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?

$\begingroup$ What is the reason to choose $5$ as compared with A046644, A256689, A256691 etc? $\endgroup$ – AHusain May 13 '16 at 5:59

$\begingroup$ The reason is very simple: A257091 coincide with A001222 in first 31 terms. $\endgroup$ – Alexander Buturlakin May 17 '16 at 5:02
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 $$(1p^{s})^{1/5}=\sum_{k=0}^\infty (1)^k\binom{1/5}k p^{ks}.$$ Thus if $n=\prod p_i^{e_i}$, coefficient of $n^{s}$ in $\zeta(s)^{1/5}$ equals $$\prod_j (1)^{e_j}\binom{1/5}{e_j}=\prod_{j}\frac{1(1+5)(1+2\cdot 5)\dots (1+(e_j1)5)}{5^{e_j} e_j!}$$ 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 $1/5t$, $t=0,1,\dots,e_j1$.) And 5 arises in the power $\sum e_k$ plus exponent of 5 in $\prod e_j!$.
The first two exceptions to your rule: $A257091(243)  A001222(243) = 6  5 = 1$ and $A257091(486)  A001222(486) = 7  6 = 1$, but $243$ and $486$ are not multiples of $32$.
EDIT: Of course $243 = 3^5$. From computing the results up to $n=10000$, it looks to me like if $n = \prod_{j} p_j^{e_j}$ is the prime factorization of $n$, $A257091(n)  A001222(n) = \sum_j \lfloor e_j/5 \rfloor$. Thus $7776 = 2^5 \cdot 3^5$ and $A257091(7776)  A257091(7776) = 12  10 = 2$.