After a small rephrasing, here it is together with the proof, hopefully correct.
PROPOSITION:
For $n \in \mathbb{N}$, let $n+1=p_0^{j_0} \ldots, p_i^{j_i}$, where $p_0, \ldots, p_i$ are the prime factors of $n+1$ in increasing order. Let $[q_0, \ldots, q_m]$ be the list of the positions in the sequence of primes starting with $2 $ of these factors with their respective multiplicities. Let $[q_1-q_0,\ldots, q_m-q_{m-1}]$ be the list of their consecutive differences, clearly all $\geq 0$. Let $b(x)=\lfloor log_2(x+1) \rfloor$.
Then $b(n) \geq \sum_{l=1}^m b(q_l-q_{l-1})$.
PROOF:
For $n=0$ equality holds, the list on the right (factoring of $1$) being empty.
For $n>0$ observe that
$\sum_{l=1}^m b(q_l-q_{l-1})
= \sum_{l=1}^m \lfloor log_2(1+q_l-q_{l-1}) \rfloor
\leq \lfloor \sum_{l=1}^m log_2(1+q_l-q_{l-1}) \rfloor
\leq \lfloor \sum_{l=1}^m log_2(q_l) \rfloor$
$= \lfloor log_2(\prod_{l=1}^m q_i) \rfloor
\leq \lfloor log_2(\prod_{l=0}^m p_i) \rfloor
= \lfloor log_2(n+1)\rfloor$ $= b(n)$
This opens a more general question: what conditions are sufficient for a bijection from $\mathbb{N}$ to finite sequences of $\mathbb{N}$ for a similar property to hold? A necessary one (that the bijection derived from primes satisfies) seems to be that all elements in the sequence corresponding to $n$ are smaller then $n$.