BEAF (Bowers' Exploding Array Function) is a notation invented by Jonathan Bowers, used for expressing certain very large numbers. It is defined like so:
{a, b} = ab
{a, b, c} = a^(c)b
{a, b, 1, d} = {a, a, {a, b - 1, 1, d}, d-1}
{a, b, c, d} = {a, {a, b-1, c, d}, c-1, d} when c > 1
Any array starting with a 1 will degenerate to 1, and any array with a 1 as the second argument will degenerate to the first argument. An array with a 1 can be truncated just before the 1 if it occurs in the second argument, but 1s in later positions serve as placeholders. Because 2^^^...^^^2 = 4 regardless of the number of arrows, any array that starts with two 2s will degenerate to 4. Below are some examples of 3-argument arrays:
{3, 2, 2} = 3^^2 = 27
{3, 3, 3} = 3^^^3 = 3^^7,625,597,484,987 (tritri)
{4, 4, 2} = 4^^4 = 2361022671...............5261392896 (8.072304*10153 digits)
{4, 4, 3} = 4^^^4 = 4^^(4^^(4^^4))
{4, 4, 4} = 4^^^^4 (tritet)
4-argument arrays are when the notation breaks into new territory. We will begin with:
{3, 3, 1, 2} = {3, 3, {3, 2, 1, 2}} = {3, 3, {3, 3, 3}} = {3, 3, 3^^^3} = 3^^^^^^^^^^^^^...^^^^^^^^^^^^^3 w/ 3^^^3 ^s
This number is called the tritriplex, analogous to the googolplex. Similarly, {3, 4, 1, 2} is equal to {3, 3, {3, 3, 1, 2}}, or 3^^^^^^^^^^^^^...^^^^^^^^^^^^^3 with {3, 3, 1, 2} arrows! Indeed, {3, 65, 1, 2} is greater than Graham's number. In fact, {a, b, 1, 2} is equal to a expanded to b (denoted a {{1}} b), or a{a{a{...{a{a{a}a}a}...}a}a}a with b copies of a. Expansion with the base of 2 will collapse to 4 even if the polyponent is greater than 2, as it would iterate the number of up-arrows between two 2s which just collapses to 4.
10{{1}}100 is the corporal. The corporal is roughly approximated by G_99 in the continued Graham's sequence.
The next of Bowers' operators, multiexpansion, is denoted a{{2}}b, or {a, b, 2, 2}. a{{2}}b is equal to a{{1}}a{{1}} ... {{1}}a{{1}}a with b copies of a.
Bowers also defined the grand tridecal, equal to {10, 10, 10, 2}.
The next Bowersism we will cover is the tetratri, equal to an array of four 3s, or 3 powerexploded to 3:
{3, 3, 3, 3} = 3{{{3}}}3
To get to the tetratri, start with 3. Call this Stage 1. Each stage after Stage 1 is 3{{3{{3...{{3{{3}}3}}...3}}3}}3 where the number of 3s is the previous stage. Stage 2 is already equal to 3{{3}}3, and Stage 3 will
The tetratri is a number that leaves Graham's number in the dust... but we are just getting started with Bowers' work!
We are now at the beginning of 5-argument arrays. The general series covered several paragraphs back is the series of megotions of 10, or {10, n, 1, 1, 2}. Megotion with the base of 2 degenerates to 4, as with all of Bowers’ operators we’ve covered, and so we will begin with megotions of 3. The term "megotion" was coined by Aarex Tiaokhiao.
{3, 2, 1, 1, 2} = {3, 3, 3, 3}
So {3, 2, 1, 1, 2} is really just the tetratri in disguise. In fact, n megoted to 2 is {n, n, n, n}. Next:
{3, 3, 1, 1, 2} = {3, 3, 3 {3, 2, 1, 1, 2}} = {3, 3, 3, tetratri}
This number is 3 {{{{{{{{{{{{{...{{{{{{{{{{{{3}}}}}}}}}}...}}}}}}}}}}}}} 3 with a tetratri pairs of brackets! By just increasing the polyponent by 1, we have blown past the breaking point of Conway chained arrow notation! Next up we have:
{3, 4, 1, 1, 2} = {3, 3, 3 {3, 3, 1, 1, 2}} = {3, 3, 3, {3, 3, 3, tetratri}}
Now we have a number that is 3 {{{{...{{{3}}}...}}}} with X brackets, where X is the previous insane number!
