Logic of Ugly Numbers in C - c

Question

Following from here, I am trying to develop my own logic to generate a sequence of ugly numbers. But every time all the numbers are printed.
I am determining if the first 3 prime factors of the number are 2, 3 and 5 and placing them in a count variable determining the total count of prime factors of a number x.
If the count is greater than 3, the number is not ugly.
Here is the code:
/* To generate a sequence of Ugly numbers
Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, …
shows the first 11 ugly numbers. By convention, 1 is included.
*/
#include<stdio.h>
#include<math.h>
int isprime(int x)
{
int i;
for(i=2;i<=sqrt(x);i++)
if(x%i==0)
return 0;
return 1;
}
int isUgly(int x)
{
int count=0; //To maintain the count of the prime factors. If count > 3, then the number is not ugly
int i;
for(i=2;i<=sqrt(x);i++)
{
if(isprime(i) && x%i==0)
{
count++;
if(count > 3)
return 0; // Not ugly
}
}
return 1;
}
int main(void)
{
int i,n=10;
printf("\n The ugly numbers upto %d are : 1 ",n);
for(i=2;i<=n;i++)
{
if(isUgly(i))
printf(" %d ",i);
}
return 0;
}

Answer 1

Here's a version of isUgly() that seems to work for me.
int isUgly(int x)
{
int i;
static int factors[] = {2, 3, 5};
// Boundary case...
// If the input is 2, 3, or 5, it is an ugly number.
for ( i = 0; i < 3; ++i )
{
if ( factors[i] == x )
{
return 1;
}
}
if ( isprime(x) )
{
// The input is not 2, 3, or 5 but it is a prime number.
// It is not an ugly number.
return 0;
}
// The input is not a prime number.
// If it is divided by 2, 3, or 5, call the function recursively.
for ( i = 0; i < 3; ++i )
{
if ( x%factors[i] == 0 )
{
return isUgly(x/factors[i]);
}
}
// If the input not a prime number and it is not divided by
// 2, 3, or 5, then it is not an ugly number.
return 0;
}

Answer 2

Try this one:
#include<stdio.h>
long int n, count=1;
void check(long int i)
{
if(i==1){
++count;
return;
}
else if(i%2==0)
check(i/2);
else if(i%3==0)
check(i/3);
else if(i%5==0)
check(i/5);
else
return;
}
void main(){
for(n=1;;n++){
check(n);
if(count==1000){
printf("%ldth no is %ld\n",count,n);
break;
}
}
}