Write a function factorial (n) that recursively calculates and returns n! n! = 1 * 2 * 3 .... (n-2) * (n-1) * n
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Make it in Python
Write a function factorial (n) that recursively calculates and returns n!
n! = 1 * 2 * 3 .... (n-2) * (n-1) * n
The mathematical function of the faculty function f (n) is as follows:
1 n=0
f(n)
n*f(n-1). n>0
The function must be recursively implemented to be correct.
Your function should not take any inputs nor make any prints it solves the function while.
use this code to test
number = int (input ())
print (factorial (number))
Step by step
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- (Program) Write a program that tests the effectiveness of the rand() library function. Start by initializing 10 counters, such as zerocount, onecount, twocount, and so forth, to 0. Then generate a large number of pseudorandom integers between 0 and 9. Each time 0 occurs, increment zerocount; when 1 occurs, increment onecount; and so on. Finally, display the number of 0s, 1s, 2s, and so on that occurred and the percentage of time they occurred.An arithmetic sequence is a sequence of values where successive values have a common difference. For example, 2,5,8,11,... is an arithmetic sequence starting at 2 with a common difference of 3. We call the starting point s and the difference d. Fill in the missing lines below to write a recursive Python function called arithmetic that takes values for s, d, and n, and returns the nth term of the arithmetic sequence. For example, given the main function: def main(): for i in range(1,6): print(arithmetic(2,3,i)) the output will be: 2 5 8 11 14 Note: No marks will be given for a non-recursive solution. Code with missing lines: # Line 1 # Line 2 # Line 3 # Line 4 # Line 5 Matching each line of code below to its proper line number (Line 1 to Line 5), indicated in the partial code above, to solve the problem. Some lines of code will not be used at all; for these you would select option 6 (i.e., Not Used) from the list of options - do not leave these blank. Question 8 options:…Using recursion, write a Python function def before(k,A) which takes an integer k and an array A of integers as inputs and returns a new array consisting of all the integers in A which come before the last occurrence of k in A, in the same order they are in A. For example, if A is [1,2,3,6,7,2,3,4] then before(3,A) will return [1,2,3,6,7,2]. If k does not occur in A, the function should return None.
- Write a recursive Python function that matches the following docstring: '''Function -- sum_fivesSums all positive multiples of five between 0 and some integer, n, inclusive.Parameters:n -- an integer.Returns:The sum of all positive multiples of five up to and including n.''' Examples: sum_fives(1) returns 0 sum_fives(5) returns 5 sum_fives(10) returns 15Make it in Python Write a function factorial (n) that recursively calculates and returns n!n! = 1 * 2 * 3 .... (n-2) * (n-1) * nThe mathematical function of the faculty function f (n) is as follows: 1 n=0 f(n) n*f(n-1). n>0 The function must be recursively implemented to be correct.Your function should not take any inputs nor make any prints it solves the function while. number = int (input ())print (factorial (number))c++ Implement a function that recursively calculates the nth number in the Fibonacci sequence of numbers. The Fibonacci sequence starts at 0 and 1. Each number thereafter is the sum of the two preceding numbers. This gives us the following first ten numbers: 0,1,1,3,5,8,13,21,34, .... Input Expected output0 01 17 139 34
- In C programming Mathematically, given a function f, we recursively define fk(n) as follows: if k = 1, f1(n) = f(n). Otherwise, for k > 1, fk(n) = f(fk-1(n)). Assume that there is an existing function f, which takes in a single integer and returns an integer. Write a recursive function fcomp, which takes in both n and k (k > 0), and returns fk(n). int f(int n);int fcomp(int n, int k){Implement a recursive C++ function which takes a character (ch) and a positive integer (n) and prints thecharacter ch, n times on the screen. The prototype of your function should be:void printChar (char ch, int n)For example, calling printChar('*',5) should display ***** on screen.Note: There should NOT be any loop in your function.The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n = 0 return a elsif n = 1 return b else for i in 1..n c <- a + b a <- b b <- c return b b) def f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a c) def f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1 return arr[n] else for i in 2..n arr[i] <- arr[i-1] + arr[i-2] return arr[n] d) def f(n): arr[0..n] <- [0, ..., n] if n <= 1 return arr[n] else a <- 0 for i in 0..n a <- a + arr[i] return a
- Write a recursive function that, given a sequence of comparable values, returns the count of elements where the current element is less than the following ( next ) element in the given sequence. See the examples given below. def count_ordered ( seq ) : """ Input : A sequence of comparable elements Output : The number of elements that are less than the following element in the sequence Example : >>> count_ordered ( [ 1 , 2 , 3 , 4 , 5 , 6 ] ) 5 >>> count_ordered ( ( 1 , 12, 7.3 , -2,4 ) ) 2 >>> count_ordered ( 'Python' ) 2 >>> count_ordered ( [ 6 ] ) 0 >>> count_ordered ( [ ] ) 0 """ In the first example above , count_ordered ( [ 1,2,3,4,5,6 ] )the returned answer is 5 because for all the first 5 numbers the current number is less than the next number. In the second example above, count_ordered ( ( 1,12,7.3 , -2,4 ) )the…The following function f uses recursion: def f(n): if n <= 1 return n else return f(n-1) + f(n-2) 5 Let n be a valid input, i.e., a natural number. Which of the following functions returns the same result but without recursion? a) def f(n): a <- 0 b <- 1 if n = 0 return a elsif n = 1 return b else for i in 1..n c <- a + b a <- b b <- c return b f(n): a <- 0 i <- n while i > 0 a <- a + i + (i-1) return a f(n): arr[0] <- 0 arr[1] <- 1 if n <= 1 return arr[n] else for i in 2..n arr[i] <- arr[i-1] + arr[i-2] return arr[n] f(n): arr[0..n] <- [0, ..., n] if n <= 1 return arr[n] else a <- 0 for i in 0..n a <- a + arr[i] return ais it correct ? Write a recursive function that returns the nth number in a fibonacci sequence when n is passed to function. The fibonacci sequence is like 0,1,1,2,3,5,8,13...... Answer: #include <iostream> using namespace std; int getFibonacci(int n) { if (n == 0 || n == 1) return n; else return getFibonacci(n - 1) + getFibonacci(n - 2); } int main() { int n = 7; int result = getFibonacci(n); cout << result; }
