I made something too large for me to understand.

The concept of this number is "repeating n times."

First, let's begin with this: SH(x)=39^x

The concept of this number is "repeating n times"(I may repeat this sentence itself n times), so let's repeat SH(x) n times. But it's the first repeat, so I write like this:


(note: we can't write SH^n(x) because there is not n)

Now you know what I will do next. Repeat that n times again.


I don't want to repeat such a formula, so I use n.


How big is this? That's easy. SH(n,x)\simeq 39 \uparrow ^{\mathrm n+1} x

It's too simple. Let's go larger.

The concept of this number is "repeating n times," that's why I repeated n times.

Now, SH(n,x) means "repeating (repeating n times) times." There are two "repeating (something) times" phrase. Now, you know where to diagonalizate. Yes. I will have n "repeating (something) times" phrase.

For more extension, I'll change the definition slightly:



This is larger than the first definition, but the number will have same number of up-arrows.

In this case, SH(n,0)has n times more repeat.

Now, I can translate "n 'repeating (something) times' phrase" in math languege.

Small letters are numbers and capital letters are "array" of numbers.




(note: N,M can be nothing and 0 in the second formula is the leftmost zero.)

Now, SH can have any number of numbers.

It's hard, but let's calculate simple one.

SH(1,n)=\underbrace {SH(0,SH(0,...SH(1,0))}_n


\therefore SH(1,n)\simeq 39\uparrow\uparrow n+1

In the almost same reason, SH(n,x)\simeq 39 \uparrow ^{\mathrm n+1} (x+n)


\simeq SH(39\uparrow\uparrow n+1,n)

\simeq 39\uparrow ^{ \mathrm 39\uparrow\uparrow n+1} (n+39\uparrow\uparrow n+1)

\geq 39\to n+1 \to 2 \to 2

It is still calculable, but now I stop calculating. "Defining a number" is important.

For my history, I will write old version of Ita-Chihaya number.

(note: I will write the Ita-Chihaya number version with Greek letters.)

Ita-Chihaya number(alpha)=SH(72,72,...(72 times)...,72,72)

Contnue to Ita-Chihaya number(2).

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