Conjecture - no natural number $k$ exists such that:

1. $P$ is the sequence of all primes starting from the $k$th prime

2. $A$ is a sequence of natural numbers such that:

• $\forall n : A_n<P_n<A_{n+1}$

• $\forall n : F(A_n) \leq F(A_{n+1})$, where $F(x)$ is the number of primes in the factorization of $x$

Take $k=3$ for example:

• $P=5,7,11,13,17,19,\dots$
• Here, because of $11$ and $13$, we must use $12 \in A$
• Then, because $F(12)=3$, we must use $16 \in A$
• Then, because of $17$ and $19$, we must use $18 \in A$
• But $F(18)<F(16)$, so we are unable to define $A$

How can I go about proving this conjecture?

The only lead I have on it, is that the sequence $A$ must include the "mid value" of every pair of twin primes (starting from the $k$th prime).

Thanks

Conjecture - no natural number $k$ exists such that:

1. $P$ is the sequence of all primes starting from the $k$th prime

2. $A$ is a sequence of natural numbers such that:

• $\forall n : A_n<P_n<A_{n+1}$

• $\forall n : F(A_n) \leq F(A_{n+1})$, where $F(x)$ is the number of primes in the factorization of $x$

Take $k=3$ for example:

• $P=5,7,11,13,17,19,\dots$
• Here, because of $11$ and $13$, we must use $12 \in A$
• Then, because $F(12)=3$, we must use $16 \in A$
• Then, because of $17$ and $19$, we must use $18 \in A$
• But $F(18)<F(16)$, so we are unable to define $A$

How can I go about proving this conjecture?

The only lead I have on it, is that the sequence $A$ must include the "mid value" of every pair of twin primes (starting from the $k$th prime).

Thanks

As a side note, I should mention that the motivation behind this, is an interest in finding a general definition for a sequence of natural numbers, such that between every two consecutive elements there exists exactly one prime number. So any insights or notions on this issue will also be appreciated...