After posting, I did some more work on this myself, and I think there may be some value in thinking of how series of n numbers are built by adding n at various places in series of n-1 numbers. For example, with n=3:

312, 132, and 123 are created by adding 3 to the 1st, 2nd, or 3rd place in the series 12.

321, 231, and 213 are created by adding 3 to the 1st, 2nd, or 3rd place in the series 21.

Here's a table describing the number of series of n numbers with x numbers out of sequence. Rows are n, starting with 1. Columns are x, starting with 0

1

1 1

1 4 1

1 9 13 1

1 16 61 41 1

To get a series of 3 numbers with 0 out of sequence, you have to add 3 to the end of a series of 2 numbers (the only one is 12) that has 0 out of sequence.

To get a series of 3 numbers with 1 out of sequence, you could either add 3 to a series of 2 numbers that has 0 out of sequence so that it is out of sequence (adding it to the 1st or 2nd place of 12), or you could add 3 to a series of 2 numbers that has 1 out of sequence (the only on is 21), so that it does NOT add to the number of out-of-sequence numbers (adding it to the 2nd or 3rd place of 21).

To get a series of 3 numbers with 2 out of sequence, you have to add 3 to a series of 2 numbers that has 1 out of sequence so that it does add to the number of out-of-sequence numbers (adding it to the 1st place of 21).

Applying the same logic in adding a 4 to series with n=3:

The only series with n=3 that has 0 numbers out of sequence is 123.

Four series have 1 number out of sequence: 312, 213, 132, 231.

The only series with n=3 that has 2 numbers out of sequence is 321.

There is only one way to add 4 to 123 so that it is NOT out of sequence.

There are three ways to add 4 to 123 so it is out of sequence. These contribute 3 of the 9 series with n=4 that have one number out of sequence.

The only way to add 4 to either 312 or 213 so that it does NOT increase the numbers out of sequence is to add it on the end. These contribute 2 more of the 9 series with n=4 that have one number out of sequence.

There are two ways to add 4 to either 132 or 231 so that it does NOT increase the numbers out of sequence (adding it to the last or second-to-last spots). These contribute 4 more of the 9 series with n=4 that have one number out of sequence.

Since there are 16 possible ways 4 could be added to the 4 series that have one number out of sequence, and 6 of them do NOT increase the numbers out of sequence, then 10 of them do increase the numbers out of sequence. These contribute 10 of the 13 series with n=4 that have two numbers out of sequence. The other 3 come from adding 4 to the series 321 at any poiint but the beginning.

Thus, any cell in the above table is arrived at by combining the number of out-of-sequence ways n can be added to any of the n-1 series that have x-1 numbers out of sequence, and the number of in-sequence ways that n can be added to any of the n-1 series that have x numbers out of sequence. 16 is 4+12, 61 is 33+28, 41 is 37+4.

Since we know that the number of series with one number out of sequence is (n-1)^2, it's easy to calculate how many n=6 series with two numbers out of sequence would be contributed by adding 6 out of sequence to n=5 series with one number out of sequence. Since 5 of the 25 n=6 series with one number out of sequence are produced by adding 6 to any of first 5 places of 12345 (612345, 162345, etc.), the other 20 must be ways that 6 can be added to the 16 n=5 series with one number out of sequence so that it does NOT increase the numbers out of sequence. And since there are six ways that 6 can be added to each of those 16 series (for a total of 96), the other 76 must produce n=6 series with two numbers out of sequence.

But of course that doesn't tell us how many n=6 series with x=2 are contributed by adding 6 in sequence to n=5 series with x=2.

Please understand that I'm not a mathematician, so don't judge me too harshly is this is all a fruitless waste of mental energy, and there's a much simpler solution.