Hello friends, my name is Too Sharp. Today, we are going to talk about question number without consecutive ones in binary presentation. The question is given: find the total number of numbers from zero to Taurus to n minus 1 such that the numbers do not have consecutive ones in the binary presentation. For example, if n is equal to 2, we have numbers from 0 to 3. So, let's see how many numbers do not have consecutive ones in their binary presentation: 1, 2, and 3. Therefore, when n is equal to 2, you should return 3. Notice how the number 3 has consecutive ones in its binary representation. Let's see if n is equal to 3. Here, we have 5 numbers that do not have consecutive ones in their binary representation, while there are 3 numbers that do have consecutive ones. So, when n is equal to 3, the answer is fewer than 5. So, how do we solve this? We'll use a very simple implementation of a Fibonacci series. Let's take a look at the binary representation for a four-bit number, which goes from 0 to 15. In this Fibonacci series, we are adding "1"s in front of the three's binary representation. These numbers are exactly the same as this number, only that "1" is added in front of them. The number of numbers that do not have consecutive ones here will be exactly the same as this class of 5. It is not going to contribute to the -1, so it will not cause any number to become a contributing 1. So, we are just looking at the representation from 0 to 111, and we have already calculated that the number is 5. The total number of numbers for this half is 4. These numbers already...