{4, 2, 1, 1, 2} = {4, 4, 4, 4}
Ditto.
{4, 3, 1, 1, 2} = {4, 4, 4, {4, 2, 1, 1, 2}} = {4, 4, 4, supertet}
Here we have 4 {{{{{{{{{{{{{{{{{{{{...{{{{{{{{{{4}}}}}}}}}}...}}}}}}}}}}}}}}}}}}}}}}} 4, where the number of bracket pairs is a supertet! Of course this is larger than {3, 3, 1, 1, 2}, because the tetratri is obviously less than the supertet.
{4, 4, 1, 1, 2} = {4, 4, 4, {4, 3, 1, 1, 2}} = {4, 4, 4, {4, 4, 4, supertet}}
This number is obviously larger than {3, 4, 1, 1, 2}, as the base is 4 instead of 3.
In general, {a, b, 1, 1, 2} = {a, a, a, {a, b-1, 1, 1, 2}}, and {a, 1, 1, 1, 2} = a.
pentatri = {3, 3, 3, 3, 3}
pentatet = {4, 4, 4, 4, 4}
superpent = {5, 5, 5, 5, 5}
hexatri = {3, 3, 3, 3, 3, 3}
hexatet = {4, 4, 4, 4, 4, 4}
superhex = {6, 6, 6, 6, 6, 6}
heptatri = {3, 3, 3, 3, 3, 3, 3}
supersept = {7, 7, 7, 7, 7, 7, 7}
superenn = {9, 9, 9, 9, 9, 9, 9, 9, 9}
iteral = {10, 10, 10, 10, 10, 10, 10, 10, 10, 10} (can also be called superdecal)
We are about to cover numbers much larger than anything we’ve seen in the entire document up to this point.
The ultatri is an array of 27 3s. This was a number that came up often when working with dimensional arrays of 3s in the old version of BEAF, because it was equal to {3, 3, 3, 3, …, 3, 3, 3} w/ {3, 3, 3} 3s, but that is now a much larger number we will encounter in a bit.
Bowers’ first dimensional array googolism is the goobol, which is {10, 100(1)2}. Written out as a linear array, it is:
{10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10}
Just imagine trying to solve that!
In general, 2-row arrays are defined like so:
{a, b(1)c} = {a, a, a, a, a, a…, a, a, a, a, a(1)c-1} w/ b copies of a (In the case of c=2, this is just a shorthand for linear arrays)
{a, b, 2(1)2} = {a, a, a, a, a, a, a, …, a, a, a, a, a, a, a}
{a, a, a, a, a, a, …, a, a, a, a, a, a}
{a, a, a, a, a, a, …, a, a, a, a, a, a}
…(b layers)…
{a, a, a, a, a, a, …, a, a, a, a, a, a}
a
(or, alternatively, {a, {a, {a, …{a, {a, a(1)2}(1)2}…}(1)2}(1)2} w/ b copies of a
The dupertri is equal to {3, {3, 3, 3}(1) 2}, or {3, 3, 2(1)2}, which is a linear array of tritri 3s! If 4 arguments was enough to crush Graham’s number, and 5 arguments left Conway chained arrow notation in the dust, imagine what this can do.
Bowers’ next 2-row array googolism, the gossol, is equal to {10, 10(1) 100}, or {10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 99}.
Bowers' next googolism, the latri, is much larger, but there is still an explanation for it; in stages. Start with 3. Call this Stage 1. Next have {3, 3, 3}. This is a power tower of 7,625,597,484,987 3s. Call this Stage 2. Next have an array of {3, 3, 3} 3s, or a dupertri. Call this Stage 3. Latri is Stage Dupertri. MIND-BLOWN!!!
