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Not to be confused with great googol.

The greagol (short for 'great googol') is equal to E100#100#100, using Hyper-E Notation.[1][2] The term was coined by Sbiis Saibian. The greagol can be described as the 100th member of the "grangol series", where the 1st member is a grangol (E100#100), the 2nd is the grangoldex (E100#grangol), the 3rd is the grangoldudex (E100#grangoldex), ... and in general the nth member of the series, call it S(n), is equal to E100#S(n-1). It follows that:

Greagol = E100#(E100#(E100#( ... ... (E100#(E100#100)) ... ... )))

w/100 "E"s in the expression.

A greagol can also be defined as follows:

Greagol = 101010......1010100 w/ 101010......1010100 w/ ... ... ... ... w/ 101010......1010100 w/100 10s 10s ... 10s

where there are 100 power towers, and the number of tens in each is equal to the power tower to the immediate right. The rightmost power tower contains exactly 100 10s and is equal to the grangol.

This number is also called towergol by Aarex Tiaokhiao.[3]

## Comparison with gaggol Edit

The greagol is greater than and comparable to the gaggol. A gaggol = E1#1#100. Therefore:

gaggol = E1#1#100 < E100#100#100 = greagol

Although it would seem that a greagol is only slightly larger than a gaggol, it can be shown that:

gaggolgaggol < greagol

Where the left sided superscript is shorthand for gaggol tetrated to the gaggol. Proving this is a little more tricky than proving that giggolgiggol < Grangol, but can still be done. To prove this we observe that:

$\text{gaggol} \uparrow\uparrow \text{gaggol} =(10 \uparrow\uparrow 10 \uparrow\uparrow\uparrow 99) \uparrow\uparrow \text{gaggol} < 10 \uparrow\uparrow (10 \uparrow\uparrow\uparrow 99 + \text{gaggol}) << 10 \uparrow\uparrow (\text{gaggol}^2)$

This follows from the fact that $$(10\uparrow\uparrow A)\uparrow\uparrow B < 10\uparrow\uparrow (A+B)$$, which follows from the Knuth Arrow Theorem.

$10 \uparrow\uparrow (\text{gaggol}^2) < 10 \uparrow\uparrow 10 \uparrow\uparrow (10 \uparrow\uparrow\uparrow 99 +1)$

This follows from the fact $$\text{gaggol}^2 = 10^{2\cdot\text{gaggol}} << 10^\text{gaggol} = 10 \uparrow\uparrow (10 \uparrow\uparrow\uparrow 99 +1)$$

Further,

1+10^^^991010 < 1+10^^^98101010 < ... < 1+10^^^110...10 w/100 10s = 1110...10 w/100 10s = E1#11#100 < E100#100#100.

Therefore, gaggolgaggol < greagol.

Furthermore, we can provide the following upper and lower bounds for gaggolgaggol:

E1#10#100 < gaggolgaggol < E1#11#100

This proof is very similar to the one comparing giggol with a grangol. Back then, giggolgiggol was bounded between E10#100 and E11#100. Now, gaggolgaggol is bounded between E1#10#100 and E1#11#100. See the similarities?

Gigangol is also larger than geegol^^^geegol, and geegol^^^geegol can be similarly bounded in Hyper-E. Notice that the hyper operators used for bounding giggol, gaggol, and geegol increase with each step (giggol^giggol < grangol, gaggol^^gaggol < greagol, and geegol^^^geegol < Gigangol).

## Approximations in other notations Edit

Notation Approximation
Up-arrow notation $$100 \uparrow\uparrow\uparrow 101$$
Chained arrow notation $$100 \rightarrow 101 \rightarrow 3$$
Hyperfactorial array notation $$102!2$$
BEAF $$\{100,101,3\}$$
Fast-growing hierarchy $$f_4(100)$$
Hardy hierarchy $$H_{\omega^4}(100)$$
Slow-growing hierarchy $$g_{\zeta_0}(100)$$