User blog comment:Eners49/The secret 0th hyper-operator?/@comment-2601:142:2:EC49:A496:CA1D:3DA2:B5DD-20180728144329/@comment-2601:142:2:EC49:A496:CA1D:3DA2:B5DD-20180728185841

Response to Eners49: Thank you! Your idea probably makes the most sense logically, although it is interesting to pick apart ?^n notation, even if it doesn't always make sense.

Response to A fandom user:

ONE QUESTION MARK AT A TIME YOU PEOPLE! XD.

To at least partly answer your question, I'm gonna try to take another look at ?? notation, (Using the first definition, not Eners' new one.) and try to see what's going on.

WARNING: ANOTHER ESSAY INCOMING. LIKE THRICE AS LONG AS THE FIRST ONE.

Here are all the two-input ?? numbers we have created so far:

2??2 = 4

And that's it.

Not to worry! We can use the x?y when y is at least 2 more than x to figure out some more, like 0?2 and -1?2. Since all we need is for y to be two, we can create INFINITELY MANY ?? numbers that have calculatable values.

Here are some more ?? two-input numbers with the new method:

0??0 = 3

-1??-1 = 3

-2??-2 = 3

No growth. No nothing.

To get a grasp on ?? notation, we need more ?? values, but to get more values, we need to get more ? values, but to do that, we need more x?2 numbers, but to do THAT we need to figure out examples of x?y that DON'T fit into our old x?y rules!

To do that, let's try to answer Eners' question about 3?4, or at least BOUND the answer. We know it's smaller than 3?5 which makes it smaller than 3?3?3 which makes it smaller than 6. By the same token, we know that it's larger than 3?3 which makes it larger than 5.

This means that 3?4 can be bounded between 5 and 6. Mystery (sort of(not really(ok not at all))) solved.

In fact, x?(x+1) can be bounded between x+2 and x+3 in the same fashion.

What about x?(some number smaller than x)?

Well, it can be bounded between x and x?x, which is x+2. So x?y when y<x can be bounded between x and x+2.

We can bound it further with some good-old 0=1 logic. We know that x is less than x?anything because x is just one repitition of the ? function, which can be written as x+1.

Huzzah for black magic mathematics! x?y when y<x can NOW be bounded between x+1 and x+2.

Now that we've created rules for bounding ? numbers, let's make a table of ? numbers!

x?y where the possibilites for x are on top and the possibilities for y are on the left: (If this doesn't turn out well in the post than oops.)

1      2      3      4      5

1      3     3-4    4-5   5-6   6-7

2     3-4     4     4-5   5-6   6-7

3      4     4-5     5     5-6   6-7

4      5       5     5-6    6     6-7

5      6       6      6     6-7    7

And now we can take all the numbers where y is two, (i.e. the 2nd row) and then plop them into x??x form and:

5??5 = 6-7

4??4 = 5-6

3??3 = 4-5

2??2 = 4

1??1 = 3-4

0??0 = 3

-1??-1 = 3

-2??-2 = 3

No obvious pattern.

Well, to get even MORE ?? values we could reuse some of the tricks Eners used to get ? values before I came up with my ? notation rules two hours ago.

He showed that 2+3 = 2?4 = 5, which we can extend downward to show that 2?3 = 2??4 giving us our first x??y number where x and y AREN'T the same.

2??4 = 4-5

We can also take the ?? numbers that have a definite value and use the ?? function on them again, then extend them up to ? notation, and get an actual value!

0??(0??0) = 0??3 and 0?3 = 4

-1??(-1??-1) = -1??3 and -1?3 = 4

-2??(-2??-2) = -2??3 and -2?3 = 4

These give us our first rule about ?? numbers: If y is at least 3, and x is 0 or less, than x??y = y+1.

'''ALL THAT FOR ONE EXTREMLY SPECIFIC RULE! A FANDOM USER, IF YOU'RE STILL HERE READING THIS, WHO KNOWS WHAT YOUR NESTED QUESTION MARK NUMBER IS! I'M NOT GONNA FIGURE IT OUT BEFORE THE NEXT POINCARE RECCURENCE TIME!'''