Eccoci.

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 $4 \le n \le 3\,000\,000$.