Shell Programming and Scripting

Hi,
I want to access a variable outside the awk program. My program is as below:- I can not access the exact value of k (See the last line of the program).

#!/usr/bin/sh
j=10
k=1
#k is declared outside awk
awk '
BEGIN {
i=1;
j1="'"$j"'"
printf("\n ## Value of j1 is %d ##", j1);
while (i<=j1)
{
k=i*i;
printf("\n The square of %d is %d", i, k);
i=i+1;
}
printf("\n");
}
'
#Accessing the variable out side the awk program
echo "The value of k is $k"

The current Output is:

## Value of j1 is 10 ##
The square of 1 is 1
The square of 2 is 4
The square of 3 is 9
The square of 4 is 16
The square of 5 is 25
The square of 6 is 36
The square of 7 is 49
The square of 8 is 64
The square of 9 is 81
The square of 10 is 100
The value of k is 1

At the last line, the value of k should be 100...How its possible..?


Second thing is: In the above program, If I delete the line (or comment it)
"printf("\n ## Value of j1 is %d ##", j1);" , then I am getting the following out put:
The square of 1 is 1
The value of k is 1

Question is: Why the output is not comming as before(ie square of first 10 numbers..)...? If I want to get all squares upto 10 without that printf statement, then what should be done..?

I am not clear why you are using awk in your program. you can very much get away without that.

rakeshawasthi, Thanks.
My requirement is not to get the square of numbers. This script is just an example to make understand the exact problem. There are some variables which I am modifying under awk and finally I need to access those from the outside of the awk. This script is a perfect example of my requirement.