Similar code to n=n+1 (n=n+x) with conditional variable to stop

Question

This may be an simple algorithm, but I can't do it.
I need to do a VI with an algorithm so similar to n=n+1 , but I need 2 variables, something like n=n+x, and other conditional variable (numerical control).
I need to enter the values n and x with a control and after the sustraction of these values but I need to print the value and after repit the process until the conditional value, and impress all the values incluiding the the first n value and the conditional value. And stop the process when the value is equal to the conditional value to get a data.
Basically
n=n+x
n=0
x=2
Conditional value (y)
y=10
and impress all the values in a data
n=0 (impress)
n=0+2 -->2 (impress)
n=2+2--->4 (impress)
n=4+2---->6 (impress)
n=6+2---->8 (impress)
n=8+2---->10 (impress)
And get a data like
0
2
4
6
8
10
The real problem is that I need enter with numerical controls all values and get the data with the n's first value and the conditional value.
If someone can help me, I appreciate it very much.
Thanks

Answer 1

ramktamu wrote:
This is what you are looking for.
The stopping problem has already been mentioned.
The "first call?" and "select" can be eliminated by initializing the shift register with "n".
Unless the "x" and "y" are allowed to vary during execution of the loop (would need very fast fingers and would generate irreproducible results!), these terminals belong before the loop. No need to go to the UI to re-read their value with every iteration of the loop.
Typically, the number of iterations can be calculated from the inputs using simple math, so a FOR loop would be much more appropriate and efficient here.
Try to simplify according to these comments and re-post.