The cyclic sequence of even digits, with odds in between Consider the following sequence:

1, 0, 1, 2, 4, 1, 6, 8, 0, 1, 2, 4, 6, 8, 1, 0, 2, 4, 6, 8, 1, 0, 2, 4, 6, 8, 0, 1, ...


The even digits start from 0 and are grouped into runs of increasing length. They are arranged cyclically, meaning that they are sorted in ascending order until 8 is reached, and then cycled back from 0. 1 separates the runs of even digits, and it also starts the sequence. Let's visualize how this sequence is formed:

                 1, 0, 1, 2, 4, 1, 6, 8, 0, 1, 2, 4, 6, 8, 1, 0, 2, 4, 6, 8, 1,  ...

                    -     ----     -------     ----------     -------------
run length:         1      2          3            4                5            ...
position of 1:   X     X        X           X              X                 X   ...
even sequence:      0,    2, 4,    6, 8, 0,    2, 4, 6, 8,    0, 2, 4, 6, 8      ...




Acceptable Input and Output methods:


Receive an integer N as input and output the Nth term of this sequence. 
Receive an integer N as input and output the first N terms of this sequence.
Print the sequence indefinitely.


You can choose either 0 or 1-indexing for the first two methods.

You can compete in any programming language, while using the standard input and output methods. Standard loopholes are forbidden. This is code-golf, so the shortest code in each language wins.
Haskell, 50 46 bytes



1#cycle[0,2..8]
n#r=1:take n r++(n+1)#drop n r


Try it online!

1#cycle[0,2..8] returns the sequence as infinite list.

-4 bytes thanks to Ørjan Johansen!