Is there any answer for this Bernoulli difference? Is there any answer for this  Bernoulli difference equation
$$B_{n+1}(x)-B_n(x)=?$$
where $B_n(x)$ is the Bernoulli polynomial defined by the exponential generating function$${ze^{xz} \over \mathrm{e}^{z} - 1} = \sum_{n = 0}^{\infty}{B_{n}(x) \over n!}\,z^{n}$$
 A: There are probably many ways one could write the right-hand side of your expression. Here's a series expansion that follows from the generating function:
Let
$$
F(t,x) = \frac{t e^{xt}}{e^t - 1}.
$$
Note that $B_{n+1}(x) - B_n(x)$ is generated by
$$
\tilde{F}(t,x) = \frac{\partial F(t,x)}{\partial t} - F(t,x) = F(t,x) \left((x-1) - \frac{1}{t}(F(t,1) - 1) \right).
$$
Note also that $B_{n+1}(x)/(n+1)$ is generated by $G(t,x) = (F(t,x) - 1)/t$ since
$$
\sum_{n \geq 0} \frac{B_{n+1}(x)}{n+1} \frac{t^n}{n!} = \frac{1}{t} \sum_{n \geq 1} B_n(x) \frac{t^n}{n!} = \frac{1}{t} \sum_{n \geq 0} B_n(x) \frac{t^n}{n!} - \frac{1}{t} = \frac{F(t,x) - 1}{t}.
$$
Thus, $\tilde{F}(t,x) = F(t,x) \tilde{G}(t,x) \equiv F(t,x) ((x-1) - G(t,1))$. Now, from the definition of $\tilde{F}$ we have
\begin{align}
\left( B_{n+1}(x) - B_n(x) \right) &= \frac{\partial^n \tilde{F}}{\partial t^n} \Big|_{t=0} \\
&= \sum_{k=0}^n \binom{n}{k} \frac{\partial^{n-k} F}{\partial t^{n-k}} \Big|_{t=0} \frac{\partial^k \tilde{G}}{\partial t^k} \Big|_{t=0} \\
&= \sum_{k=0}^n \binom{n}{k} B_{n-k}(x) \frac{\partial^k \tilde{G}}{\partial t^k} \Big|_{t=0}.
\end{align}
Finally, recalling that
$$
\frac{\partial^k \tilde{G}}{\partial t^k} \Big|_{t=0} = 
\begin{cases}
x - 1 - b_1 \quad \text{ if $k=0$} \\
-\frac{b_{k+1}}{k+1} \qquad \quad \text{otherwise},
\end{cases}
$$
where $b_k = B_k(1)$ is the $k$th Bernoulli number of the second kind, we find that
$$
B_{n+1}(x) - B_n(x) = B_n(x) \left( x - \frac{3}{2} \right) - \sum_{k=1}^n \binom{n}{k} \frac{b_{k+1}}{k+1} B_{n-k}(x),
$$
where we have used that $b_1 = B_1(1) = 1/2$.
