$a_{2n-1}$$a_{2n}$=$a_2$$a_{2n-3}$+ ??????$(a_2a_{2n-3}+a_4a_{2n-5}.....+a_{2n-2}a_2)$ for $n$ $\geq 2$ ?????? I think there's something wrong.(it should be$(a_2a_{2n-3}+a_4a_{2n-5}.....+a_{2n-2}a_1)$I guess)

Yeah I just corrected it

$$a_{2k+1}=1$$and $$a_{2k}=2^k$$use mathematical induction I think.

Same thing everyone got on the contest. Can u fill in the details

$$P: a_{2k+1}=1,a_{2k}=2^k$$we supposed P is right for $$a_1,a_2,...a_k$$(1)k is odd then we have $$a_{k+1}=a_ka_{k+1}=a_2a_{k-2}+(a_2a_{k-2}+a_4a_{k-4}+...+a_{k-1}a_1)=a_2+a_2+a_4+...+a_{k-1}=2^1+2^1+2^2+...2^\frac{k-1}{2}=2^\frac{k+1}{2}$$(2)else then we have $$na_ka_{k+1}=a_2a_{k-2}+(a_2a_{k-2}+a_4a_{k-4}.....+a_{k-2}a_2)=n2^\frac{k}{2}$$so $$a_{k+1}=1$$