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As one can easily prove http://math.stackexchange.com/questions/20564/sums-of-square-free-numbers-is-this-conjecture-equivalent-to-goldbachs-conjec every integer greater than $1$ is a sum of two squarefree numbers.

Can we have bounds for the length of these numbers? I write $(n,m)$ to denote the sum of a squarefree number of n prime factors to one of m prime factors, when $n$ or $m$ $=0$ then i mean that the summand is $1$.

Chenn's theorem asserts that for large enough even numbers the length $(2,1)$ is enough Goldbach's conjecture says that $(1,1)$ would be enough too.

CONJECTURE: Every odd number can be written as a sum of two squarefree numbers of length at most $(2,1)$ (meaning as a sum of a prime and a double of a prime or a sum of a prime plus 2 or as a sum of 1 plus a double of a prime)

Questions

1 Is there any easy counterexample?

2 do i really need the prime plus 2 or the 1 plus the double of a prime in order to have all the odd numbers? It seems too difficult to me to prove that i do not need them.

3 What is the relation of this conjecture to Goldbach's conjecture? does the one implies the other?

I apologise for the elementary style of my question , i think that this conjecture is well known but i haven't met it. If it is well known maybe it is known the relation to the Goldbach Conjecture. Maybe i miss something obvious...

NOTE: From the second question we have one new conjecture

CONJECTURE:Every prime $p$ is $p=p1+2(p2-1)$ and $p=p3+1/2(p4-1)$ for some primes $p1,p2,p3,p4$ . Of course someone could ask many questions about these form for the consecutive primes ,etc.

EDIT:after asking this question i found this related article en.wikipedia.org/wiki/Lemoine%27s_conjecture at wikipedia.

ADDED i think that one can easily see that if every even number is the sum of a prime and a Sophie Germain prime or his pair (meaning a prime of the form $2p+1$ ) this would be too strong to implie both do we have a counterexample to this??

As one can easily prove https://math.stackexchange.com/questions/20564/sums-of-square-free-numbers-is-this-conjecture-equivalent-to-goldbachs-conjec every integer greater than $1$ is a sum of two squarefree numbers.

Can we have bounds for the length of these numbers? I write $(n,m)$ to denote the sum of a squarefree number of n prime factors to one of m prime factors, when $n$ or $m$ $=0$ then i mean that the summand is $1$.

Chenn's theorem asserts that for large enough even numbers the length $(2,1)$ is enough Goldbach's conjecture says that $(1,1)$ would be enough too.

CONJECTURE: Every odd number can be written as a sum of two squarefree numbers of length at most $(2,1)$ (meaning as a sum of a prime and a double of a prime or a sum of a prime plus 2 or as a sum of 1 plus a double of a prime)

Questions

1 Is there any easy counterexample?

2 do i really need the prime plus 2 or the 1 plus the double of a prime in order to have all the odd numbers? It seems too difficult to me to prove that i do not need them.

3 What is the relation of this conjecture to Goldbach's conjecture? does the one implies the other?

I apologise for the elementary style of my question , i think that this conjecture is well known but i haven't met it. If it is well known maybe it is known the relation to the Goldbach Conjecture. Maybe i miss something obvious...

NOTE: From the second question we have one new conjecture

CONJECTURE:Every prime $p$ is $p=p1+2(p2-1)$ and $p=p3+1/2(p4-1)$ for some primes $p1,p2,p3,p4$ . Of course someone could ask many questions about these form for the consecutive primes ,etc.

EDIT:after asking this question i found this related article en.wikipedia.org/wiki/Lemoine%27s_conjecture at wikipedia.

ADDED i think that one can easily see that if every even number is the sum of a prime and a Sophie Germain prime or his pair (meaning a prime of the form $2p+1$ ) this would be too strong to implie both do we have a counterexample to this??

As one can easily prove http://math.stackexchange.com/questions/20564/sums-of-square-free-numbers-is-this-conjecture-equivalent-to-goldbachs-conjec every integer greater than $1$ is a sum of two squarefree numbers.

Can we have bounds for the length of these numbers? I write $(n,m)$ to denote the sum of a squarefree number of n prime factors to one of m prime factors, when $n$ or $m$ $=0$ then i mean that the summand is $1$.

Chenn's theorem asserts that for large enough even numbers the length $(2,1)$ is enough Goldbach's conjecture says that $(1,1)$ would be enough too.

CONJECTURE: Every odd number can be written as a sum of two squarefree numbers of length at most $(2,1)$ (meaning as a sum of a prime and a double of a prime or a sum of a prime plus 2 or as a sum of 1 plus a double of a prime)

Questions

1 Is there any easy counterexample?

2 do i really need the prime plus 2 or the 1 plus the double of a prime in order to have all the odd numbers? It seems too difficult to me to prove that i do not need them.

3 What is the relation of this conjecture to Goldbach's conjecture? does the one implies the other?

I apologise for the elementary style of my question , i think that this conjecture is well known but i haven't met it. If it is well known maybe it is known the relation to the Goldbach Conjecture. Maybe i miss something obvious...

NOTE: From the second question we have one new conjecture

CONJECTURE:Every prime $p$ is $p=p1+2(p2-1)$ and $p=p3+1/2(p4-1)$ for some primes $p1,p2,p3,p4$ . Of course someone could ask many questions about these form for the consecutive primes ,etc.

EDIT:after asking this question i found this related article en.wikipedia.org/wiki/Lemoine%27s_conjecture at wikipedia.

