# print the pattern 1 121 12321 1234321

/* The following solution is in Python. Here , i get the n value from the user and generate the pattern for n times. */

n = input()
L = []
for i in range(1,n+1):
L.append(str(i))
print ''.join(L)+''.join(L[::-1][1:])

Sample Input 1:

5

Sample output 1:

```1
121
12321
1234321
123454321```

Sample Input 2:

15

Sample Output 2:

```1
121
12321
1234321
123454321
12345654321
1234567654321
123456787654321
12345678987654321
12345678910987654321
123456789101110987654321
1234567891011121110987654321
12345678910111213121110987654321
123456789101112131413121110987654321
1234567891011121314151413121110987654321```

#include<iostream>
using namespace std;
int main()
{
int i,j,k,t,n,c,a[20];
cin>>t;
for(k=0;k<20;k++)
a[k]=(k+1);
while(t--)
{
cin>>n;
for(i=0;i<n;i++)
{
c=i-1;
for(j=0;j<(2*i+1);j++)
{
if(i>=j)
cout<<a[j];
else
{
cout<<a[c];
c--;
}
}cout<<" ";
}cout<<endl;
}
return 0;
}

import java.util.Scanner; public class pattern { public static void main(String...a) { int n,i,j=1,k,l,m; System.out.println("enter the number"); Scanner s=new Scanner(System.in); n=s.nextInt(); if(n!=0) System.out.println(1); for(i=2;i<=n;i++) { j=j*10+i; k=j; m=j; m=m/10; while(m>0) { l=m%10; k=k*10+l; m=m/10; } System.out.println(k); } } }
answered Sep 8, 2016 by 95peeyush jain 1 flag

#include<stdio.h>
int main()
{
int i,j,n,k;
printf("enter the number :");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
for(k=n;k>i;k--)printf(" ");
for(j=1;j<=i;j++)
{
printf("%d",j);
}
if(i>1)
{
for(j=i;j>1;j--)
{
printf("%d",j-1);
}
}
printf("\n");
}
return 0;
}

answered Oct 13, 2016 by Rishu

#include<stdio.h>
void main()
{
int n,i,j;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
for(j=1;j<=i;j++)
printf("%d",j);
for(j=i-1;j>=1;j--)
printf("%d",j);
printf("\n");
}

}

Simple Python solution for small inputs using pure maths!!!

for i in range(1,int(raw_input())+1):

print pow(((pow(10,i)-1)/9),2)

answered Oct 20, 2016 by gsomani
```#include <iostream>
using namespace std;

int main() {
int n = 4;
for(int i=1;i <=n ;i++){
for(int j=1;j <=i ;j++)
cout << j;
for(int j=i-1;j >=1 ;j--)
cout << j;
cout<< endl;
}
return 0;
}```