous question, if we take $2$ arithmetic progressions for each $p_i>M$, instead of $1$ with the same condition then we have something stronger than the twin prime conjecture and polignac's conjecture in general, If we take $3$ arithmetic progressions we have something stronger than the problem that infinitely many times exists every logical form with $3$ primes ( for example p, p+2,p+6), etc. To these problems it is enough to take infinitely many $M$, not all...

Secondly it is one of the easiest questions to this spirit and i want to know if we can say something about this kind of questions with known techniques. So

What kind of mathematical techniques we could use to reach this kind of problems?

Can we compute the function of the length of the intervals that we can cover using the primes not dividing $M$ below some $x$ and compare it to the growth function of the number of the primes below some $x$ ?

Notice that even for the primes below $M$ the arithmetic progressions can start from $1$ , the first term of every $A_i$ for $p_i < M$ can be less than $p_i$ this changes and the first term of the $A_i's$ becomes larger and larger comparing with the $p_i's$

I am waiting for any help to this direction, thank you.

Let $M$ be a natural number $M>1$. For every prime $p_i$ not dividing $M$ take an arithmetic progression $A_i=k_i+np_i$ , $n\geq 0$ such that $k_i>p_i^2/M$. Is there any $M$ and some choice of the $k_i's$ such that $\cup A_i= \mathbb{N}$ ?

CONJECTURE: There is not such choice for any $M$

MOTIVATION: It is not hard to see that if there is not such a choice for any $M$ Dirichlet's theorem in arithmetic progressions is an easy consequence. Of course this is much more general. We can modify many well known problems in this way.

For example ,to the previous question, if we take $2$ arithmetic progressions for each $p_i>M$, instead of $1$ with the same condition then we have something stronger than the twin prime conjecture and polignac's conjecture in general, If we take $3$ arithmetic progressions we have something stronger than the problem that infinitely many times exists every logical form with $3$ primes ( for example p, p+2,p+6), etc. To these problems it is enough to take infinitely many $M$, not all...

Secondly it is one of the easiest questions to this spirit and i want to know if we can say something about this kind of questions with known techniques. So

What kind of mathematical techniques we could use to reach this kind of problems?

Can we compute the function of the length of the intervals that we can cover using the primes not dividing $M$ below some $x$ and compare it to the growth function of the number of the primes below some $x$ ?

Notice that even for the primes below $M$ the arithmetic progressions can start from $1$ , the first term of every $A_i$ for $p_i < M$ can be less than $p_i$ this changes and the first term of the $A_i's$ becomes larger and larger comparing with the $p_i's$

I am waiting for any help to this direction, thank you.

Let $M$ be a natural number $M>1$. For every prime $p_i$ not dividing $M$ take an arithmetic progression $A_i=k_i+np_i$ , $n\geq 0$ such that $k_i>p_i^2/M$. Is there any $M$ and some choice of the $k_i's$ such that $\cup A_i= \mathbb{N}$ ?

CONJECTURE: There is not such choice for any $M$

MOTIVATION: It is not hard to see that if there is not such a choice for any $M$ Dirichlet's theorem in arithmetic progressions is an easy consequence. Of course this is much more general. We can modify many well known problems in this way.

For example ,to the previous question, if we take $2$ arithmetic progressions for each $p_i>M$, instead of $1$ with the same condition then we have something stronger than the twin prime conjecture and polignac's conjecture in general, If we take $3$ arithmetic progressions we have something stronger than the problem that infinitely many times exists every logical form with $3$ primes ( for example p, p+2,p+6), etc. To these problems it is enough to take infinitely many $M$, not all...

Secondly it is one of the easiest questions to this spirit and i want to know if we can say something about this kind of questions with known techniques. So

What kind of mathematical techniques we could use to reach this kind of problems?

Can we compute the function of the length of the intervals that we can cover using the primes not dividing $M$ below some $x$ and compare it to the growth function of the number of the primes below some $x$ ?

Notice that even for the primes below $M$ the arithmetic progressions can start from $1$ , the first term of every $A_i$ for $p_i < M$ can be less than $p_i$ this changes and the first term of the $A_i's$ becomes larger and larger comparing with the $p_i's$

I am waiting for any help to this direction, thank you.

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edited Dec 1, 2011 at 19:31

Asterios Gkantzounis

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Let $M$ be a natural number $M>1$. For every prime $p_i$ not dividing $M$ take an arithmetic progression $A_i=k_i+np_i$ , $n\geq 0$ such that $k_i>p_i^2/M$. Is there any $M$ and some choice of the $k_i's$ such that $\cup A_i= \mathbb{N}$ ?

