It give $\frac{(n+2)a_{n-1}}{4a_na_{n+1}}+\frac{1}{a_na_{n+1}a_{n+2}}=\frac{(n+3)a_n}{4a_{n+1}a_{n+2}},n\ge 3,$ therefore (1) $a_{n+2}=\frac{(n+3)a_n^2-4}{(n+2)a_{n-1}},n\ge 3$ and (2) $\frac{16}{a_1}+4=5a_2^2\Longrightarrow a_1=1,a_2=2.$ From (1) $a_6a_3=18$ and $a_5=\frac{3a_3^2-2}{5}$. Therefore or 1.$a_3=2,a_5=2,a_6=9$ or 2. $a_3=3,a_5=5,a_6=6$ or 3. $a_3=18,a_5=194,a_6=1$. From (1) (n=6) we have $a_8=\frac{9a_6^2-4}{8a_5}$ 1. and 3. don't work. Therefore a_n=n for n<7. By induction chec, that $a_{n+2}=\frac{(n+3)n^2-4}{(n+2)(n-1)}=n+2.$ It give $a_n=n.$.

Rust wrote: (2) $\frac{16}{a_1}+4=5a_2^2\Longrightarrow a_1=1,a_2=2.$ Why not $a_1=16$ and $a_2=1$ ?

It is contraditon with $a_4=4$.

sorry, I don't see contradiction with $a_4=4$. well you still may have $a_1=16$ $a_2=1$ $a_3=3$ $a_4=4$ $a_5=10$ $a_6=6$ and the first three sums will be ok. but then a_7 will not be an integer, which is the contradiction

hello to all

imortal wrote: sorry, I don't see contradiction with $a_4=4$. well you still may have $a_1=16$ $a_2=1$ $a_3=3$ $a_4=4$ $a_5=10$ $a_6=6$ and the first three sums will be ok. but then a_7 will not be an integer, which is the contradiction In this way we should try all possible versions for contradictions and come to the conclusion $a_n=n$

yep, but the only two versions are $a_1=1$, $a_2=2$ or $a_1=16$, $a_2=1$ so it is not a problem