Question 1195898
<font color=black size=3>
If the sequence is
u_(n+1) = 2u_(n+1) + u_(n)
then you will arrive at the conclusion that @ikleyn reached: in that the terms should alternate in sign. 
However, assuming that u3 = 9 and u6 = 128 are correct, then there's no way the above sequence works.


--------------------------------------------------------------------------


I noticed that your link mentioned this instead
u_(n+2) = 2u_(n+1) + u_(n)


The n+1 portion at the front was changed to n+2


If that's the case, then we can say the following two equations
u6 = 2*u5 + u4
u5 = 2*u4 + u3


Let,
x = u5
y = u4
and furthermore we are given,
u3 = 9
u6 = 128


So those two equations mentioned update to
128 = 2*x + y
x = 2*y + 9
after plugging in the given knowns and unknowns.


Apply substitution to solve for y.
128 = 2*x + y
128 = 2*(2y+9) + y
128 = 4y+18 + y
128 = 5y+18
5y+18 = 128
5y = 128-18
5y = 110
y = 110/5
y = 22


Let's find x based on that.
x = 2y+9
x = 2*22+9
x = 44+9
<font color=red>x = 53</font>


Since x = 53 and y = 22, we can then say: <font color=red>u5 = 53</font> and u4 = 22



Check:
u3 = 9
u4 = 22
<font color=red>u5</font> = 2*u4+u3 = 2*22+9 = 44+9 = <font color=red>53</font>
u6 = 2*u5+u4 = 2*53+22 = 106+22 = 128
The answer is fully confirmed.


Again this all hinges on the assumption that the sequence is u_(n+2) = 2u_(n+1) + u_(n)
</font>