A recursive sequence

Let $b_k=\frac{a_k}{2^k}$ . Then $b_{k+1}=\frac{a_{k+1}}{2^{k+1}}=\frac{2a_{k}+k2^k}{2^{k+1}}=b_k+\frac{k}{2}$ . Take the summation from $k=1$ to $k=n$ for both sides, wee obtain $b_{n+1}=b_1+\frac{n(n+1)}{4}$ . Since $b_1=\frac{1}{2}$ , we have $\frac{a_{n+1}}{2^{n+1}}=b_{n+1}=\frac{1}{2}+\frac{n(n+1)}{4}$ . After some manipulation, $a_n=2^{n-2}\left(n^2-n+2\right)$ .

Note that $4096=2^{12}<89^2=7921<2^{13}=8192$

When $n=89$ , $a_n=2^{87}\left(89^2-89+2\right)< 2^{87}\left(89^2\right)< 2^{87}2^{13}=2^{100}$ .

When $n=90$ , $a_n=2^{88}\left(90^2-90+2\right)=2^{88}\left(89^2+2(89)+1-90+2\right)=2^{88}\left(89^2\right)>2^{88}2^{12}=2^{100}$ .

So the answer is $\boxed{90}$ .