It seems to be true that OEIS sequences A001222 and A257091 are closely related. First one is the number of prime divisors of n counted with multiplicity. The second one is the logarithms to the base 5 of the denominators of the Dirichlet series of zeta(s)^(1/5). Actually first terms of these sequences satisfy A257091(n)-A001222(n)= 1 if n is a multiple of 32, and 0 otherwise. Is it true, obvious, known or somthing else? 



The number of prime factors of a natural number
Robert Israel's guess is almost correct, actually for $n=\prod p\_j^{e\_j}$ we have
$$
A257091(n)−A001222(n)=\sum\_j \nu\_5(e\_j!)=\sum\_j \sum\_{m=1}^{\infty} \lfloor e\_j/5^m\rfloor
$$
(where $\nu\_5(N)$ denotes the largest $k$ for which $5^k$ divides $N$).


Indeed, expanding as binomial each Euler multiple in $\zeta^{1/5}$ we get $$(1-p^{-s})^{-1/5}=\sum\_{k=0}^\infty (-1)^k\binom{-1/5}k p^{-ks}.$$
Thus if $n=\prod p\_i^{e\_i}$, coefficient of $n^{-s}$ in $\zeta(s)^{1/5}$ equals $$\prod\_j (-1)^{e\_j}\binom{-1/5}{e\_j}=\prod\_{j}\frac{1(1+5)(1+2\cdot 5)\dots (1+(e\_j-1)5)}{5^{e\_j} e\_j!}$$
This rational number have only 5's in the denominator (because for other prime $p$ we may replace $-1/5$ to some integer like $(p^{4n}-1)/5$ without changing $p$-adic valuation of all multiples $-1/5-t$, $t=0,1,\dots,e\_j-1$.) And 5 arises in the power $\sum e\_k$ plus exponent of 5 in $\prod e\_j!$.


