Estimating $\sum\limits_{d\mid n}{d+a\choose b}$ Is there any way of estimating a sum like
$$\sum_{d\mid n}{d+a\choose b},$$
for positive integers $a$ and $b$? For example, in the OEIS we find that
$$\begin{align*}
\sum_{d\mid n}{d+1\choose 2}
&=\frac{1}{2}(\sigma_1(n)+\sigma_2(n)),\\
\sum_{d\mid n}{d+2\choose 3}
&=\frac{1}{6}(2\sigma_1(n)+3\sigma_2(n)+\sigma_3(n)),\\
\sum_{d\mid n}{d+3\choose 4}
&=\frac{1}{24}(6\sigma_1(n)+11\sigma_2(n)+6\sigma_3(n)+\sigma_4(n)).
\end{align*}$$
I fail to spot the pattern for $d+4$. Anyway, what I'm looking for is more of a simple asymptotic expression (even though the above examples are kinda cute!).
 A: $\sum_{d\mid n}{d+a\choose b}
$
The unsigned Stirling numbers
of the first kind,
$c(n, k)$
are defined by
$\prod_{k=0}^{n-1} (x+k)
=\sum_{k=0}^n c(n, k)x^k
$.
(see https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind)
Therefore
$\begin{array}\\
\rho_{a, b}(n) 
&=\sum_{d\mid n}{d+a\choose b}\\
&=\sum_{d\mid n}{d+a\choose b}\\
&=\dfrac1{b!}\sum_{d\mid n}\dfrac{(d+a)!}{(d+a-b)!}\\
&=\dfrac1{b!}\sum_{d\mid n}\prod_{k=a-b+1}^a (d+k)\\
&=\dfrac1{b!}\sum_{d\mid n}\prod_{k=0}^{b-1} (d+a-b+1+k)\\
&=\dfrac1{b!}\sum_{d\mid n}\sum_{k=0}^{b}(d+a-b+1)^kc(b, k)\\
&=\dfrac1{b!}\sum_{d\mid n}\sum_{k=0}^{b}c(b, k)\sum_{j=0}^k\binom{k}{j}d^j(a-b+1)^{k-j}\\
&=\dfrac1{b!}\sum_{d\mid n}\sum_{k=0}^{b}c(b, k)\sum_{j=0}^k\binom{k}{j}d^jr^{k-j}
\quad r = a-b+1\\
&=\dfrac1{b!}\sum_{k=0}^{b}c(b, k)\sum_{j=0}^k\binom{k}{j}r^{k-j}\sum_{d\mid n}d^j\\
&=\dfrac1{b!}\sum_{k=0}^{b}c(b, k)\sum_{j=0}^k\binom{k}{j}r^{k-j}\sigma_j(n)\\
&=\dfrac1{b!}\sum_{j=0}^n\sigma_j(n)\sum_{k=j}^{b}c(b, k)\binom{k}{j}r^{k-j}\\
&=\dfrac1{b!}\sum_{j=0}^n\sigma_j(n)\sum_{k=0}^{b-j}c(b, k+j)\binom{k+j}{j}r^{k}\\
&=\dfrac1{b!}\sum_{j=0}^n\sigma_j(n)u(a, b, j)\quad\text{where } u(a, b, j)=\sum_{k=0}^{b-j}c(b, k+j)\binom{k+j}{j}r^{k}\\
\end{array}
$
If $b=a+1$,
as in the examples,
$r=0$
so
$u(a, a+1, j)
=\sum_{k=0}^{b-j}c(a+1, k+j)\binom{k+j}{j}r^{k}
=c(a+1, j)\binom{j}{j}
=c(a+1, j)
$
so that
$\rho(a, a+1, n)
=\dfrac1{(a+1)!}\sum_{j=0}^n\sigma_j(n)u(a, a+1, j)
=\dfrac1{(a+1)!}\sum_{j=0}^n\sigma_j(n)c(a+1, j)
$.
A: This is a complement to @marty cohen answer. We can also give an identity for the case $a=-1$.
With the notation ${n\brack k}=c(n,k)$, we have
$$ \sum_{d\mid n}{d-1\choose b} = \frac{1}{b!}\sum_{j=0}^b{b+1\brack j+1}(-1)^{b-j}\sigma_j(n). $$
