Occurance of Prime Factor 3 Let $P^{3}(a, b)$ return the sum of the frequency of Prime Factor 3 of all Integers in $[a, b]$, e.g: $P^{3}(5, 10) = 3, P^{3}(12, 20) = 4, ...$ 
Then

$\forall x>10 \in \mathbb N: P^{3}(2^{x}, 2^{x+1}) > P^{3}(1, 2^{x}) + 1$ 

?

Partial
Notice that:
$$P^3(a,b)=P^3(b,1)-P^3(a,1)$$
So your question becomes:
$$P^3(2^{k+1},1)>2P^3(2^k,1)+1$$
There is a standard formula for $P^3(N,1)$:
$$ P^3(N,1)=\sum_{i=1}^{\lfloor \log_3(N) \rfloor} \lfloor \frac{N}{3^i} \rfloor $$
So:
$$\sum_{i=1}^{\lfloor \log_3(2^{k+1}) \rfloor} \lfloor \frac{2^{k+1}}{3^i} \rfloor \geq 2\sum_{i=1}^{\lfloor \log_3(2^{k}) \rfloor} \lfloor \frac{2^{k}}{3^i}\rfloor+1$$
Notice that $\lfloor 2 f(x) \rfloor \geq 2\lfloor f(x) \rfloor$ so:
$$\sum_{i=1}^{\lfloor \log_3(2^{k+1}) \rfloor} \lfloor \frac{2^{k+1}}{3^i} \rfloor \geq 2 \sum_{i=1}^{\lfloor \log_3(2^{k+1}) \rfloor} \lfloor \frac{2^{k}}{3^i} \rfloor \geq 2 \sum_{i=1}^{\lfloor \log_3(2^{k}) \rfloor} \lfloor \frac{2^{k}}{3^i}\rfloor $$
And this is a pretty good partial result. 
:/
