This is a modified version of the kolakoski sequence.
Refer to the problem https://www.mathworks.com/matlabcentral/cody/problems/2371 for the original one.
for this case,
Given a particular initial vector a and length x, generate the kolakoski sequence (first x terms) from that vector.
For example if a=[2,1] and x=10 ; the sequence would look like --
2 2 1 1 2 1 2 2 1 2 2
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So it's kind of like Kolakoski; the number still gives the length at pos n, like A000002. However, the following number is decided according to a base named "a". So, if a=[4,2,3], the remainder 0,1 or 2 (since a has length 3) will be mapped to 4,2,3 in this order.
I recommend solving the problem in the description if you haven't already. You can reuse your code here. You will need to make some changes, but it will be easier than solving this one from zero. Numberphile has a video about it if you are still stuck.
I fail to understand some of the examples, where the initial terms of the output are not the same as the input (a) -
a = [1,3,1,2];
x=30;
y=[1, 3, 3, 3,...] %first 4 terms of y not equal to a
a = [2,1];
x=10;
y=[2, 2, 1, ...] %first 2 terms of y not equal to a
Although I highly doubt Dyuman's question is still open, for the benefit of anyone else wondering the same thing: a is not the beginning of the sequence, but rather the elements that get repeated in the sequence over and over again (in runs of varying lengths).