What you are computing is a binomial transform (see on Wikipedia or MathWorld)
If your original sequence is $u_0,\dots,u_n$, you compute successive differences of your list, and you keep only the first term of these differences.
The differences are
$$\begin{matrix}
19 & 77 & 265 & 715 & 1607 & 3169\\
58 & 188 & 450 & 892 & 1562\\
130 & 262 & 442 & 670\\
132 & 180 & 228\\
48 & 48\\
0
\end{matrix}$$
So the list of first terms is $(v_k)=(19,58,130,132,48)$, with last nonzero term being here $v_4=48$, and you can then write, with $m=4$,
$$u_n=\sum_{k=0}^{m} {n \choose k}v_k$$
That is
$$u_n=19{n \choose 0}+58{n \choose 1}+130{n \choose 2}+132{n \choose 3}+48{n \choose 4}\\
=2n^4+10n^3+21n^2+25n+19$$
And by this method, your next term is $u_6=5677$.
Now, the reason why this works is that, given a sequence $u_n$, if you define, for all $n\geq0$,
$$v_n=(-1)^n\sum_{k=0}^n (-1)^k{n \choose k}u_k$$
Then you have, for all $n\geq0$,
$$u_n=\sum_{k=0}^n {n \choose k}v_k$$
I won't give a proof since you want to keep things simple, but at least you can see where the first sum comes from. If you compute the first term of successive differences of $u_n$, you have:
$$u_0$$
$$u_1-u_0$$
$$(u_2-u_1)-(u_1-u_0)=u_2-2u1+u_0$$
$$(u_3-2u_2+u_1)-(u_2-2u_1+u_0)=u_3-3u_2+3u_1-u_0$$
$$(u_4-3u_3+3u_2-u_1)-(u_3-3u_2+3u_1-u_0)=u_4-4u_3+6u_2-4u_1+u_0$$
And you should see a pattern.
Also, since
$${n \choose k}=\frac{n(n-1)\cdots(n-k+1)}{k!}$$
You can thus write your second sum as a polynomial in $n$.
An interesting feature of binomial transform is that if you original sequence is a polynomial of degree $m$, then the successive differences vanish after the $m$th, and you recover your polynomial in terms of binomial coefficients (it's just a change of basis).
Another way to view this is, if you have a finite sequence (of length $m$), you can compute all possible differences, that is up to the $(m-1)$th, and assuming all the following differences would be zero, you reconstruct the interpolating polynomial $P$ (then of degree $m-1$), such that $P(0)$ is you first term, $P(1)$ the second, etc. It's a practical way to compute the interpolating polynomial when absissas are the integers $0,1,\dots m-1$. Notice that it's a special case of Newton interpolation.
I wrote that with a list of length $m$ you get a polynomial of degree $m-1$, actually it's of degree at most $m-1$. You may have noticed that in your case, the last difference is zero, so the polynomial has degree $m-2=4$.
Seeing such a question, you may actually answer any number you wish for the next one, as you can always (at least) find an interpolating polynomial that would give this number. For example, you just have to add it to your list, and compute the binomial transform with the expanded list, to recover a new polynomial. Of course, it's not what is expected, but such an exercise is not very interesting, from the mathematical viewpoint.
However, it happens in practice that you have a sequence that you want to identify, and the binomial transform may be extremely useful to find a pattern, though it does not work in all cases. And another very useful tool is OEIS. It's just a database, but it's a huge one.
Suppose that we have a sequence:
$$a_0,a_1,...a_k$$
And we want to find the function of $n$ that defines $a_n$.
To do this we start by letting $a_{n+1}-a_n=\Delta a_n$ and we call this operation on $a_n$ the forward difference. Then given $\Delta a_n$ we can find $a_n$. Sum both sides of the equation from $n=0$ to $x-1$, and note that we have a telescoping series:
$$\sum_{n=0}^{x-1} \Delta a_n=\sum_{n=0}^{x-1} (a_{n+1}-a_n)=a_{x}-a_{0}$$
Hence $a_n=a_0+\sum_{i=0}^{n-1} \Delta a_i$. Also $\Delta a_n=\Delta (0)+\sum_{i=0}^{n-1} \Delta^2 a_n$...and so forth. Using this we must have if the series converges:
$$a_n=a_0+\Delta (0) \sum_{x_0=0}^{n-1} 1+\Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} 1+\Delta \Delta \Delta (0) \sum_{x_0=0}^{n-1} \sum_{x_1=0}^{x_0-1} \sum_{x_2=0}^{x_1-1} 1+\cdots$$
Where $\Delta^i (0)$ denotes the first term ($n=0$) of the $i$ th difference sequence of $a_n$.
Through a combinational argument, If we take $\Delta^0 (0)=a_0$ and ${n \choose 0}=1$ we may get:
$$a_n=\sum_{i=0}^{\infty} \Delta^i(0) {n \choose i}$$
If it is the case you want the sequence to start with $a_1$ we need to shift this result to the right one:
$$a_n=\sum_{i=0}^{\infty} \Delta^i(1) {n-1 \choose i}$$
Note when you re-define $a_0$ to be $a_1$ by shifting it's index to the right one, you're re-defining $a_1-a_0=\Delta^1(0)$ to be $a_2-a_1=a_{1+1}-a_{1}=\Delta^1(1)$.
Best Answer
There's a nice trick for recursive sequences of this type, where $u_{n+1}$ is a linear function of $u_n$: find a constant $r$ such that $u_{n+1}-r$ is a constant multiple of $u_n-r$. In this case, the multiplication factor in the linear function is $3$, so we're searching for an $r$ with $u_{n+1}-r = 3(u_n-r)$. Solving for $r$: \begin{align*} u_{n+1}-r &= 3(u_n-r) \\ (3u_n+1)-r &= 3u_n-3r \\ 1+2r &= 0 \\ r &= -\tfrac12. \end{align*}
Why does this help us? Because $u_{n+1}+\frac12 = 3(u_n+\frac12)$, it's easy to prove by induction that $u_n+\frac12 = 3^{n-1}(u_1+\frac12)$ for all $n\ge1$. (In other words, this slightly shifted version of the sequence really is geometric.) Solving for $u_n$: \begin{align*} u_n+\tfrac12 &= 3^{n-1}(u_1+\tfrac12) \\ u_n &= 3^{n-1}(3+\tfrac12)-\tfrac12 \\ u_n &= \tfrac12(7\cdot3^{n-1}-1). \end{align*} (Of course there's some formula we could memorize that gives the answer immediately, which is fine ... but it's always best to know how such formulas are derived in the first place.)