I'm pondering the following sequence:
$$\begin{equation} \begin{split} a_1 & = b \\ a_{n+1} & = c\frac{1}{n}\sum_{k=1}^{n}a_k = c \times \text{mean of } \{a_1,\dots,a_n\} \end{split} \end{equation}$$
For example, setting $b=1,c=1/2$, the sequence goes
$$1, \frac{1}{2}, \frac{3}{8}, \frac{5}{16}, \frac{35}{128}, \frac{63}{256}, \frac{231}{1024}, \frac{429}{2048}, \frac{6435}{32768}, \frac{12155}{65536}, \frac{46189}{262144}, \dots$$
Does the general term $a_n=a(n,b,c)$ of this sequence have a closed form in terms of $b$ and $c$? Closed forms do exist for some values of $c$:
$$\begin{equation} \begin{split} a(n,b,2) & = nb \\ a(n,b,3) & = \frac{n(n+1)}{2}b \end{split} \end{equation}$$
Since those formulae contain the closed-form expressions for the sums of the $0$th and $1$st powers of the first $n$ integers, one might be tempted to assume
$$a(n,b,4) \stackrel{\text{?}}{=} \frac{n(n+1)(2n+1)}{6}b$$
in analogue with the formula for the sum of squares. However, this turns out to be false.
The recurrence is equivalent to $$ a_{n+1}=\frac{\left(n-1\right)a_{n}+ca_{n}}{n}=\frac{c+n-1}{n}a_{n}. $$ This expression for the running mean (modulo the $c$ term) is used in numerical analysis for stability reasons. This recurrence has solution $$ a_{n+1}=\frac{b\left(c\right)_{n}}{n!} $$ where $\left(\cdot\right)_{n}$ is the Pochhammer symbol (a.k.a. rising factorial).
To verify this expression's correctness, here's Wolfram computing your answers using the formula above for $b=1,c=1/2$: