Olympic Rings Sequence Challenge:

Given an index integer n, either output the n'th item in this sequence, or output the sequence up to and including index n:

25,25,7,28,29,20,21,22,23,14,35,26,7,28,29,20,16,29,12,15,28,21,14,17,30,13,16,29,12,15,28,21,10,6,12,18,15,11,7,13,19,17,13,9,15,21,18,14,10,16,22,19,15,11,17,23,20,16,12,18,24,21,17,13,19,25,23,19,15,21,27,24,20,16,22,28,25,21,17,23,29,16,13,9,15,21,18,14,10,16,22,20,16,12,18,24,21,17,13,19


How does this sequence work?

NOTE: In this explanation, index n is 1-indexed.
Put the numbers 1 through x on two lines of length n*6 - 1, where x depends on the current iteration and the length of the numbers used, and then sum the digits of the n'th/right-most Olympic Rings of those two lines.

The first number in the sequence is calculated as follows:

The length of the lines are 5 (because 1*6 - 1 = 5):
12345
67891(0)

Then leave the digits in an Olympic Rings pattern:
1 3 5
 7 9

And sum them:
1+3+5+7+9 = 25


So n=1 results in 25.

The second number in the sequence is calculated as follows:

The length of the lines are 11 (because 2*6 - 1 = 11):
12345678910
11121314151(6)

Then leave the digits in the second/right-most Olympic Rings pattern:
      7 9 0 
       4 5

And sum them:
7+9+0+4+5 = 25


So n=2 results in 25.

The third number in the sequence is calculated as follows:

The length of the lines are 17 (because 3*6 - 1 = 17):
12345678910111213
14151617181920212(2)

Then leave the digits in the third/right-most Olympic Rings pattern:
            1 2 3
             0 1

And sum them:
1+2+3+0+1 = 7


So n=3 results in 7.

etc.

Challenge rules:


When you output the n'th item in the sequence, you are allowed to take the input as 0-indexed instead of 1-indexed, but keep in mind that the calculations of n*6 - 1 will then become (n+1)*6 - 1 or (n+1)*5 + n.
Single numbers of more than one digit can be split up at the end of the first line when we've reached the length n*5 + n-1, so it is possible that a number with 2 or more digits is partially the trailing part of line 1, and partially the leading part of line 2.


General rules:


This is code-golf, so shortest answer in bytes wins.
Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
Standard rules apply for your answer, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
Default Loopholes are forbidden.
If possible, please add a link with a test for your code.
Also, please add an explanation if necessary.


Test cases:

Here is a paste-bin of the test cases 1-1,000, so feel free to choose any of them.

Some additional higher test cases:

1010:   24
1011:   24
2500:   19
5000:   23
7500:   8
10000:  8
100000: 25
Retina, 70 68 62 bytes

.+
10**
.
$.&gt;`
~(`.+
6*$+*
)`.(.+)
L`.{$.1}
%,-6`.

,2,9`.
*
_


Try it online!

Explanation

Let's call the input n, and we'll use 3 as an example.

.+
10**


The 10** is short for 10*$&amp;*_ which replaces the input with a string of 10n underscores.

.
$.&gt;`


Now we replace each underscore with the length of the string up to and including that underscore. So this just results in the number from 1 to 10n all concatenated together (10n is always enough to fill up two lines of the required length).

~(`.+
6*$+*


Eval! This and the next stage will generate the source code of another program, which is then run against that string of concatenated integers.

To generate that program, this stage first replaces the integers with a string of 6n underscores ($+ refers to the program's original input).

)`.(.+)
L`.{$.1}


Then replace those underscores with L`.{…}, where … is 6n-1 (the length of the lines we're looking at). So we've generated a regex, whose quantifier depends on the original input.

When this program gets eval'ed it matches chunks of length 6n-1, of which there will be at least two. For our example input 3, we end up with:

12345678910111213
14151617181920212
22324252627282930


Now we just need to extract the relevant digits.

%,-6`.



First, on each line (%) we remove all but the last five digits (,-6). That gives us

11213
20212
82930


Finally:

,2,9`.
*


We expand every other digit (2) in the first ten (9, this is 0-based) in unary. Those happen to be those in the Olympic Rings positions.

_


And we count the number of resulting underscores, to sum them and convert the result to decimal.