And this is all just the latri, which is one of Bowers’ smallest dimensional array googolisms! The latri can also be expressed as {3, 3 (1) 3}. Below we will try to solve {3, 3, 3 (1) 3}, which Sbiis Saibian describes as “vastly larger than any number of 3s you could think of using only a single row”.
{3, 3, 3 (1) 3}
= {3, {3, 2, 3(1)3}, 2(1)3}
= {3, {3, {3, 1, 3(1)3}, 2(1)3}, 2(1)3}
= {3, {3, {3, 3, 2(1)3}, 2(1)3}, 2(1)3}
= {3, {3, {3, {3, 2, 2(1)3}, 1(1)3}, 2(1)3}, 2(1)3}
= {3, {3, {3, {3, {3, 1, 2(1)3}, 1(1)3}, 1(1)3}, 2(1)3}, 2(1)3}
= {3, {3, {3, {3, 3, 1(1)3}, 1(1)3}, 2(1)3}, 2(1)3}
= {3, {3, {3, {3, 3(1)3}, 1(1)3}, 2(1)3}, 2(1)3}
= {3, {3, {3, latri(1)3}, 2(1)3}, 2(1)3}
= {3, {3, {3, 3, 3, 3, … 3, 3, 3, 3(1)2}, 2(1)3}, 2(1)3} w/ latri 3s in the middle expression
As if that wasn’t enough, we can have multiple arguments on the second row:
{5, 5, 3(1)3, 5}
=
The smallest non-trivial 3-row array is {a, b(1)(1)c}, which solves to {a, a, a, …, a, a, a(1)a, a, a, …, a, a, a} w/ b copies of a on either side of the (1).
{3, 2(1)(1)2}
= {3, 3(1)3, 3}
The cyperal is an unusual googolism that I encountered on the internet around the time of writing this, which was on the Googology Wiki but was deleted for having no sources. Its array form is:
{65, 1010¹⁰⁰ (1) 1000, 10100 (1) 7 x 101,623, 42}
which, in planar form, is...
|65, 1010^100 |
|1000, 10100 |
|7x101623, 42|
Yikes! The cyperal is a number so large that it requires three rows to represent! It is, in fact, the only googolism with a base between 11 and 99 that I did not invent. In fact, we can compute its last digits because it is a mind-bogglingly vast power tower of 65s.
It is easy to show that the cyperal must end with ...25. Since it must be a super massive power of 65, and 65 is one more than 26, all powers of 65 must end with 25 since 65 x 65 = 4225, and 25 x 65 = 1625.
Finding more terminating digits of the cyperal is not as easy. In fact, we can find the last six digits of the cyperal based on the fact that 65 = 26 + 1. Since 653 is 274,625, and 625 x 65 is 40625, so all powers of 65 end with 625, and so the cyperal ends with ...625.
The same fact can be used to show that the cyperal ends with ...0625. In fact, 655 ends with 90625, and 90625 x 65 ends with 90625, and so the cyperal ends with ...90625.
656 ends with 890625, and 890625 x 65 is 57890625, which also ends with 890625, and this means that the cyperal ends with ...890625.
Finding the millions place digit of the cyperal is not so easy, but the last 7 digits of powers of 65 alternate between 7890625 and 2890625 after 656. So, we have:
65N mod 10000000 = 2890625 when N is even and N > 7
65N mod 10000000 = 7890625 when N is odd and N ≥ 7
The cyperal can be imagined as:
6565^65^65^65^65^65^65^65^65^65...65^65^65^65^65^65^65^65^65^65
where the number of 65s in the tower is an integer which, on this scale, is just under a cyperal. We can eventually reduce this to the form:
6565^N
where N is some unknown integer (which is itself a vast power tower of 65s). Despite that all we know about that integer is that it is just under the cyperal and that it is itself a very large power of 65, we don't need to know anything about it to determine the last 7 digits of the cyperal.
In 6565^N, the number n in 65n must be odd, and so 6565^N ends with 7890625. So we have proven that the cyperal ends with ...7890625. Can we do better? It turns out that we can.
