Code:
def collatz(num):
if ( num %2 ==0 ) :#if number is even below calculation
proceed
Number=num//2 #num value in with divied by two and assign new
number variable
print(Number);#print value
if Number!=1:#if Number is not equal to 1
collatz(Number);#function are called
else:
return (Number)#else return number
elif (num % 2 == 1) : #else number is odd below calculation
proceed
Number=3*num+1 #num value in with add one and multy by three and
assign new number variable
print (Number)#print value
collatz(Number)#function are called
return Number;#return number
#user input and validation while loop
while True:
try:
num = int(input("Enter number:")) #user input and store value in
num
except ValueError:
print("Enter a valid integer")
continue
else:
if num < 1:#condition check
print("enter a valid positive integer")
continue
break
collatz(num)
Output:
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CODE NEEDS TO BE IN PYTHON. OLA 5: Collatz Sequence Function 12 17 34 Due: Fri...
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Language is Python 3, please include comments Question 1 Write function collatz, that takes a positive integer x as input and prints the Collatz sequence starting at x. A Collatz sequence is obtained by repeatedly applying this rule to the previous number x in the sequence: 3/2 if x is even 3.2 +1 if x is odd. Your function should stop when the sequence gets to number 1. Note: It is an open question whether the Collatz sequence of every...
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