Over a TCP connection, suppose host A sends two segments to host B, host B sends an acknowledgement for each segment, the first acknowledgement is lost, but the second acknowledgement arrives before the timer for the first segment expires. What will host A do

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.