There's a clear pattern. Each term, you add two times the previous difference. As such, 33's difference from 17 was 16. 16*2=32, which you then add to 33, to get 65.
[The following part is a little more complicated. You don't have to look at this.]
Interestingly enough, each term is also 1 more than a multiple of 2. 33=2^5+1, 17=2^4+1, 9=2^3+1, 5=2^2+1. Does this pattern hold for every number, however? We can show this through some induction.
We only need this much information: The first term is 2^2+1, the difference to the next term is 2^2, and the differences are multiplied by 2 every time.
2^2+1+2^2 is 2^2+2^2+1, through rearrangement, and then through factoring we get 2(2^2)+1, which equals 2^3+1.
Since we shown that this produces the next power of two for the first case, does it hold for variable n? 2^n+1+2^n = 2^n+2^n+1 = 2(2^n)+1 = 2^n+1.
Yes it does, therefore your next answer will always be one more than a power of 2, in this case 2^6+1=64+1=65.