Sure, According to the sequence, we notice that an element is represented as $$$xb$$$ or $$$a+xb$$$ where $$$x$$$ is any positive integer. noticing that, let $$$m =\lfloor\frac{a}{b}\rfloor$$$, there are $$$m$$$ multiples of $$$b$$$ before $$$a$$$ which surely means that the first $$$m$$$ numbers in the sequence are these multiples because an element can be represented as $$$xb$$$. We now reach a state where every two elements in the sequence become $$$a(m+x)$$$ and $$$a+xb$$$. you will choose which one is the $$$Kth$$$ element depending on the parity of the remaining steps. Replace this $$$x$$$ by the remaining steps divided by two to obtain the answer. Well, to explain more I'll just explain an example. $$$A = 6, B = 2, K = 7$$$. here there a $$$3$$$ multiples of $$$2$$$, $$$[2,4,6]$$$ and with the $$$0$$$ as $$$F_0$$$, the sequence is now $$$[0,2,4,6]$$$, subtract our steps by $$$4$$$, $$$K = K - 4 = 3$$$, now every two elements will be $$$6+2x$$$ and $$$2(3+x)$$$. The sequence will be $$$[0, 2, 4, 6, 6, 8, 8, 10, 10,\dots]$$$. The $$$Kth$$$ element is $$$2(3+\lfloor\frac{3}{2}\rfloor) = 8$$$. The implementation will vary but the idea is the same, if anything is not clear please ask because I don't explain my solutions that much so it might not be clear.