What numbers are these, and why do they show up here? 
This came up when trying to evaluate $\sum n^3/2^n$. In order to do this I had to repeatedly differentiate and multiply by $x$ to make the function match the summation. What are these numbers, and why do they show up here? I would appreciate some more insight as to how exactly these coefficients of the expansion come about through this iterated process
 A: A small and useful trick.
Consider, for the time being
$$S=\sum_{n=1}^p n^3\, x^n$$ and write
$$n^3=n(n-1)(n-2)+3n(n-1)+n$$
$$S=\sum_{n=1}^p n(n-1)(n-2)\, x^n+3\sum_{n=1}^p n(n-1)\, x^n+\sum_{n=1}^p n\, x^n$$
$$S=x^3\sum_{n=1}^p n(n-1)(n-2)\, x^{n-3}+3 x^2\sum_{n=1}^p n(n-1)\, x^{n-2}+x\sum_{n=1}^p n\, x^{n-1}$$
$$S=x^3\Bigg[\sum_{n=1}^p x^n \Bigg]'''+3x^2\Bigg[\sum_{n=1}^p x^n \Bigg]''+x\Bigg[\sum_{n=1}^p x^n \Bigg]'$$ and
$$\Bigg[\sum_{n=1}^p x^n \Bigg]=\frac{x \left(x^p-1\right)}{x-1}=1+\frac{ \left(x^p-1\right)}{x-1}$$
When finished, let $x=\frac 12$
A: Partial Answer:
A check on the OEIS, entry $A000278$ suggests these coefficients are the (reflected triangle of the) Stirling numbers of second kind.
The Stirling numbers of the second kind (Wikipedia) give the number of ways to partition an $n$-member set into $k$ nonempty and unlabeled subsets. One may algebraically define
$$\newcommand{\S}[2]{{#1 \brace #2}}
\S n k := \frac{1}{k!} \sum_{i=0}^k (-1)^i \binom k i (k-i)^n $$
Wikipedia offers a table of some values:

Notice how if you left-align your triangle, strip away the below table's first column, and then reverse the order of entries in your triangle (or Wikipedia's), you get a match.

I have no particular insight as to why these numbers pop up in your summation, however.

Possible Alternative?
Further checking of the OEIS suggests sequence $A213735$ (link) may be related, as it seems to match the same pattern, but with extra negative factors.
This sequence's definition involves the falling factorial; denoted $(x)_k$, it is like a factorial but only goes back enough to get $k$ factors:
$$(x)_k = x(x-1)(x-2) \cdots (x-(k-1)) = \frac{x!}{(x-k)!}$$
The sequence suggests a relation to the expansion of $(x+k)_k$, and expressing $x^n$ as
$$x^n = \sum_{i=n-1}^1 a_i (x+i)_i$$
(where the weird bounds ensure we sum in decreasing order, like in the OEIS). The $a_i$ give the members of the sequence/triangle we form.
Something about this feels more likely, but I'm not entirely sure how.
