The number of prime factors of a natural number

Question

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?

Answer 1

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 $\prod e_j!$.

Answer 2

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 $A257091(7776) - A257091(7776) = 12 - 10 = 2$.