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

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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...