# [Rust], score \$>f_4(126) / 100^3\$ 

<!-- language-all: lang-rust -->

    fn main(){let b='~' as usize;let mut n=b+b;for _ in b..n{n<<=n;for _ in b..n{n<<=n;print!("{}",n)}}}

[Try it online!][TIO-kgs2cxda]

Note that the code's validity is dependent on the assumption that a computer with infinite memory will have an arbitrary precision pointer to index the infinite memory. I also attempted to write a version that wouldn't overflow instantly with arbitrary precision integers but it still overflowed on the second bitshift. 

## (Incomplete) Size analysis:
The size of \$n\$, the final output of the code, it has \$f_4(126)\$ as a lower bound because the bitshifts are equivalent to \$f_2(n)\$ and are recursively applied like in the fast growing hierarchy. However, I do not have the faintest clue how to calculate the actual output, since it is the result of concatenating every value of \$n\$ that passes through the inner loop. For now, I'll just list my score as \$>f_4(126)/100^3\$ until I can come up with better bounds.

[Rust]: https://www.rust-lang.org/
[TIO-kgs2cxda]: https://tio.run/##bchLCoAgFAXQrbycaBRuQF1LJCgIeQs/k8S2bjTvDE@quYzhQXEPEHM7XCFr@MNpz1RzuJ36KtZCMHaxyp@JNgogKyUatDb4vSsFlEmw1tmKufc@xgs "Rust – Try It Online"