To find the last 8 digits of the cyperal we need to go up another level of abstraction. This time we look at the modulo 4 of the exponent.
65N mod 100000000 = 12890625 when N is divisible by 4 and N ≥ 8
65N mod 100000000 = 37890625 when N mod 4 = 1 and N ≥ 8
65N mod 100000000 = 62890625 when N is even but not divisible by 4 and N ≥ 8
65N mod 100000000 = 87890625 when N mod 4 = 3 and N ≥ 8
However, in 6565ⁿ, N always leaves a remainder of 1 when divided by 4. What this means is that the last 8 digits of the cyperal are ...37890625.
Using modular arithmetic, it is possible to obtain many more terminating digits of the cyperal. Below are the last 308 digits of the cyperal.
...90910226371387483205853844566554438559558443336328810534248414126709046020881625058145449546742497844939457039077907342702945277369631189580823223007998281862726982878913986604434116952564558971473499957685769695150699189279248080802346630929306365756474625825674638381068692893904881202615797519683837890625
The dimentri is equivalent to a 3x3x3 cubic array of 3s.
dimentri = {3, 3, 3 (1) 3, 3, 3 (1) 3, 3, 3 (2) 3, 3, 3 (1) 3, 3, 3 (1) 3, 3, 3 (2) 3, 3, 3 (1) 3, 3, 3 (1) 3, 3, 3}
The dimentri is an unspeakably huge number, much larger than anything you could ever think of by merely iterating the number of 3s in a linear array (that would merely be {3, N, 2(1)2}. It can also be represented as {3, 3 (3) 2}.
It turns out that as we solve the cube, we would eventually get it down to two dimensions, but it would be an incomprehensibly large plane! And of course we would have to solve it into a linear array, and we finish with some finite, but indescribably large array of 3s.
Well, this number pales in comparison to what is coming...
We can, in fact, define more dimen-x numbers to fill the gap between the dimentri and Bowers' other low-level dimensional array googolism, the dimendecal.
We can define the dimentet as a four-dimensional array of 256 4s, or a 4x4x4x4 tesseract of 4s, where a tesseract is the 4-dimensional analog of a cube. Its full array form is:
{4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (3) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (3) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (3) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (2) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4 (1) 4, 4, 4, 4}
This number is much larger than the dimentri! In fact, it can be represented as 44 & 4, which means that there are 44, or 256 4s in the array, and the 4s are arranged into a 4x4x4x4 tesseract.
The dimenpent is a 5x5x5x5x5 penteract of 5s, where a penteract is the 5-dimensional analog of a cube. It can be represented as 55 & 5, indicating that there are 3,125 5s in the array, and also how the 5s are arranged.
The dimenhex is a 6x6x6x6x6x6 hexeract of 6s, where a hexeract is the 6-dimensional analog of a cube. It can be expressed as 66 & 6, which indicates that there are 46,656 6s in the array arranged into a 6x6x6x6x6x6 6-dimensional hypercube. Storing the array for this number would take 40 pages in a Word document.
Next, we have the dimendecal. It is impossible to type its full array form here (which would have 10 billion 10s), as it would take several gigabytes of space, and also it would require millions of pages, but it would start and end as:
{10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (2) 10, 10, 10, 10............
............10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (2) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10 (1) 10, 10, 10, 10, 10, 10, 10, 10, 10, 10}
The gongulus is a 100-dimensional array of a googol 10s. Bowers says about this number, it never comes to an end when trying to solve. It is given as an example of a number in superclass 5 on the Googology Wiki article about the concept of superclasses devised by Robert Munafo.
At first, a 100-dimensional cube doesn't sound like that much, but once again, there are a googol 10s in it, and when we get into the centeract, we will find that looks are really deceiving with numbers of this magnitude.
To get a feel for the centeract of 10s, you would first split it into ten 99-dimensional cubes. Then, you would split each 99-dimensional cube into ten 98-dimensional cubes. Each 98-dimensional cube consists of ten 97-dimensional cubes. Then, you would split each 97-dimensional cube into ten 96-dimensional cubes. This means that there are 100 98-dimensional cubes, and 1000 97-dimensional cubes.
