Sequence of numbers whose factorial on prime factorisation contains prime powers of prime numbers, whose power is greater than $1$ or contains multiplicity of one for all prime numbers less than equal to number.
Sequence: $1, 2, 3, 4, 5, 8, 14, \ldots$
If $k$(prime) is to be checked for its presence in sequence, then if $k-1$ is in sequence, then $k$ is also in sequence.
$a(3)=4$ as $4!=(2^3)\cdot(3^1)$. Since multiplicity of $2$ is $3$, which is prime, and multiplicity of $3$ is $1$, so $4$ finds it's presence in the sequence.
For all large enough $k$, there will exist a prime $p$ with $4p\le k<5p\ll p^2$). Then we will have $p^4\mid k!$, but $p^5\nmid k!$ and $k$ is not in your sequence. The existence of such $p$ follows asymptotically from the Prime Number Theorem. However, we even have some explicit bounds such as [Dusart 2010] $$\frac x{\ln x-1}<\pi(x)<\frac x{\ln x-1.1}\qquad\text{if }x\ge 60184. $$ This makes $\pi(\tfrac k4)>\pi(\tfrac k5)$ at least for all $k$ with ($k\ge 5\cdot 60184$ and) $$ \frac{\frac k4}{\ln\frac k4-1}>\frac{\frac k5}{\ln\frac k5-1.1}$$ and ths inequality is equivalent to $$ k>\frac{5^5e^{1.5}}{4^4}\approx 55.$$ Hence your sequence is finite and all terms in it are $< 300920$. By explicitly testing the few $k$ in the remaining range, we find that $\pi(\frac k4)>\pi(\frac k4)$ already holds for all $$ k>115.$$ Meanwhile the range is short enough to testing your condition completely. As it turns out, the full sequence is $$ 1,2,3,4,5,8,14.$$