show that number theory inequality $a_{1}+a_{2}+\cdots+a_{k}<\frac{5}{2}n$

Question

Let $n,a_{1},a_{2},\cdots,a_{k}$ be postive integers and at least greater than $1$,and such $$(a_{1})!\cdot(a_{2})!\cdots(a_{k})!|n!$$

show that $$a_{1}+a_{2}+\cdots+a_{k}<\dfrac{5}{2}n$$

I have know prove $k=2$ see Erdos 1968problem But unfortunately I am looking for much smaller bound Any idea would be helpful.

I think use $$a_{1}+a_{2}+\cdots+a_{k}-s_{2}(a_{1})-s_{2}(a_{2})-\cdots-s_{k}(a_{k})<n-s_{2}(n)<n$$ where $s_{p}(n)$ is the sum of the digits of $n$ when written in base $p$

Add it. the constant $\dfrac{5}{2}$ is best?

Answer 1

If there is a counterexample $a_1≥a_2≥⋯≥a_k$ and $a_1≥4$, then new sequence generated by replacing $a_1$ with $a_1−2$ and $2$ is also counterexample. Therefore, we can assume every $a_i$ to be $2$ or $3$.

Because the exponent of $2$ in the factorization of $n!$ is less than $\sum_{i=0}^\infty\frac{n}{2^i}=n$, $k< n$.

Because the exponent of $3$ in the factorization of $n!$ is less than $\sum_{i=0}^\infty\frac{n}{3^i}=\frac{n}{2}$, less than $\frac{n}{2}$ of $a_i$s are equal to or larger than $3$.

Therefore, we have$$\sum_{i=1}^ka_i=2k+\text{#}(a_i=3)<2n+\frac{n}{2}=\frac{5n}{2}$$and it is done!