Consider a positive integer sequence $S=(s_1,s_2,\dots ,s_n)$.

Let $f(S)$ be the perimeter of the maximum-area quadrilateral whose side lengths are $4$ elements $(s_i,s_j,s_k,s_l)$ of $S$ (all $i,j,k,l$ distinct). If there are many quadrilaterals with the same maximum area, then choose the one with the largest perimeter.

For example, if $S=(8,9,14,9,27)$, then we can take the elements $(9,14,9,27)$ and form an isosceles trapeziumAn isosceles trapezium (US: trapezoid) is a quadrilateral where one pair of opposite sides are parallel and of different lengths, and the other pair has the same length. with parallel side lengths $14$ and $27$ and both leg lengths $9$. The area of this quadrilateral is $127.611470879\cdots$ It can be shown that this is the largest area for any quadrilateral that can be formed using side lengths from $S$. Therefore, $f(S)=9+14+9+27=59$.

Let $u_n=2^{B(3n)}+3^{B(2n)}+B(n+1)$, where $B(k)$ is the number of $1$ bits of $k$ in base $2$.
For example, $B(6)=2$, $B(10)=2$ and $B(15)=4$, and $u_5=2^4+3^2+2=27$.

Also, let $U_n$ be the sequence $(u_1,u_2,\dots ,u_n)$.
For example, $U_{10}=(8,9,14,9,27,16,36,9,27,28)$.

It can be shown that $f(U_5)=59$, $f(U_{10})=118$, $f(U_{150})=3223$.
It can also be shown that $\sum f(U_n)=234761$ for $4\le n\le 150$.
Find $\sum f(U_n)$ for $4\le n\le 3000000$.