Why does program count one extra value (cycle "while")?

As explained, you're checking n then changing n, you want to change n then check before continuing.

You can use the walrus operator to assign n and check it's value all in the while statement. (requires Python 3.8+)

N = int(input())
n = 1
k = 1

while (n := k**2) < N:
    print(n)
    k += 1

This essentially assigns n to k**2 then checks if that result is <N before continuing.

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49

Your program outputs 1 4 9 16 25 36 49 64 if your input is 50 because the `while`` loop is checking the value before you increase it. Once in the loop, you increase it, calculate the square and then print.

If you want it to terminate, try setting calculating n as the last step in the loop:

N = int(input())

n = 1
 
k = 1

while n < N:

    
    print(n)
    k = k + 1
    n = k ** 2

You're checking whether you reached the limit before you calculate the square and print it. So you're checking the previous value of n, not the one that's about to be printed.

Move the check inside the loop.

while True:
    n = k ** 2
    if n >= N:
        break
    print(n)
    k += 1

The n < N is evaluated after you've changed (and printed) n.

n = 1
k = 1

N=50

while 1:
    n = k ** 2
    if n > N:
        break
    print(n)
    k = k + 1

To fix this, break before you print, moving the evaluation inside the loop rather than after the last update of n

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49

With the condition of your code, for example, when n = 49, The condition is fulfilled because 49 < 50 therefore it will continue to process the value and print the new one. But once n = 64 which is > 50, it stops. This is a possible solution:

N = int(input())
n = 1
k = 1

while True:
    if n >= N:
        break

    n = k ** 2
    print(n)
    k = k + 1

This will continuously execute the code but once the condition is met that n >= N, it will stop executing.