Fibonacci numbers are a sequence of integers, starting with 1, where the value of each number is the sum of the two previous numbers, e.g. 1, 1, 2, 3, 5, 8, etc. Write a function called fibonacci that takes a parameter, n, which contains an integer value, and have it return the nth Fibonacci number. (There are two ways to do this: one with recursion, and one without.)
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Fibonacci numbers are a sequence of integers, starting with 1, where the value of each number is the sum of the two previous numbers, e.g. 1, 1, 2, 3, 5, 8, etc. Write a function called fibonacci that takes a parameter, n, which contains an integer value, and have it return the nth Fibonacci number. (There are two ways to do this: one with recursion, and one without.)
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- This is in Python This has two parts. This is a problem on recursion. Part a: Write a recursive function that accepts an integer argument, n. The user needs to be asked for the number n. The function should display n lines of asterisks on the screen, with the first (the top) showing 1 asterisk, the second from the top showing two asterisks, up to the nth line which shows n asterisks. Part b. Write a recursive function that accepts an integer argument, n. The user needs to be asked for the number n. The function should display n lines of asterisks on the screen, with the first (the top) showing n asterisks, the second from the top showing n-1 asterisks, up to the nth line which shows 1 asterisk. Submit the two files and two sample outputs for each of the parts.please code in python Here is another recursion example, but with less guidance. Write a function log2(x), which gives an integer approximation of the log base 2 of a positive number x, but it does so using recursion. Use a base case where you return 0 if x is 1 or less. In the general case, add 1 to the answer and divide x by 2. (This is purposely vague so you can figure it out yourself.)A 5-digit positive integer is entered through the keyboard, write a function to calculate multiplication of digits of the 5-digit number: (1) Without using recursion (2) Using recursion
- The goal is to rewrite the function, below, such that passes in a different list of parameters, particularly eliminating the need to pass low and high for each recursive call to binary_search. defbinary_search(nums,low,high,item): mid=(low+high)//2iflow>high:returnFalse #The item doesn't exist in the list!elifnums[mid]==item:returnTrue# The item exists in the list!elifitem<nums[mid]:returnbinary_search(nums,low,mid-1,item)else:returnbinary_search(nums,mid+1,high,item) The new function should be prototyped below. The number of changes between the given version, and the one requested is not significant. defbinary_search(nums,item):pass# Remove this and fill in with your code Tip: If you consider that high and low are used to create a smaller version of our problem to be processed recursively, the version requested will do the same thing, just through a different, more Pythonic technique.Write a recursive function that takes as a parameter a nonnegative integer and generates the following pattern of stars. If the nonnegative integer is 4, then the pattern generated is:********************Also, write a program that prompts the user to enter the number of lines in the pattern and uses the recursive function to generate the pattern. For example, specifying 4 as the number of lines generates the above pattern.Write a recursive function, reverseDigits, that takes an integer as a parameter and returns the number with the digits reversed. Also, write a program to test your function.
- Computer scientists and mathematicians often use numbering systems other than base 10. Write a program that allows a user to enter a number and a base and then prints out the digits of the number in the new base. Use a recursive function baseConversion (num, base) to print the digits. Hint: Consider base 10. To get the rightmost digit of a base 10 number, simply look at the remainder after dividing by 10. For example, 153 % 10 is 3. To get the remaining digits, you repeat the process on 15, which is just 153 // 10. This same process works for any base. The only problem is that we get the digits in reverse order (right to left). The base case for the recursion occurs when num is less than base and the output is simply num. In the general case, the function (recursively) prints the digits of num // base and then prints num % base. You should put a space between successive outputs, since bases greater than 10 will print out with multi-character "digits." For example, baseConversion(1234,…Write a recursive function that returns true if the digits of a positive integer are in increasing order; otherwise, the function returns false. Also, write a program to test your function.Using recursion, write a Python function def countOdds(A) which which takes an array of integers A as input and returns the number of elements in the array that are odd. For example, if A is [1,2,5,3,6,5,3,5,5,4] the the function should return 7. If A is empty then it should return 3. Hint: To check if n is odd it suffices to check if dividing it by 2 gives remainder 1 (in Python, this remainder is calculated by n%2).