CONJECTURE: There is not such choice for any $M$

MOTIVATION: It is not hard to see that if there is not such a choice for any $M$ Dirichlet's theorem in arithmetic progressions is an easy consequence. Of course this is much more general. We can modify many well known problems in this way.

For example ,to the previous question, if we take $2$ arithmetic progressions for each $p_i$$p_i>M$, instead of $1$ with the same condition then we have something stronger than the twin prime conjecture and polignac's conjecture in general, If we take $3$ arithmetic progressions we have something stronger than the problem that infinitely many times exists every logical form with $3$ primes ( for example p, p+2,p+6), etc. To these problems it is enough to take infinitelyinfinitely many $M$, not for all...and more

Does anyone needs a proof for these?

Secondly it is one of the easiest questions to this spirit and i want to know if we can say something about this kind of questions with known techniques. So

What kind of mathematical techniques we could use to reach this kind of problems?

Can we compute the function of the length of the intervals that we can cover using the primes not dividing $M$ below some $x$ and compare it to the growth function of the number of the primes below some $x$ ?

Notice that even for the primes below $M$ the arithmetic progressions can start from $1$ , the first term of every $A_i$ for $p_i < M$ can be less than $p_i$ this changes and the first term of the $A_i's$ becomes larger and larger comparing with the $p_i's$

I am waiting for any help to this direction, thank you.

Let $M$ be a natural number $M>1$. For every prime $p_i$ not dividing $M$ take an arithmetic progression $A_i=k_i+np_i$ , $n\geq 0$ such that $k_i>p_i^2/M$. Is there any $M$ and some choice of the $k_i's$ such that $\cup A_i= \mathbb{N}$ ?

CONJECTURE: There is not such choice for any $M$

MOTIVATION: It is not hard to see that if there is not such a choice for any $M$ Dirichlet's theorem in arithmetic progressions is an easy consequence. Of course this is much more general. We can modify many well known problems in this way.

For example if we take $2$ arithmetic progressions for each $p_i$ then we have something stronger than the twin prime conjecture and polignac's conjecture in general, If we take $3$ arithmetic progressions we have something stronger than the problem that infinitely many times exists every logical form with $3$ primes etc. To these problems it is enough to take infinitely many $M$, not for all...and more

Does anyone needs a proof for these?

Secondly it is one of the easiest questions to this spirit and i want to know if we can say something about this kind of questions with known techniques. So

What kind of mathematical techniques we could use to reach this kind of problems?

Can we compute the function of the length of the intervals that we can cover using the primes not dividing $M$ below some $x$ and compare it to the growth function of the number of the primes below some $x$ ?

Notice that even for the primes below $M$ the arithmetic progressions can start from $1$ , the first term of every $A_i$ for $p_i < M$ can be less than $p_i$ this changes and the first term of the $A_i's$ becomes larger and larger comparing with the $p_i's$

I am waiting for any help to this direction, thank you.

Let $M$ be a natural number $M>1$. For every prime $p_i$ not dividing $M$ take an arithmetic progression $A_i=k_i+np_i$ , $n\geq 0$ such that $k_i>p_i^2/M$. Is there any $M$ and some choice of the $k_i's$ such that $\cup A_i= \mathbb{N}$ ?

CONJECTURE: There is not such choice for any $M$

MOTIVATION: It is not hard to see that if there is not such a choice for any $M$ Dirichlet's theorem in arithmetic progressions is an easy consequence. Of course this is much more general. We can modify many well known problems in this way.

For example ,to the previous question, if we take $2$ arithmetic progressions for each $p_i>M$, instead of $1$ with the same condition then we have something stronger than the twin prime conjecture and polignac's conjecture in general, If we take $3$ arithmetic progressions we have something stronger than the problem that infinitely many times exists every logical form with $3$ primes ( for example p, p+2,p+6), etc. To these problems it is enough to take infinitely many $M$, not all...

Secondly it is one of the easiest questions to this spirit and i want to know if we can say something about this kind of questions with known techniques. So

What kind of mathematical techniques we could use to reach this kind of problems?

Can we compute the function of the length of the intervals that we can cover using the primes not dividing $M$ below some $x$ and compare it to the growth function of the number of the primes below some $x$ ?

Notice that even for the primes below $M$ the arithmetic progressions can start from $1$ , the first term of every $A_i$ for $p_i < M$ can be less than $p_i$ this changes and the first term of the $A_i's$ becomes larger and larger comparing with the $p_i's$

I am waiting for any help to this direction, thank you.

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