Answer: $n = 1$ By induction, we can show that $a_n = \dfrac{4.2^{n} - 3}{2^{n}+3} \ \ \ \ \forall n \geq 2$ So the sequence is integer when $n = 1$ Note: the formula above can be found by letting $a_n = \dfrac{x_n}{y_n} $ for $(x_n,y_n)=1; x_n,y_n$ are postive integer sequences and doing some calculations to show that $x_{n+1} + y_{n+1} = 2(x_n + y_n)$ and $5y_{n+1} = x_n + 6y_n \forall n \geq 2$

But one can also easily solve without guessing the explicit formula: We can guess the limit $4$ by plugging in $a_n=a_{n+1}=a$ and solving for $a=4$. Then writing $b_n=4-a_n$, we get $b_{n+1}=\frac{5b_n}{10-b_n}$ from where it immediately follows that $b_n$ is decreasing (and positive). Since $b_1=3, b_2=\frac{15}{7}$ and $b_3=\frac{15}{11}$ and $b_4=\frac{15}{19}<1$, there can be no more integers in the sequence.