program to print/display 11111,1111,111,11,1 to nth term.

//program to print/display ...11111,1111,111,11,1 to nth term.
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
int m=1,n,number,denom;
printf("enter value of 'n' as a number of terms \n");
scanf("%d",&n);
printf("now enter number having %d digits(1)\n",n)
scanf("%d",&number);while(n>=m)
   {
            printf("%d\n",number);
      denom=pow(10,n-1);      
         number=number/denom;
      
      n--;
   }
getch();
}
..................................
logics in mind:

if n= 6(It means a number with 6 ones i.e. 111111)
111111,11111,1111,111,...
can be shown as
first 11111
second dividing above number by 10th power of entered number -1
as the second iteration goes , it has value with 1 less digit.
it continues until the value lasts to 1.

program to print/display 1,11,111,1111,11111,.... to nth term.

//program to print/display 1,11,111,1111,11111,.... to nth term.
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
int k=1,m=1,n;
printf("enter value of 'n'\n");
scanf("%d",&n);
while(m<=n)
   {
      printf("%d\n",k);
      
      k=k+pow(10,m);
      m++;
   }
getch();
}

logic applied:

1,11,111,1111...
can be written as
1,10+1,100+11,1000+111....
can be written as
1,10 ¹+1,10 ²+11,10 ³+111
converting it into formula
k=k+pow(10,m);

program to print/display 1,6,11,16....nth term.

//program to print/display 1,6,11,16....nth term.
#include<stdio.h>
#include<conio.h>
void main()
{
int k=1,n;
printf("enter value of 'n'\n");
scanf("%d",&n);
while(k<=n)
   {
      printf("%d,",k);
      k=k+5;
   }
getch();
}