Now to solve the gongulus array, you would first decompose the array into ten incomprehensibly large 99-dimensional cubes. But the next step would require solving a very large number of 98-dimensional cubes, and solving each 98-dimensional cube would require solving many 97-dimensional cubes. Then, you would have to solve many 96-dimensional cubes, and even more 95-dimensional cubes, and still more 94-dimensional cubes, and yet more 93-dimensional cubes, and even more 92-dimensional cubes, and then you would have to get down to the hepteracts, then solve a large number of hexeracts, and then solve an even larger number of penteracts, and then a still larger number of tesseracts, and then a yet larger number of cubes. Each of all these cubes consists of 10 squares, and it would require lots of solving to decompose the squares, and then the lines! And that is all just the first 98-dimensional cube of the first 99-dimensional cube!
You would do the same thing again, and repeat the process an extremely large number of times! After the first 99-dimensional cube is finished, you would have to solve a second 99-dimensional cube, which involves solving a huge amount of 98-dimensional cubes, and an even bigger amount of 97-dimensional cubes, and then more 96-dimensional cubes, and an insane amount of 95-dimensional cubes, and in between that and the squares there will be 52-dimensional cubes, and then 51-dimensional cubes, and then 50-dimensional cubes..., 34-dimensional cubes, 33-dimensional cubes, 32-dimensional cubes,..., 21-dimensional cubes, 20-dimensional cubes, 19-dimensional cubes, 18-dimensional cubes..., but you will have done this once before. Then you would have to solve the hepteracts, then the hexeracts, then the penteracts, and then the tesseracts, and then the cubes, the squares, and finally the lines, once again! Repeat this solving of 99-dimensional cubes eight more times. Then you will eventually end up with a very long linear array of 10s, and before we can even get it down to a dozen fewer terms, it has gone beyond what we can really describe.
And then, you would have to get this down to a three-entry array, which you would have to convert to an up-arrow expression, where we would take away one up-arrow and iterate that operator so that the expression has ten 10s, and then solve 10 ↑N - 1 (10 ↑N - 1 (10 ↑N - 1 (10 ↑N - 1 (10 ↑N - 1 (10 ↑N - 1 (10 ↑N - 1 (10 ↑N - 1 10))))))), where N is the insane number that appears at the end of the three-argument array that we reached a bit earlier. Then, take that insane number that came from the up-arrow expression a bit earlier in this paragraph, take away one up-arrow after the first 10, and then have that number of 10s in the expression! Then we would have to get it down to a hendecation of 10, and then a decation of 10, then an ennation of 10, then an octation of 10, then a heptation of 10, and then a hexation of 10, and then a pentation of 10, and then a tetration of 10, and then a very large power tower of 10s which we would have to solve like a normal power tower. Once the power tower has been reduced to two levels, multiply N 10s together, where N is the number that results from the power tower of 10s with the bottommost 10 chopped off. The result will be a gongulus.
A full explanation of Graham's number would be nothing compared to this overview of the computation procedure of a gongulus.
We have now worked our way up to the triakulus, which is Bowers' smallest pentational array googolism. It is equal to 3 ↑↑↑ 3 & 3. Its array is difficult to even begin to imagine. You would first have to imagine what a pentational array looks like, and then imagine filling that array with tritri 3s.
The triakulus can also be expressed as {3, 3, 3} & 3, which means it lies on an interesting spot: the space that the array is in can be represented using array notation. This also places the triakulus at the beginning of {X, X, X} arrays.
The triakulus is also the smallest ill-defined Bowersism, because arrays beyond (X^^X)2 have two different interpretations. Bowers himself strongly supports the stronger interpretation, known as the “climbing-method interpretation”, which we will also use, and which I suggest adopting as standard since Bowers himself supports it.
The kungulus is equal to 10 ↑↑↑ 100 & 10. Its array has a gaggol 10s, and is in pentational space. It can also be represented as {10, 100, 3} & 10. Before long, we will reach arrays that are so large that the space they are in needs to be represented with the various array structures!
