This is not a proof!!!!! $ a_1 = 0$, $ a_2 = 2k+1$ $ a_{n+1}=\frac{(n+1)}{n-1} a_n-3$ $ \therefore$ $ \begin{array}{c} a_3=6 k \\ a_4=12 k-3 \\ a_5=20 k-8 \\ a_6=30 k-15 \end{array}$ $ \therefore$ $ a_n=k (n-1) n-(n-3) (n-1)$ $ a_{2009}=4034072 k-4028048$ $ (4034072 k-4028048\equiv 2 k+2002\equiv 0) \bmod 2010$ $ (2 k\equiv -2002\equiv 8) \bmod 2010$ $ k= 4 + 1005 m$ $ a_n\equiv 3 (n-1) (335 m n+n+1)$ ?$ m=0$ $ \therefore$ $ \left(3 (n-1) (n+1)\equiv 3 n^2+2007\right) \bmod 2010$ $ \left(3 n^2\equiv 3\right) \bmod 2010$ ?$ \left(n^2\equiv \right) \bmod 670$ $ \cdots$

EulerIntegral wrote: This is not a proof!!!!! $ a_1 = 0$, $ a_2 = 2k + 1$ $ a_{n + 1} = \frac {(n + 1)}{n - 1} a_n - 3$ $ \therefore$ $ \begin{array}{c} a_3 = 6 k \\ a_4 = 12 k - 3 \\ a_5 = 20 k - 8 \\ a_6 = 30 k - 15 \end{array}$ $ \therefore$ $ a_n = k (n - 1) n - (n - 3) (n - 1)$ $ a_{2009} = 4034072 k - 4028048$ $ (4034072 k - 4028048\equiv 2 k + 2002\equiv 0) \bmod 2010$ $ (2 k\equiv - 2002\equiv 8) \bmod 2010$ $ k = 4 + 1005 m$ $ a_n\equiv 3 (n - 1) (335 m n + n + 1)$ ?$ m = 0$ $ \therefore$ $ \left(3 (n - 1) (n + 1)\equiv 3 n^2 + 2007\right) \bmod 2010$ $ \left(3 n^2\equiv 3\right) \bmod 2010$ ?$ \left(n^2\equiv \right) \bmod 670$ $ \cdots$ i think $ n=669$

∫FaiLurE∮ wrote: EulerIntegral wrote: This is not a proof!!!!! $ a_1 = 0$, $ a_2 = 2k + 1$ $ a_{n + 1} = \frac {(n + 1)}{n - 1} a_n - 3$ $ \therefore$ $ \begin{array}{c} a_3 = 6 k \\ a_4 = 12 k - 3 \\ a_5 = 20 k - 8 \\ a_6 = 30 k - 15 \end{array}$ $ \therefore$ $ a_n = k (n - 1) n - (n - 3) (n - 1)$ $ a_{2009} = 4034072 k - 4028048$ $ (4034072 k - 4028048\equiv 2 k + 2002\equiv 0) \bmod 2010$ $ (2 k\equiv - 2002\equiv 8) \bmod 2010$ $ k = 4 + 1005 m$ $ a_n\equiv 3 (n - 1) (335 m n + n + 1)$ ?$ m = 0$ $ \therefore$ $ \left(3 (n - 1) (n + 1)\equiv 3 n^2 + 2007\right) \bmod 2010$ $ \left(3 n^2\equiv 3\right) \bmod 2010$ ?$ \left(n^2\equiv \right) \bmod 670$ $ \cdots$ i think $ n = 669$ 269 < 669 n=269 also satisfies n^2=1 ( mod 670 )

let $ a_n = b_n + 3n - 3$ we attain $ b_{n + 1} = \frac {n + 1}{n - 1}b_n$ so we determine : $ b_n = \frac {n(n - 1)}{2}a_2$ so $ a_n = \frac{n(n - 1)}{2}a_2 + 3n - 3$ now, it is esay !!

I hope there isn't any careless mistake $ (n - 1)a_{n + 1} = (n + 1)a_n + 3 - 3n$ $ \frac {a_{n + 1}}{(n + 1)n} = \frac {a_n}{n(n - 1)} - \frac {3}{n(n + 1)}$ $ \frac {a_{n + 1}}{(n + 1)n} - \frac {3}{n + 1} = \frac {a_n}{n(n - 1)} - \frac {3}{n}$ Therefore, $ \frac {a_n}{n(n - 1)} - \frac {3}{n} = \frac {a_2}{2} - \frac {3}{2}$ This implies $ a_n = \frac {n(n - 1)}{2}a_2 + 3(n - 1) - \frac {3n(n - 1)}{2}$ Put n = 2009, and mod 2010 $ - 1004a_2 - 6 + 3 \times 1004 \equiv 0 (mod 2010)$ $ 1004(a_2 - 3) \equiv - 6 (mod 2010)$ $ 2008(a_2 - 3) \equiv - 12 (mod 2010)$ $ a_2 - 3 \equiv 6 (mod 1005)$ $ a_2 \equiv 9 (mod 1005)$ But $ a_2$ is odd, so $ a_2 \equiv 9 (mod 2010)$ Put this into the formula of $ a_n$, $ a_n = \frac {n(n - 1)}{2}9 + 3(n - 1) - \frac {3n(n - 1)}{2} = 3n(n - 1) + 3(n - 1) = 3(n^2 - 1) (mod 2010)$ 2010 = 2*3*5*67, so to have $ a_n \equiv 0 (mod 2010)$, we need $ n^2 \equiv 1 (mod 670)$ $ 670 | (n - 1)(n + 1)$ Divisible by 2 means $ n$ is odd. Divisible by 5 means $ n \equiv 1 or - 1 (mod 5)$ Same for 67 Obviously, the smallest possible $ n$ would then be 66.

There is a mistake at the last part - everything up to n^2=1 (mod670) are alright, but your answer (66) doesn't satisfy it as it should be odd =.= Anyway, the smallest answer should be 269 following your proof