User blog:Tetramur/Pentational arrays - definition

I will define some words:

1. The parametre of structure is the dimension of the external level of structure and the length of one side of structure's hypercube.

2. BS is basic structure - X^^X.

3. Homogeneous structure is such a structure that doesn't have a finite number in the end. So, X^^X and X^^^X + X^^X^^3 + X^^7 + X^5 are both homogeneous. The structure that is not homogeneous is called inhomogeneous. For example, X^^X + 91 and X^^^X + X^^X^^3 + X^^7 + X^5 + 187 are inhomogeneous.

4. External level of the product of structures is the last factor in a product.

Prime block function
That is some special function that will define prime blocks of structures recursively.

1. The prime block of one entry is an entry itself.

2. When you add new structure after existing (in descending or equal order - smaller structure must follow larger), the prime blocks will be glued to each other. In another words, you can not add larger structure after smaller. You can add equal structures to each other, too.

3. When you have an infinite external level of structure, you must cut infinite parameters to p, where p is prime entry.

So, I will give examples.

The prime block of BS we consider to be known.

So, what is prime block of X^^X + 1? This is prime block of BS and plus one entry left.

X^^X * 2 is split into the sum: X^^X + X^^X. The two basic prime blocks are glued.

I have a look into my table of results and I see that X^^X * 2 is of the same ordinal type on both systems. Moving on...

X^^X * X is infinite line and infinite sum: X^^X + X^^X + X^^X + ... + X^^X. Leaving only p of these 'cause I cut infinite side of this line to p elements, by rule 3: X^^X * p. Departing into the deep sees of ill-defined which can be well-defined...

X^^X * X^2 is infinite square. Cutting structure to p tetrational hypercubes by each side, we have prime block, by using rule 3.

(X^^X)^2 is a product of two BS's: X^^X * X^^X. External level is X^^X. I cut the sides and dimensions to p: X^^X * p^^p or (p^^p)^2. Oh, this is a miracle. But the next level I will expain more thoroughly.

(X^^X)^2 * 2 is a sum of two squares: X^^X * X^^X + X^^X * X^^X. The prime block are glued.

(X^^X)^2 * X^^X. External level is X^^X, infinite structure. Therefore I must cut the external level to p elements by each side - this is tetrational hypercube. But that is (X^^X)^3.

Why ordinal of a larger structure, when added, receives ordinal of a smaller structure as the multiplier?

For structures smaller than (X^^X)^2 it follows directly from the properties of structures. I don't know the way that it can be proven for higher structures, but it is crucial for following research.

Why ordinal of a larger structure, when multiplied, receives ordinal of a smaller structure as the exponent?

It would directly follow from first question when correctly answered.

So, (X^^X)^X is followed. We cut p levels in this structure, so it will leave p elements on each side: (X^^X)^p.

Whoa, epsilon-one is reached, I think. I call (X^^X)^X "superior BS" or "SBS".