This is a fascinating problem, and there are still a few questions I hope to explore further.

In short, yes, most definitely it requires the retrograde to be a palindrome with, potentially, a reversed “seam” in the middle (more on that in a bit.)

Here’s my reasoning. Without losing generality, we can assume that the northwest corner is PC 0, and that will always fill the NW–SE diagonal. For now, let’s call the final PC of the first row x—and this is the value we want to have go down at least the first part of the NE–SW diagonal. So far, our first two rows are:

0 _ _ _ _ _ _ _ _ _ _ x
_ 0 _ _ _ _ _ _ _ _ x _

So let’s be quite general about the second note of our P0 and call it y. Here’s the thing: we know that the second row will be exactly y less than the top row. Let’s fill some of that in:

0 y _ _ _ _ _ _ _ _ _ x
-y 0 _ _ _ _ _ _ _ _ x (x-y)

Thus, the second to last note of our P0 row needs to be y larger than x. Right, because everything in the second row needs to be y less than the top row? So we have:

0 y _ _ _ _ _ _ _ _ (x+y) x
-y 0 _ _ _ _ _ _ _ _ x (x-y)

Thus, the penultimate note is precisely as much bigger than the last note as the second note was bigger than the first (and remember, we’re in a mod-12 universe, so it is perfectly general to talk about adding numbers, since any subtracting is the same as addition by the mod-12 complement).

This logic will continue the whole way through, the third note of the row will be precisely as much bigger than the first as the third-to-last will be bigger than the last. Etc.

Now, another thing I realized is that there is a way to make the NE–SW diagonal the same the whole way, but only with 6 as the pitch class. This is because the value has to be its own complement, and only 0 and 6 have that property. Additionally, (and maybe relatedly) this is also the only way to make (any) tone row a true palindrome. Since the prime order move from the sixth to seventh note has to be the same as the the retrograde move from the seventh to the sixth while somehow being an “increase” in one direction while being a “decrease” in the other. The only interval that goes to the same pitch class either up or down is 6.

Unanswered questions that I hope to revisit:

What other restrictions are there in the note content? You definitely can’t use any row that starts in 0 and ends on 6 and get this effect.

How related is this to all-combinatorial hexachords? I have some wondering that there might be a relation.

The fact that the true palindrome version must have a tritone relation between the first and last notes, and a tritone relation in the middle definitely calls all-interval rows to mind. Coincidence?

Anyway, let me know if I need to clarify.

This is a fascinating problem, and there are still a few questions I hope to explore further.

In short, yes, most definitely it requires the retrograde to be a palindrome with, potentially, a reversed “seam” in the middle (more on that in a bit.)

Here’s my reasoning. Without losing generality, we can assume that the northwest corner is PC 0, and that will always fill the NW–SE diagonal. For now, let’s call the final PC of the first row x—and this is the value we want to have go down at least the first part of the NE–SW diagonal. So far, our first two rows are:

0 _ _ _ _ _ _ _ _ _ _ x
_ 0 _ _ _ _ _ _ _ _ x _

So let’s be quite general about the second note of our P0 and call it y. Here’s the thing: we know that the second row will be exactly y less than the top row. Let’s fill some of that in:

0 y _ _ _ _ _ _ _ _ _ x
-y 0 _ _ _ _ _ _ _ _ x (x-y)

Thus, the second to last note of our P0 row needs to be y larger than x. Right, because everything in the second row needs to be y less than the top row? So we have:

0 y _ _ _ _ _ _ _ _ (x+y) x
-y 0 _ _ _ _ _ _ _ _ x (x-y)

Thus, the penultimate note is precisely as much bigger than the last note as the second note was bigger than the first (and remember, we’re in a mod-12 universe, so it is perfectly general to talk about adding numbers, since any subtracting is the same as addition by the mod-12 complement).

This logic will continue the whole way through, the third note of the row will be precisely as much bigger than the first as the third-to-last will be bigger than the last. Etc.

Now, another thing I realized is that there is a way to make the NE–SW diagonal the same the whole way, but only with 6 as the pitch class. This is because the value has to be its own complement (note how, in your Lutyens example, the two PCs that form the other diagonal are t and 2—complements. This will always be the case), and only 0 and 6 have the property of being self-complementary. Additionally, (and maybe relatedly) this is also the only way to make (any) tone row a true palindrome. Since the prime order move from the sixth to seventh note has to be the same as the the retrograde move from the seventh to the sixth while somehow being an “increase” in one direction while being a “decrease” in the other. The only interval that goes to the same pitch class either up or down is 6.

Unanswered questions that I hope to revisit:

What other restrictions are there in the note content? You definitely can’t use any row that starts in 0 and ends on 6 and get this effect.

How related is this to all-combinatorial hexachords? I have some wondering that there might be a relation.

The fact that the true palindrome version must have a tritone relation between the first and last notes, and a tritone relation in the middle definitely calls all-interval rows to mind. Coincidence?

