Determine the time complexity function of the program

Can someone explain how the output of this recursive function is 18? Recursion is confusing to me. def R(n):    if n>=5:        return 10    return R(n+1) + 2 print(R(1))

The following code is used to derive Fibonacci algorithm with this sequence: (0, 1, 1, 2, 3, 5, 8, 13, 21, …) where each number is “add” of two previous ones. int fib(int n){ if (n==0) return 0; else if (n == 1) return 1; else return fib(n−1) + fib(n−2); } Explain why the following RISC-V assembly code works, you need to explain all the details using the comments as hints?   # IMPORTANT! Stack pointer must remain a multiple of 16!!!!   addi x10,x10,8                # fib(8), you can change it fib: beq x10, x0, done             # If n==0, return 0 addi x5, x0, 1 beq x10, x5, done             # If n==1, return 1 addi x2, x2, -16              # Allocate 2 words of stack space sd x1, 0(x2)                  # Save the return address sd x10, 8(x2)                 # Save the current n addi x10, x10, -1             # x10 = n-1 jal x1, fib # fib(n-1) ld x5, 8(x2) # Load old n from the stack sd x10, 8(x2) # Push fib(n-1) onto the stack addi x10, x5, -2 # x10 = n-2 jal x1, fib # Call fib(n-2)…