KAME06 wrote: We define a sequence of numbers $a_n$ such that $a_0=1$ and for all $n\ge0$: \[2a_{n+1} ^3 + 2a_n ^3 = 3 a_{n +1} ^2 a_n + 3a_{n+1}a_n^2\]Find the sum of all $a_{2023}$'s possible values. Note that $\frac{a_{n+1}}{a_n}$ is root of $2x^3-3x^2-3x+2=0$ and so is $-1,2,\frac 12$ So $a_n=2^p\frac 1{2^q}(-1)^r$ where $p,q,r$ are nonnegative integers such that $p+q+r=n$ So $a_n$ can be $(-1)^{n-p-q}2^{p-q}$ where $p,q$ are nonnegative integers such that $p+q\le n$ Which is $(-1)^n(-2)^{p-q}$ and so $(-1)^n(-2)^k$ where $k\in\{-n,-n+1,...,n-1,n\}$ And sum of possible values for $a_n$ is $(-1)^n\sum_{k=-n}^{+n}(-2)^k$ $=\boxed{\frac{2^{n+1}+2^{-n}}3}$ And, for $n=2023$ : $\boxed{\frac{2^{2024}+2^{-2023}}3}$