As one can easily prove http://math.stackexchange.com/questions/20564/sums-of-square-free-numbers-is-this-conjecture-equivalent-to-goldbachs-conjec every integer greater than $1$ is a sum of two squarefree numbers.

Can we have bounds for the length of these numbers? I write $(n,m)$ to denote the sum of a squarefree number of n prime factors to one of m prime factors, when $n$ or $m$ $=0$ then i mean that the summand is $1$.

Chenn's theorem asserts that for large enough even numbers the length $(2,1)$ is enough Goldbach's conjecture says that $(1,1)$ would be enough too.

CONJECTURE: Every odd number can be written as a sum of two squarefree numbers of length at most $(2,1)$ (meaning as a sum of a prime and a double of a prime or a sum of a prime plus 2 or as a sum of 1 plus a double of a prime)

Questions

1 Is there any easy counterexample?

2 do i really need the prime plus 2 or the 1 plus the double of a prime in order to have all the odd numbers? It seems too difficult to me to prove that i do not need them.

3 What is the relation of this conjecture to Goldbach's conjecture? does the one implies the other?

I apologise for the elementary style of my question , i think that this conjecture is well known but i haven't met it. If it is well known maybe it is known the relation to the Goldbach Conjecture. Maybe i miss something obvious...

NOTE: From the second question we have one new conjecture

CONJECTURE:Every prime $p$ is $p=p1+2(p2-1)$ and $p=p3+1/2(p4-1)$ for some primes $p1,p2,p3,p4$ . Of course someone could ask many questions about these form for the consecutive primes ,etc.

EDIT:after asking this question i found this related article en.wikipedia.org/wiki/Lemoine%27s_conjecture at wikipedia.

ADDED i think that one can easily see that if every even number is the sum of a prime and a Sophie Germain prime or his pair (meaning a prime of the form $2p+1$ ) this would be too strong to implie both do we have a counterexample to this??

As one can easily prove http://math.stackexchange.com/questions/20564/sums-of-square-free-numbers-is-this-conjecture-equivalent-to-goldbachs-conjec every integer greater than $1$ is a sum of two squarefree numbers.

Can we have bounds for the length of these numbers? I write $(n,m)$ to denote the sum of a squarefree number of n prime factors to one of m prime factors, when $n$ or $m$ $=0$ then i mean that the summand is $1$.

Chenn's theorem asserts that for large enough even numbers the length $(2,1)$ is enough Goldbach's conjecture says that $(1,1)$ would be enough too.

CONJECTURE: Every odd number can be written as a sum of two squarefree numbers of length at most $(2,1)$ (meaning as a sum of a prime and a double of a prime or a sum of a prime plus 2 or as a sum of 1 plus a double of a prime)

Questions

1 Is there any easy counterexample?

2 do i really need the prime plus 2 or the 1 plus the double of a prime in order to have all the odd numbers? It seems too difficult to me to prove that i do not need them.

3 What is the relation of this conjecture to Goldbach's conjecture? does the one implies the other?

I apologise for the elementary style of my question , i think that this conjecture is well known but i haven't met it. If it is well known maybe it is known the relation to the Goldbach Conjecture. Maybe i miss something obvious...

NOTE: From the second question we have one new conjecture

CONJECTURE:Every prime $p$ is $p=p1+2(p2-1)$ and $p=p3+1/2(p4-1)$ for some primes $p1,p2,p3,p4$ and of course a similar question for the Goldbach conjecture is leading to this conjecture: every prime p is $p=p5+(p6-1)$ for some primes $p5,p6$ (except from the obvious counterexamples for the first primes) . Of course someone could ask many questions about these form for the consecutive primes ,etc.

EDIT:after asking this question i found this related article en.wikipedia.org/wiki/Lemoine%27s_conjecture at wikipedia.

As one can easily prove http://math.stackexchange.com/questions/20564/sums-of-square-free-numbers-is-this-conjecture-equivalent-to-goldbachs-conjec every integer greater than $1$ is a sum of two squarefree numbers.

Can we have bounds for the length of these numbers? I write $(n,m)$ to denote the sum of a squarefree number of n prime factors to one of m prime factors, when $n$ or $m$ $=0$ then i mean that the summand is $1$.

Chenn's theorem asserts that for large enough even numbers the length $(2,1)$ is enough Goldbach's conjecture says that $(1,1)$ would be enough too.

CONJECTURE: Every odd number can be written as a sum of two squarefree numbers of length at most $(2,1)$ (meaning as a sum of a prime and a double of a prime or a sum of a prime plus 2 or as a sum of 1 plus a double of a prime)

Questions

1 Is there any easy counterexample?

2 do i really need the prime plus 2 or the 1 plus the double of a prime in order to have all the odd numbers? It seems too difficult to me to prove that i do not need them.

3 What is the relation of this conjecture to Goldbach's conjecture? does the one implies the other?

I apologise for the elementary style of my question , i think that this conjecture is well known but i haven't met it. If it is well known maybe it is known the relation to the Goldbach Conjecture. Maybe i miss something obvious...

NOTE: From the second question we have one new conjecture

CONJECTURE:Every prime $p$ is $p=p1+2(p2-1)$ and $p=p3+1/2(p4-1)$ for some primes $p1,p2,p3,p4$   . Of course someone could ask many questions about these form for the consecutive primes ,etc.

EDIT:after asking this question i found this related article en.wikipedia.org/wiki/Lemoine%27s_conjecture at wikipedia.