Ah, I see. That doesn't matter in big-O notation, though, because log_32(n) = log_2(n) / log_2(32), that is the difference between two bases is a constant facto

Ah, I see. That doesn't matter in big-O notation, though, because log_32(n) = log_2(n) / log_2(32), that is the difference between two bases is a constant factor. I was a bit confused because O(log_32(n)) = O(log(n)) = log(32)+log(n) = O(log(32n)), so no matter how I parsed it I would just get O(log(n)) :)