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...



Conjecture about a sequence of natural numbers, such that, $\forall n : A_n<P_n<A_{n+1}$
Indeed, as Tony Huynh expects, your conjecture follows from the prime triples conjecture. That conjecture imples that there exist infinitely many $k$ such that $k,6k-1,6k+1$ are all prime, and infinitely many $m$ such that $12m-1,12m+1$ are both prime. Whenever $P\_n$ is one of these $6k+1$ primes, you're forced to take $A\_n = 6k$, so $F(A\_n)=3$; similarly, whenever $P\_n$ is one of the $12m+1$ primes, we have $F(A\_n) = F(12m) \ge 4$.