A tridecatrix is equal to 10 ↑↑↑↑↑↑↑↑↑↑ 10 & 10, and so it is a dodecational array of a tridecal 10s. It is too difficult to even explain. And yet we still have a long way to go before we hit the next number in the -akulus series!
The tridecatrix can also be expressed as {10, 10, 10} & 10, which means it is also a linear array array of a tridecal 10s. Now we will move on to Bowers' next googolism, the homungulus.
The homungulus is equal to {10, 10, 100} & 10. It is a linear array array of a boogol 10s. It marks the beginning of a “Great Silence” – Bowers’ next googolism (the golapulus) just completely leaves this number in the dust! The famous TREE(3), which we will cover later, lies in this gap.
Extended Hyper-E notation reaches about this far. The largest Saibianism is still less than the next Bowerism.
We can now define googolisms to fill in the gap between the homungulus and the next Bowerism. We will define the hyperdimentri as 33 & 3 & 3, or a dimensional array array of 3s, where there are a dimentri 3s.
The dimentri is already insanely large, so a structural array of that many 3s would be even more absurdly huge. To imagine this number, you would first have to envision a dimensional array array, and then imagine filling that structure with a dimentri 3s. It is even harder to imagine than the triakulus.
Then, you would have to imagine solving the dimensional array array and going through the horrendous process of decomposition of the array, and then in the end, you would get a huge power of 3.
Bowers' next googolism is the golapulus. Remember the long gongulus paragraphs? Well, the gongulus is the mere number of entries in the structural array for this number! The golapulus can be expressed as 10100 & 10 & 10. It is a dimensional array array number, and it is the first Bowerism to be larger than TREE(3).
The hypertriakulus, a number defined in part 6 of the Pointless Gigantic List of Large Numbers, is equal to 3 ↑↑↑ 3 & 3 & 3, or a pentational array array of a triakulus 3s. This number is unspeakably huge, for even just the space the array is in is a pentational array, which is difficult to imagine!
Bowers' first legion array googolism is the big boowa. It is so large that it cannot be written out using the array of operator, but it can be represented using legion arrays as {3, 3, 3 / 2}. Below is an explanation of how this number is computed.
First, start with 3. Now have this many threes in an array of expression. The result is a triakulus, which we saw earlier. This is equal to 3 ↑↑↑ 3 & 3. One would first have to imagine what a pentational array might look like, and then fill that with a tritri 3s! Then, go through the nightmare of solving the triakulus array, and it would be a seemingly endless amount of time before simplifying it to a massive trimensional array, then a superdimensional array, then a dimensional array with a huge number of dimensions.
Then, get this down to a penteract, then a tesseract, then a cube, then a square, and then you would reach an insanely long linear array, and after loads and loads of solving, you would reach an array of the form {3, 3, N}, where N is an integer that is just under a triakulus. Then, you would convert this form to an up-arrow notation, where you would have to iterate the values until you get a very large power of 3.
As you saw, 3 & 3 & 3 (or Stage 2) is already frighteningly large. But the next stage is ... hold your breath now ... 3 & 3 & 3 & 3 & 3 ... 3 & 3 & 3 & 3 & 3 where the number of 3s is a triakulus! This array-of expression would stretch for the diameter of the observable universe, and unfathomably far beyond!
To compute Stage 3, first start with 3. Call this Step 1. Then have an array of three 3s. Call this Step 2. Since the result is a pentational number, now have a pentational array where the number of 3s is Step 2! This number is triakulus, or Step 3. And now keep going -- Step 4 is hypertriakulus, Step 5 is a pentational array array array of 3s with a hypertriakulus 3s, Step 6 is a pentational array array array array of 3s where there are Step 5 3s, and continue to Step 3 ↑↑↑ 3 & 3. That is stage 3! And yet this is just the beginning of the big boowa's computation...
