Yes, as you can see in the code, the loop executes once for each bit to the right of the leftmost set bit or equivalently ⌊log₂(n)⌋ times. Naively this suggest

Yes, as you can see in the code, the loop executes once for each bit to the right of the leftmost set bit or equivalently ⌊log₂(n)⌋ times. Naively this suggest a runtime of O(log(n)) but the numbers become huge pretty quickly and then all the arithmetic operations are far from O(1). The actual runtime is probably more like O(n²) or O(n² log(n)) after one accounts for the runtime of the big integer operations.