How do get combination sequence formula? What would be a closed-form formula that would determine the ith value of the sequence
1, 3, 11, 43, 171...
where each value is one minus the product of the previous value and 4?
Thanks!

The sequence can be written using the recurrence formula:  $y_n = a+by_{n-1}$ . Then using the first 3 terms, one gets $y_1=1, a=-1, b=4 $
Sometimes it is easy to convert a recurrence formula to a closed-form, by studding the structure of the math relations. 
$y_1 = 1$ 
$y_2 = a+by_1$ 
$y_3 = a+by_2 = a+b(a+by_1) = a+ab+b^2y_1$ 
$y_4 = a+by_3 = a+b(a+by_2) = a+ab+ab^2+b^3y_1$ 
$y_5 = a+ab+ab^2+ab^3+b^4y_1$ 
there is a geometric sequence  $a+ab+ab^2.. = a(1+b+b^2...) $ 
$ (1+b+b^2...b^n) = (1-b^{n+1}) / (1-b) $ 
therefore the general formula is: 
$y_n = a+ab+ab^2+...+ab^{n-2}+b^{n-1}y_1$ 
$y_n = a(1-b^{n-1}) / (1-b) + b^{n-1}y_1 $ 
Using the above parameters, $y_1=1 , a=-1,   b=4, $ 
$y_n = -1(1-4^{n-1})/(1-4) +4^{n-1}y_1$ 
$y_n = -\frac{4^{n-1}}{3} +\frac13 +4^{n-1}y_1$ 
$y_n = \frac{2}{3}4^{n-1} + \frac{1}{3} $
