Is this normal?

#include <stdio.h>

#define MAXCHAR 256

long long fibonacci(int n);
	
void main(void)
	{
		int n = 0;
		int x = 0;
		
		//declaration variables
		FILE *filename; //input filename handle
		char *fname = NULL;
		int n = 0; //used for prompting user for size of Fibonacci series
		fname = (char *)malloc(MAXCHAR * sizeof(char)); //allocate memory for size of filename (char = 1 bit)
		
		//
		//prompt user for file name and continue to do so until a valid file name is given
		printf("Enter file name: ");
		scanf("%s", fname);
		
		//filename is just the handle to that file. have to go through filename for IO operation
		filename = fopen(fname, "w"); 
		
		//promt user for size of Fibonacci series
		printf("Enter the size of the Fibonacci series to calculate: ");
		scanf("%d", &n);
		
		//print series to cmd prompt
		for (x; x <= n; x++)
			printf("%lld ", fibonacci(x));
			
			
	}
	
long long fibonacci(int n)
	{
		//special case
		if (n == 0)
			{
				return 0;
			}
		//special case	
		else if (n == 2 || n == 1)
			{
				return 1;
			}
		//recursive algorithm to return sum of previous 2 numbers in series	
		else
			{
				return fibonacci(n - 1) + fibonacci(n - 2);
			}
	}

fib(0)=0 (calls=1)
fib(1)=1 (calls=1)
fib(2)=1 (calls=1)
fib(3)=2 (calls=3)
fib(4)=3 (calls=5)
fib(5)=5 (calls=9)
fib(6)=8 (calls=15)
fib(7)=13 (calls=25)
fib(8)=21 (calls=41)
fib(9)=34 (calls=67)
fib(10)=55 (calls=109)
fib(11)=89 (calls=177)
fib(12)=144 (calls=287)
fib(13)=233 (calls=465)
fib(14)=377 (calls=753)
fib(15)=610 (calls=1219)
fib(16)=987 (calls=1973)
fib(17)=1597 (calls=3193)
fib(18)=2584 (calls=5167)
fib(19)=4181 (calls=8361)
fib(20)=6765 (calls=13529)

long long int f2(int n) {
    long long int result = 0;
    long long int previous = 1;
    int i;
    for ( i = 0 ; i < n ; i++ ) {
        long long int next = result + previous;
        previous = result;
        result = next;
    }
    return result;
}

int count = 0;
void foo ( ) {
    count++;
}

   mp    NB. matrix product
$:~ :(+/ .*)
   
   F     NB. (Fib(2) Fib(1)) ,: (Fib(1) Fib(0))
1 1
1 0

   NB. The "anti-diagonal" is Fib(1), Fib(2), ........................................... Fib(16)
   <"2 F mp^:(<16) F    NB. creeping solution
┌───┬───┬───┬───┬───┬────┬─────┬─────┬─────┬─────┬──────┬───────┬───────┬───────┬───────┬──────  ──┐
│1 1│2 1│3 2│5 3│8 5│13 8│21 13│34 21│55 34│89 55│144 89│233 144│377 233│610 377│987 610│1597 987│
│1 0│1 1│2 1│3 2│5 3│ 8 5│13  8│21 13│34 21│55 34│ 89 55│144  89│233 144│377 233│610 377│ 987 610│
└───┴───┴───┴───┴───┴────┴─────┴─────┴─────┴─────┴──────┴───────┴───────┴───────┴───────┴──────  ──┘
   NB. The "anti-diagonal" is Fib(1), Fib(2), Fib(4), Fib(8), Fib(16) {see agreement with above}, Fib(32 = 2^5)
   <"2 mp^:(<6) F       NB. high matrix powers
┌───┬───┬───┬─────┬────────┬───────────────┐
│1 1│2 1│5 3│34 21│1597 987│3524578 2178309│
│1 0│1 1│3 2│21 13│ 987 610│2178309 1346269│
└───┴───┴───┴─────┴────────┴───────────────┘