CODEWITHSUNDEEP
c
chanu
chanu

Reputation: 33

I need print out up to the N'th term and the N'th term give given that 1, 2, 5, 26, 677,.

given that 1, 2, 5, 26, 677,... such that the nth term of the series equals to (n-1)th ^2 +1 and the first term of the series is 1. Write a program using recursive function named f to compute the nth term.

I can't get anything to print out. How can i Fix this? 

#include <stdio.h>
#include <math.h>
int f(int n)
{
if(n==1)
   {
   return 1;
   }
   else
   {
     return f((n-1)^2)+1;
   }

}


int main()
{
    int N,i;
    printf("Enter the number of terms: ");
    scanf("%d",&N);
    int u = f(N);
    for(i=0;i<N;i++)
    {
    printf("%d ",u);
    }


    return 0;
}```

Upvotes: 0

Views: 165

Answers (1)

William Pursell
William Pursell

Reputation: 212248

The most direct way (IMO) to do this is something like:

#include <stdio.h>
#include <stdlib.h>
long
f(int n)
{
        return n == 1 ? 1 : f(n-1) * f(n-1) + 1;
}


int
main(int argc, char **argv)
{
        int n = argc > 1 ? strtol(argv[1], NULL, 10) : 2;
        for(int i=1; i < n; i++) {
                printf("%d: %ld\n", i, f(i));
        }
        return 0;
}

There are a lot of improvements that should be made above, but that gives the basic idea. But it seems like you're trying to do something more along the lines of:

#include <stdio.h>
#include <stdlib.h>
long
f(int n)
{
        long r= 1;
        if(n > 1) {
                r = f(n-1);
                r = r * r + 1 ;
        }
        printf("%d: %ld\n", n, r);
        return r;
}


int
main(int argc, char **argv)
{
        int n = argc > 1 ? strtol(argv[1], NULL, 10) : 2;
        f(n);
        return 0;
}

You should probably include some error checking to prevent overflow and return non-zero from main if that happens, but that will depend on how you want to handle overflow.

Upvotes: 1

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