Let the function $g:\Bbb{N}\to\Bbb{N}$ be defined as $$g(n)=\sum_{d|n \; d\ne n}g(d)$$ with $g(1)=1$, how can we evaluate a sum like $$\sum_{i=0}^\infty{g(15^i)\over15^i} \tag1$$ Need we find a closed form for $g(n)$? I have tried to do this with the Möbius inversion formula, but it has not gotten me anywhere. If we add $g(n)$ to both sides of the definition $$2g(n)=\sum_{d|n}g(d) $$ I'm unsure if it is acceptable to use the Möbius inversion formula on $2g(n)$, but if it is then $$g(n)=\sum_{d|n} \mu(d)\cdot2g\left(\frac nd\right) \Rightarrow g(n)=-2\sum_{d|n \; d\ne n}\mu(d) \,g\left(\frac nd\right)$$ but I don't think this is very useful. It also doesn't help that $g$ is not multiplicative and I can discern no easy pattern from a table of its values here. Clearly $g(p)=1$ and the more divisors a number has, the greater the output is. How can I quantify this exactly?
EDIT
I have found that $g(n)$ is the same as the number of ordered factorizations of $n$, so in $(1)$ $$\begin{align} g(15^i)&=g(3^i5^i)=2^{2i-1}\sum_{k=0}^i {i\choose k}^22^{-k}\\ \Rightarrow 1+\sum_{i=1}^\infty{g(15^i)\over15^i} &=1+\sum _{n=1}^{\infty } 2^{2 n-1} 15^{-n} \, _2F_1\left(-n,-n;1;\frac{1} {2}\right) \end{align} $$ Where $F_1$ is a hypergeometric function. Based on some partial sums I think this goes to $\frac{11}{7}$ How can we show this by hand?
Just the start of an idea, really.
Define $h(1)=1$ and $h(n)=-1$ if $n>1$. Then:
$$(g*h)(n)=\begin{cases}1&n=1\\0&n>1\end{cases}=I(n)$$
where $I(n)$ is the identity under the Dirichlet convolution.
Now, as it turns out $h(n)=2I(n)-\mathbf 1$.
So, $g*(2I-\mathbf 1)=I$ or $g*(I-\mathbf 1/2)=\frac{1}{2}I$.
Let $f_0=I$ and $f_{k+1}(n)=(1*f_k)(n)=\sum_{d\mid n} f_k(d)$.
Then $f_k$ is multiplicative, by induction, and we can show that for $k>1$:
$$f_k(p^m)=\binom{m+k-1}{k-1}$$
Let $u = \sum_{k=0}^{\infty} \frac{1}{2^k}f_k$.
Then $u*(I-\mathbf 1/2)=I$, by the usual power series argument, when the series converges, so $(u/2)*(2I-\mathbf 1)=I$, so $u/2=g$
For example:
$$\begin{align}g(p^m)&=\frac{1}{2}\sum_{k=1}^\infty \frac{1}{2^{k}}\binom{m+k-1}{k-1}\\ &=\frac{1}{4}\sum_{k=0}^{\infty} \frac{1}{2^k}\binom{m+k}{k}\\ &=\frac{1}{4}\frac{1}{\left(1-\frac12\right)^{m+1}}\\ &=2^{m-1} \end{align} $$
Which is the value you see when you just compute by hand.
Using the above formula for $g$, since $f_k$ is multiplicative, $f_k(15^i)=f_k(3^i)f_k(5^i)$, and we have:
$$g(15^i)=\frac{1}{4}\sum_{k=0}^\infty \frac{1}{2^{k}}\binom{i+k}{k}^2$$
This means you might want to look at the double power series:
$$\sum_{i,k=0}^\infty \binom{i+k}{k}^2 x^iy^k$$
where $x=\frac{1}{15}$ and $y=\frac{1}{2}$.
But more generally, that series for $g(n)$ is not symmetric for all $n$.
For example:
$$\sum \frac{g(12^i)}{12^i} = \sum_{i,k=0}^\infty \binom{i+k}{k}\binom{2i+k}{k} \frac{1}{2^k12^i}$$
I'm not sure that gets you any closer to a value for the series you want...