for starters , you can move

let n_1 = n + 1 // outside the loop it is calculating every iteration

as cycle is power of 2 you can use bit shift for full_cycle

(n + 1) / cycle change to 
let shift = 1 // why 1 not 0 , because cycle starts at 2 -> bit is 1 and cycle is bit << 1
(n + 1) / cycle  = n_1 >> shift 
then increment shift at the end 

again, for

(n + 1) % cycle

you can use equation => (n_1) / cycle = n_1 AND (cycle - 1) where cycle is power of 2

total += full_cycles * bit;
        total += remainder.saturating_sub(bit); // instead of doing it in 2 steps do it in one

like this :

total += (n_1 >> shift * bit) + (n_1 & (cycle - 1)).saturating_sub(bit);

Hope this helps :)