With Dirichlet's Divisor summatory function the mean divisor growth can be determined.
$$D(x)=\frac{1}{x} \sum_{n=1}^x d(x)= \log(x)+2 \gamma -1+ \mathcal{O}\!\left(\frac{1}{\sqrt{x}}\right)$$
Where $\gamma$ is the Euler's constant and $d(n)$ is the divisor function. Do there exist expressions where the Divisor summatory function is expressed in: $D(x=odd)$ and $D(x=even)$?
Background for question error growth "Wave Divisor Function":
Not sure if that's the question but if one wants the expressions for $d_{odd}(x)=\sum_{2k+1 \le x}d(2k+1)$ and $d_{even}$, one can easily get them from the recurrence:
$d_{even}(x)=2d_r(x/2)-d_r(x/4), d_{odd}(x)=d_r(x)-2d_r(x/2)+d_r(x/4)$ where $d_r(x)=xD(x)$ the usual divisor sum - these follow from the relation $d(4q)=2d(2q)-d(q)$ for any integer $q \ge 1$
In particular one gets that $D_{odd}=\frac{1}{4}\log x+\frac{1}{4}(2\gamma -1 + \log 4)+ O(1/\sqrt x)$ and $D_{even}=\frac{3}{4}\log x+\frac{3}{4}(2\gamma -1) -\frac{1}{4}\log 4+ O(1/\sqrt x)$
One can also apply Perron to get the principal terms - for the usual divisor sum it is the residue at $1$ of $\zeta^2(s)x^s/s$, while here one can immediately notice that $\zeta_{odd}(s)=(1-2^{-s})\zeta(s)$ so squaring $\zeta_{odd}(s)^2x^s=\zeta^2(s)x^s-2\zeta^2(s)(x/2)^s+\zeta^2(s)(x/4)^s$ from which the above follow easily too