Let $a(n)$ be the $k$-ary tuple of the exponents of the prime factorization of $n$. For example,


$$a(5184)=a(2^{6}⋅3^{4})=(6, 4), a(65536)=a(2^{16})=(16).$$


Formally, let $p\_{1}^{a\_{1}}, p\_{2}^{a\_{2}}, \ldots, p\_{k}^{a\_{k}}$ be the prime factorization of a positive integer $n$, then


$$a(n) = a(p\_{1}^{a\_{1}}, p\_{2}^{a\_{2}}, \ldots, p\_{k}^{a\_{k}}) = (a\_{1}, a\_{2}, \ldots, a\_{k}).$$


Let $|\ldots|$ be the bit-length function. For example,


$$|3| = 2,\qquad |10| = 4,\qquad |256| = 9.$$


We define the bit-length of a $k$-ary tuple $(a\_{1}, a\_{2}, \ldots, a\_{k})$ as $\sum\_{i = 1}^{k}(|a\_{i}| + 1)$. 


Question: Is $|a(n)| = O(log(|n|))$?


Thanks in advance!



Upper bound for tuple of exponents of prime factorization
It is unfortunate to use the notation $|\cdot|$ for the bit-length function, because in number theory (where this post belongs) it denotes the absolute value function. At any rate, in standard notation, you are asking if
$$ \sum\_{i=1}^k\lfloor 1+\log\_2 a\_i\rfloor\ll\log\log n. $$
The answer is no, even if we replace the terms on the left hand side by $\log\_2 a\_i$. Indeed, let $p\_2<p\_3<\dots$ be the sequence of primes, and consider $n=p\_1^2\dots p\_k^2$. Then by the Prime Number Theorem we have
$$ \log n=2(\log p\_1+\dots+\log p\_k)\sim 2p\_k\sim 2k\log k,$$
hence
$$ \sum\_{i=1}^k\log\_2 a\_i = k \sim \frac{\log n}{2\log\log n},$$
which grows much faster than $\log\log n$.


