std::floor was made constexpr in C++23, which is pretty recent as far as C++ standards go. It's possible the author didn't think using C++23 was worth the constraints it places on who could use the code.
That's a mathematical expression, not a C++ expression. And floor here isn't the C++ floor function, it's just describing the usual integer division semantics. The challenge here is that you need 128-bit integers to avoid overflowing.
Ah, you're right. I saw that the expression in the comment and in the code was the same and assumed that the commented bit was valid C++ code. You got me to look again and it's obvious that that isn't the case. I had even gone looking through the codebase to see if std::floor was included, and still missed the incorrect `^`.
I guess in that case as long as the 128-bit type supports constexpr basic math operations that should suffice to replace the hardcoded constants with their source expressions.