I'm saying that $435$ is the answer to the question. Why?
Consider the polynomial $$ p(x)=-\frac{3x^5}{2}+\frac{55x^4}{2}-\frac{375x^3}{2}+\frac{1175x^2}{2}-786x+525$$

Here is a table of values of $p(x)$ on consecutive $x$.

$\begin{array}{|l|c|c|c|c|c|c|}\hline
x & 1 & 2& 3& 4& 5&6\\ \hline
p(x)&165&195&255&285&345&\color{red}{435}\\\hline
\end{array}
$

My friend is saying that $390$ is the answer. Why? Consider the polynomial
$$g(x)=-\frac{15x^5}{8}+\frac{265x^4}{8}-\frac{1755x^3}{8}+\frac{5375x^2}{8}-\frac{3555x}{4}+570$$

Here is a table of values of $g(x)$ for $x=1,2,3,4,5,6$

$\begin{array}{|l|c|c|c|c|c|c|}\hline
x & 1 & 2& 3& 4& 5&6\\ \hline
g(x)&165&195&255&285&345&\color{red}{390}\\\hline
\end{array}
$

**How did I calculate the polynomials?**

We are calculating a polynomial $p(x)$ that attains values $165,195,255,285,345,k$ (here $k$ is any number number) when $x=1,2,3,4,5,6$.
I used a principle known as Lagrange Interpolation and the tool Wolframalpha interpolation calculator.
Similarly one can construct even more complex relations using various interpolation techniques.

**Conclusion**: There is no unique "next term of the sequence", since for arbitrary number $\lambda$, you can always form a relation in which $\lambda$ should be the next term, although some relations may look more natural than others.