The finite automaton below: The finite automaton below: accepts no word of le
accepts no word of length zero, no word of length one, and only two words of length two (01 and 10). There is a fairly simple recurrence equation for the number N(k) of words of length k that this automaton accepts. Discover this recurrence and demonstrate your understanding by identifying the correct value of N(k) for some particular k. Note: the recurrence does not have an easy-to-use closed form, so you will have to compute the first few values by hand. You do not have to compute N(k) for any k greater than 14.
a) N(13) = 84
b) N(14) = 16
c) N(12) = 10
d) N(11) = 682

Create a function (prob3_6) that will do the following: input a positive scalar integer x. if x is odd, multiply it by 3 and add 1. if the given x is even, divide it by 2. repeat this rule on the new value until you get 1, if ever. your program will output how many operations it had to perform to get to 1 and the largest number along the way. for example, start with the number 3: because 3 is odd, we multiply by 3 and add 1 giving us 10. 10 is even so we divide it by 2, giving us 5. 5 is odd so we multiply by 3 and add one, giving us 16. we divide 16 (even) by two giving 8. we divide 8 (even) by two giving 4. we divide 4 (even) by two giving 2. we divide 2 (even) by 2 to give us 1. once we have one, we stop. this example took seven operations to get to one. the largest number we had along the way was 16. every value of n that anyone has ever checked eventually leads to 1, but it is an open mathematical problem (known as the collatz conjectureopens in new tab) whether every value of n eventually leads to 1. your program should include a while loop and an if-statement.