Thread: Loop invariants

Consider the following while loop:
n:=1
while n<b, do
print n
n:=4n-2

a.) Write the first three numbers printed;
b.) Find all values of b for which the last number printed will be 342.

I am at a complete loss as to how to figure this out. I feel like just plugging in 1 for n is far too easy, and if that is the case, it will take far too long to get to 342.

1)
n:=1
while n<b, do
print n
n:=4n-2

2)

n:=4(1)-2 = 2
while n<b, do
print n
n:=4n-2

3)

n:=4(2)-2 = 6
while n<b, do
print n
n:=4n-2


Output will be 1,2,6


Also

n : 1, 2, 6, 22, 86

4n-2: 2, 6, 22, 86, 342