Continue to stage 4, which is 3 & 3 & 3 & 3 & 3 ... 3 & 3 & 3 & 3 & 3, where the number of 3s is the previous insane number! You would have to solve this long expression, but you would have gone through the solving process once before. First solve the first few 3s at the beginning and go through the sequence of 3, tritri (step 2), triakulus (step 3), hypertriakulus (step 4)..., and then, go out to step triakulus to reach stage 3, and keep going -- go to step googol, step Graham's number, step tridecatrix, step homungulus, step TREE(3), until you get to step Stage 3. Now that is a real super giant. You will now have to solve the pentational array array array ... array array, where you say array {3, triakulus / 2} - 2 times!
So you would have to get it all the way down to a pentational array array array, then a pentational array array, then an explosional array, then an expandal array, then a much bigger pentational array, and then a yet bigger tetrational array, and then an absurdly huge trimensional array, then a superdimensional array which is even more mind-boggling, then a dimensional array where the number of dimensions is mind-crushingly huge! Then, you would have to get this down to a quartic array, then a cube, and you will finally have an insanely huge planar array! You will now have to get it down to a linear array, which leads to insane levels of decomposition, and eventually a huge power of 3!
Then go to stage 5, which is 3 & 3 & 3 & 3 ... 3 & 3 & 3 & 3, where there are stage 4 (the insane number we just reached) threes! Here, we would solve the array-of expression again, but this time we would go out X steps, where X is the insane value we reached in the last paragraph. Here, you would have to get the pentational array array array array ... array array (where you say array M - 2 times, where M is Stage 4) down to a detonational array, then an explosional array, then an expandal array, and in between there will be powerexplodal arrays, multiexplodal arrays, explodopentational arrays, explodotetrational arrays ... expandopentational arrays, expandotetrational arrays, powerexpandal arrays, multiexpandal arrays, and then finally, you would reach a huge hyperoperational array. Then, you would have to decompose this down to a pentational array, then a tetrational array, then an insanely big trimensional array, and then an even bigger superdimensional array, and then a
Now go to Stage 6, which is equal to 3 & 3 & 3 & 3 ... 3 & 3 & 3 & 3, where the number of 3s is the result of Stage 5, and decompose the expression into a pentational array array array array ... array array where you would say array N - 2 times, where N is the value of Stage 5! Then you would have to get it down to a very large detonational array, then an even larger powerexplosional array, then a still larger multiexplosional array. Then you would have to decompose this into an explosional array, then an expandal array, then a hyperoperational array. It would seem to never come to an end, but now, after this, you would continue to feed back the number of 3s in the array-of expression.
Keep going to the tenth stage, the hundredth stage, the thousandth stage, the millionth stage, the googolth stage, the googolplexth stage, the giggolth stage, the tridecalth stage, the boogolth stage, the Graham's Numberth stage, the dimentrith stage, the gongulusth stage ... ... ... ... ...
Ready for it? Big boowa is equal to Stage Stage 3 -- this number just leaves even something like TREE(3) way behind! So it is impossible to just write out the number of times we would have to iterate the number of 3s in the array-of expression in terms of an array-of expression! In fact, you would have to go through the whole array-of solving fun zillions of gazillions of times before you reach the big boowa!
And that is not quite the full picture. Incredibly, this is still not the largest number Bowers has coined.
The great big boowa is equal to {3, 3, 4 / 2}. And, the third and final member of the big boowa group is the grand boowa, which is equal to {3, 3, {3, 3, 3 / 2} / 2}. Now, we can extend the big boowa group, with the following definitions:
n-ex-great big boowa = {3, 3, 3+n / 2}
n-ex-grand boowa = fn(big boowa), where f(n) = {3, 3, n / 2}
The largest of Bowers' googolisms is the meamealokkapoowa oompa, which is defined as {{L100, 10}10,10&L, 10}10,10, or a meamealokkapoowa-sized array of Ls. The meamealokkapoowa oompa is one of the largest numbers known, but it is very ill-defined, and estimates of the meamealokkapoowa oompa in the fast-growing hierarchy range from fθ(Ω^Ω + Ω)+1(10) (just slightly above the large Veblen ordinal) to fΨ(Ψ_I(0))(10) where “I” is the inaccessible cardinal.