Anyway, let me know if I need to clarify.

This is a fascinating problem, and there are still a few questions I hope to explore further.

In short, yes, most definitely it requires the retrograde to be a palindrome with, potentially, a reversed “seam” in the middle (more on that in a bit.)

Here’s my reasoning. Without losing generality, we can assume that the northwest corner is PC 0, and that will always fill the NW–SE diagonal. For now, let’s call the final PC of the first row x—and this is the value we want to have go down at least the first part of the NE–SW diagonal. So far, our first two rows are:

0 _ _ _ _ _ _ _ _ _ _ x
_ 0 _ _ _ _ _ _ _ _ x _

So let’s be quite general about the second note of our P0 and call it y. Here’s the thing: we know that the second row will be exactly y less than the top row. Let’s fill some of that in:

0 y _ _ _ _ _ _ _ _ _ x
-y 0 _ _ _ _ _ _ _ _ x (x-y)

Thus, the second to last note of our P0 row needs to be y larger than x. Right, because everything in the second row needs to be y less than the top row? So we have:

0 y _ _ _ _ _ _ _ _ (x+y) x
-y 0 _ _ _ _ _ _ _ _ x (x-y)

Thus, the penultimate note is precisely as much bigger than the last note as the second note was bigger than the first (and remember, we’re in a mod-12 universe, so it I perfectly general to talk about adding numbers, since any subtracting is the same as addition but the mod-12 complement).

This logic will continue the whole way through, the third note of the row will be precisely as much bigger than the first and the third-to-last will be bigger than the last. Etc.

Now, another thing I realized is that there is a way to make the NE–SW diagonal the same the whole way, but only with 6 as the pitch class. This is because the value has to be its own complement, and only 0 and 6 have that property. Additionally, (and maybe relatedly) this is also the only way to make (any) tone row a true palindrome. Since the prime order move from the sixth to seventh note has to be the same as the the retrograde move from the seventh to the sixth while somehow being and “increase” in one direction while being a “decrease” in the other. The only interval that goes to the same pitch class either up or down is 6.

Unanswered questions that I hope to revisit:

What other restrictions are there in the note content? You definitely can’t use any row that starts in 0 and ends on 6 and gets this effect.

How related is this to all-combinatorial hexachords? I have some wondering that there might be a relation.

The fact that the true palindrome version must have a tritone relation between the first and last notes, and a tritone relation in the middle definitely calls all-interval rows to mind. Coincidence?

Anyway, let me know if I need to clarify.

This is a fascinating problem, and there are still a few questions I hope to explore further.

In short, yes, most definitely it requires the retrograde to be a palindrome with, potentially, a reversed “seam” in the middle (more on that in a bit.)

Here’s my reasoning. Without losing generality, we can assume that the northwest corner is PC 0, and that will always fill the NW–SE diagonal. For now, let’s call the final PC of the first row x—and this is the value we want to have go down at least the first part of the NE–SW diagonal. So far, our first two rows are:

0 _ _ _ _ _ _ _ _ _ _ x
_ 0 _ _ _ _ _ _ _ _ x _

So let’s be quite general about the second note of our P0 and call it y. Here’s the thing: we know that the second row will be exactly y less than the top row. Let’s fill some of that in:

0 y _ _ _ _ _ _ _ _ _ x
-y 0 _ _ _ _ _ _ _ _ x (x-y)

Thus, the second to last note of our P0 row needs to be y larger than x. Right, because everything in the second row needs to be y less than the top row? So we have:

0 y _ _ _ _ _ _ _ _ (x+y) x
-y 0 _ _ _ _ _ _ _ _ x (x-y)

Thus, the penultimate note is precisely as much bigger than the last note as the second note was bigger than the first (and remember, we’re in a mod-12 universe, so it is perfectly general to talk about adding numbers, since any subtracting is the same as addition by the mod-12 complement).

This logic will continue the whole way through, the third note of the row will be precisely as much bigger than the first as the third-to-last will be bigger than the last. Etc.

Now, another thing I realized is that there is a way to make the NE–SW diagonal the same the whole way, but only with 6 as the pitch class. This is because the value has to be its own complement, and only 0 and 6 have that property. Additionally, (and maybe relatedly) this is also the only way to make (any) tone row a true palindrome. Since the prime order move from the sixth to seventh note has to be the same as the the retrograde move from the seventh to the sixth while somehow being an “increase” in one direction while being a “decrease” in the other. The only interval that goes to the same pitch class either up or down is 6.

Unanswered questions that I hope to revisit:

What other restrictions are there in the note content? You definitely can’t use any row that starts in 0 and ends on 6 and get this effect.

How related is this to all-combinatorial hexachords? I have some wondering that there might be a relation.

The fact that the true palindrome version must have a tritone relation between the first and last notes, and a tritone relation in the middle definitely calls all-interval rows to mind. Coincidence?

Anyway, let me know if I need to clarify.