You can look at the consecutive differences until you reach a constant difference.
$$5,\; 6,\; 10, \;19,\; 35$$
$$1,\; 4, \;9,\; 16$$
$$3,\; 5,\; 7$$
$$2, \;2.$$
Using the fact that the $n$-th differences of a sequence $\{s_n\}$ are constant and non-zero if and only if $\{s_n\}$ is generated by an $n$-th degree polynomial, we can conclude that $s_n = an^3 + bn^2 + cn + d$. We immediately have $d = s_0 = 5$. Now use the fact that if a sequence is generated by an $n$-th degree polynomial with leading coefficient $c_n$, the $n$-th differences are constantly $(c_n)n!$. In our case, $3!\,a = 2$, so $a = \frac{1}{3}.$ To solve for $b$ and $c$, we use the initial data.
$$s_1 = a + b + c + d = 6 \Rightarrow b + c = \frac{2}{3}$$
$$s_2 = 8a + 4b + 2c + d = 10 \Rightarrow 4b + 2c = \frac{7}{3}.$$
Some algebra later, we have $b = \frac{1}{2}$ and $c = \frac{1}{6}$. Thus the $(n+1)$-th term in the sequence is given by
$$s_n = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n + 5,$$
and the $n$-th term is given by
\begin{align*}
s'_n &= \frac{1}{3}(n-1)^3 + \frac{1}{2}(n-1)^2 + \frac{1}{6}(n-1) + 5 \\ &= \frac{1}{3}n^3 - \frac{1}{2}n^2 + \frac{1}{6}n + 5.
\end{align*}
This approach doesn't use standard calculus, but rather calculus of differences. Some of the analogs are pretty clear: the $n$-th differences resemble $n$-th derivatives, in the sense that both are computed by subtracting "close-by" terms and that the $n$-th difference/derivative is constant and equal to $a\, n!$. This is a much more elegant solution than trying to interpolate a 3rd degree polynomial much more generally (though some may argue it's effectively the same by forcing the $y$-intercept to be the constant term and using differentiation to find the leading coefficient).
N.B. The commenters on your post are perfectly correct in that this sequence really could be anything (who knows, maybe the next several terms are all 0). The approach I have chosen begins with the hunch that we will eventually reach constant differences and that the constant differences are the same throughout the entire sequence.