Find general term of a recurrent sequence If $a_1=6$ and $a_{n+1}= \frac{1}{18}(2a_n-1+\sqrt{8a_n+1}), n\geq1$ expressed $a_n$ in the function of $n$.
My attempt: We calculated the first 5 terms of the series, but I was unable to determine the desired formula and then apply mathematical induction.
 A: We note that the terms are (perhaps surprisingly) all rational. As such, we posit that the square root will always evaluate to a rational number, i.e. that $8a_n+1=b_n^2$, where $b_n$ is rational. Let's work through the algebra of this substitution:
$$\frac{b_{n+1}^2-1}{8}=\frac{2(\frac{b_{n}^2-1}{8})-1+b_n}{18}$$
$$b_{n+1}^2-1=\frac{2(b_n^2-1)-8+8b_n}{18}$$
$$b_{n+1}^2=\frac{2b_n^2+8b_n+8}{18}=\frac{b_n^2+4b_n+4}{9}=(\frac{b_n+2}{3})^2$$
So, $b_{n+1}=\frac{b_n+2}{3}\implies b_{n+1}-1=(b_n-1)/3$. Some backtracking tells us that $b_1-1=7-1=6$, whence we see that $b_n-1=\frac{6}{3^{n-1}}$. We then see:
$$b_n=1+\frac{18}{3^n}$$
Some algebra using $a_n=\frac{b_n^2-1}{8}$ ultimately gets us to $a_n=\frac{3^{2-n}+3^{2(2-n)}}{2}$, and we are done here.
A: After a bit of trying around, you can define $b_n=\sqrt{8a_n+1}-1$ which yields:
$$
\frac{(b_{n+1}+1)^2-1}{8}=\frac{1}{18}\left(\frac{(b_n+1)^2-1}{4}+b_n\right)\iff\\
(b_{n+1}+1)^2-1=\frac{1}{9}\left(b_n^2+6b_n\right)\iff (b_{n+1}+1)^2=\frac{1}{9}(b_n+3)^2\implies b_{n+1}=\frac{1}{3}b_n
$$
And now, the recursion is easy to handle.
