User blog comment:KthulhuHimself/Plexation, TaN, and such/@comment-5529393-20151014102617/@comment-27014275-20151014110018

(n,,n,,c) should come next, and it resolves like this, ((n,n,n,...,n),,c-1), (n copies of n) now, as I stated, we IGNORE the ,,c-1 bit until we resolve ALL the previous orders, as in, suppose we eventually have ((a,b),,c-1) (a and b being whatever last two orders we're left with), once we get rid of b (meaning, turn it to 0), we get (x,,c-1)=((x,x,x,...,x),,c-2) (x being whatever last order we get then), and then just repeat the process.

You can proceed from there to (n,,n,,n), and (n,,,1), and (n,,,n) and so on.

I've calculated these ordinals meticulously, and don't think you and I had calculated them the same. I'll soon post the process leading to them.