I was reviewing Deedlit's awesome explanation for how the rules of planar arrays work at
How can the number $\left\langle \matrix {3&3\\3&3}\right\rangle $ be described?
as well as
and
https://mrob.com/users/chrisb/
in an effort to better understand the effect of the second entry on the second row; in particular, in the case of (3,3(1)3,3) as asked in the first link.
I was able to get as far as determining that
(3,3(1)1,2) =
(3,3,3(1)(3,3(1)3)) =
(3,3,3(1)Latri)
and now I'm wondering if it's possible to at least work backward from the original array, substituting a downward incremented array for the first entry on the second row. (P.S. as far as I understand, there is no copilot, correct?)
In {3,3 (1) 3,3}, the third 3 is the pilot, so there is no copilot. So we decrement the pilot by 1, and replace the first row with the passengers, which is {first number} repeated {second number} times, i.e. a sequence of 3 3's. So
{3,3 (1) 3,3}
= {3,3,3 (1) 2,3}
Reducing again, we now have the third 3 in the first line as the pilot, so the copilot is the second 3. The rule then is to decrement the pilot and replace the copilot with the entire array, except with the prime (also the second) entry decremented by 1, so we get
= {3, {3,2,3 (1) 2,3}, 2 (1) 2,3}
Then the same rule again:
= {3, {3, {3,2,3 (1) 2,3} - 1, 2 (1) 2,3}, 1 (1) 2,3}
= {3, {3, {3,2,3 (1) 2,3} - 1, 2 (1) 2,3} (1) 2,3}
Now the first row is down to two entries, so the pilot is again the first number in the second row, so we get
= {3, 3, 3, 3, ..., 3 (1) 1,3} where there are {3, {3,2,3 (1) 2,3} - 1, 2 (1) 2,3} 3's in the first row
At this point we will need an enormous number of reductions to reduce the first row. After an enormous number of steps, the array will eventually reduce to something of the form
{3, X, (1) 1,3}
where X is hard to distinguish from {3, 3, 3, 3, ..., 3 (1) 1,3} itself, for example it will be bigger than {3, 2, 3, 3, ..., 3 (1) 1,3}, since the second entry will already get that big after a single reduction of {3, 3, 3, 3, ..., 3 (1) 1,3}.
So now we go to
{3, 3, 3, ..., 3 (1) {3, X-1 (1) 1,3}, 2} where there are X 3's
which is somewhat close to what you were asking. The array will eventually resolve into {3, Y (1) {3, X-1 (1) 1,3}, 2}, where Y is even more enormous, but it will never actually reduce to something of the form {3,3 (1) N, 2}.
Edit: I don't think your example is quite correct. We have{3,3 (1) 1,2}
= {3,3,3 (1) {3,2 (1) 1,2}, 1}
= {3,3,3 (1) {3,3,3 (1) {3,1 (1) 1,2} 1} }
= {3,3,3 (1) {3,3,3 (1) 3} }