Question

Implement the following function. Do not use any local variables or loops.
void pattern(unsigned int n)
// Precondition: n > 0;
// Postcondition: The output consists of lines of integers. The first line
// is the number n. The next line is the number 2n. The next line is
// the number 4n, and so on until you reach a number that is larger than
// 4242. This list of numbers is then repeated backward until you get back
// to n.
/* Example output with n = 840:
840
1680
3360
6720
6720
3360
1680
840

Answer 1

as it is mentioned you cannot use loops, the only other option to "imitate" a loop is to use recursion. The following code is in python and considering how python code is quite similar to a code in "english" you might get some hints from it:

def pattern(n):
print n
if (n > 4242):
print n
return
pattern(2*n)
print n

Answer 2

I didn't see FoolAround's solution before devising my own, so mine is slightly different.

void pattern(int n) {
cout << n << endl;
if (n < 4242)
pattern(2*n);
cout << n << endl;
}

Answer 3

Oh, that should be "if (n <= 4242)"