  ## Solution - identify, SORT, index, solve

by Derek Kisman

This is a "standard" ISIS (Identify, Sort, Index, Solve) puzzle. However, unusually, most of the challenge actually lies in the second step: the Sort.

Step #1 is to Identify. The images all clue words sounding like XXXlion (ie, fake big-number words).

Step #2 is to Sort, and this is a doozy. The words are all assigned to particular numbers. Enormous numbers. Really, really, really, REALLY enormous numbers. The only way to have a hope of getting the Sort right is to bring out some mathematical tools. When defining the numbers I stuck to terminology that you'll find in mainstream mathematics (thus on Wikipedia), so you won't see BEAF or Extensible-E numbers. But you should quickly stumble on the Googology Wiki, a rather mind-blowing site that will still come in handy.
Wolfram Alpha is also an invaluable tool, trivializing some of the comparisons.

Here are the identified images (including their final sorted position):

 1 TRILLIAN = 2 up^(10^100) 2 3 REPTILIAN = Number of particles in the known Universe 10 GILLIAN = 43^45^47^49^41^3 15 HYLIAN = 6 inside square 25 MEDALLION = 3->2->3->3 13 COTILLION = 3^2^2^4^35^15 23 BULLION = Graham's number 19 PERIHELION = 3^4^5^6^7^...^(10^100) 24 PYGMALION = 2->3->3->3 7 AUSTRALIAN = Largest named number in the Avatamsaka Sutra 14 JULIAN = ((((10^100)!)!)!)! 6 SICILIAN = 13^315760124882724518 5 TRISKELION = 20^270354175698445357 8 VERMILION = A(4, 5) 18 ORWELLIAN = 2->(10^100)->2 20 REBELLION = A(10^100, 10^100) 9 PAVILION = 2 up up 8 28 ECTHELION = BB(100) 2 APHELION = Number of permutations of these numbers 26 SCALLION = Loader's number 4 RIGELLIAN = Largest named number in "The Sand Reckoner" 16 NAPOLEON = A(5, 1) 11 CHAMELEON = 68^75^96^92^39^3 12 STALLION = 2^38^64^20^57^13 17 SMULLYAN = (10^100)^...^7^6^5^4^3 27 BRAZILIAN = A(BB(99), BB(99)) 29 EMIL LEON = BIG FOOT + 1 22 CORELLIAN = Moser's number 21 FILLION = 3 up^(10^100) 3

Here's the correct ordering, with notes as to why:
1. TRILLIAN = 2 up^(10^100) 2
-- 2 up^x 2 is always 4.
2. APHELION = Number of permutations of these numbers
3. REPTILIAN = Number of particles in the known Universe
-- Estimates range from 10^78 to 10^82. Even if we're off by many orders of magnitude, there's no confusion with the next one.
4. RIGELLIAN = Largest named number in "The Sand Reckoner"
-- ((10^8)^(10^8))^(10^8), or around 10^10^17.
5. TRISKELION = 20^270354175698445357
6. SICILIAN = 13^315760124882724518
-- These two are tricky. They're both 10^10^17.54622039124297 on Wolfram Alpha. But if you divide 3157... by 2703..., you can see it's slightly larger than ln(20)/ln(13). In the 38th decimal digit. :) (You can generate lots of cases like this by approximating logs on the Stern-Brocot tree.)
7. AUSTRALIAN = Largest named number in the Avatamsaka Sutra
-- The "incalculable", 2^2^108 or around 10^10^37. Varies slightly depending on the translation.
8. VERMILION = A(4, 5)
-- Note that there's an easy way to convert from A to arrow notation (fortunately!). This is 2 up up 8 - 3, or around 10^10^10^10^10^4.30. Wolfram Alpha automatically simplifies things to a power-of-10 tower; this is useful for comparing the next few numbers.
9. PAVILION = 2 up up 8
-- 2^2^2^2^2^2^2^2. Yes, that's a gap of 3. LOL.
10. GILLIAN = 43^45^47^49^41^3
-- 10^10^10^10^10^5.066289492652996.
11. CHAMELEON = 68^75^96^92^39^3
-- 10^10^10^10^10^5.066289492653197. Yikes!
12. STALLION = 2^38^64^20^57^13
-- 10^10^10^10^10^22.94066043321802.
13. COTILLION = 3^2^2^4^35^15
-- 10^10^10^10^10^22.94066043326378. This would be a bitch without WA.
14. JULIAN = ((((10^100)!)!)!)!
-- Stirling's approximation means each ! more or less adds a power tower level. When dealing with towers of powers, the base doesn't tend to matter much; (n/e)^n and 10^n are, surprisingly, roughly the same. In general it's the height of the tower and the top exponent that matter. Wolfram Alpha confirms this is around 10^10^10^10^10^10^2.
15. HYLIAN = 6 inside square
-- This is Moser's notation. As above, taking (n^n)^(n^n) isn't all that different from n^n^n. So this simplifies to 6^6^6^6^6^6^7, or about 10^10^10^10^10^10^5.34.
16. NAPOLEON = A(5, 1)
-- 2 up^3 4 becomes 2 up up 65536. Tower height is 65536.
17. SMULLYAN = (10^100)^...^7^6^5^4^3
-- A tower height of 10^100-2. The base matters very little.
18. ORWELLIAN = 2->(10^100)->2
-- Hmm. This tower has height 10^100. However, we need to compare tops. 4^3 << 2^2^2^2 << (10^100-1)^(10^100), so this one's in between.
19. PERIHELION = 3^4^5^6^7^...^(10^100)
-- Tower height of 10^100-2 but this one's top-heavy.
20. REBELLION = A(10^100, 10^100)
-- Uh oh. We've left tower heights, and even iterated towers, far behind, and now we're counting arrows. We have 10^100-2 here...
21. FILLION = 3 up^(10^100) 3
-- ...and 10^100 here. Arrows are everything now, even with a top of 3 vs. 10^100+3.
22. CORELLIAN = Moser's number
-- A 2 in a Mega-gon. According to the Googology article, n in a k-gon is less than n up^(2k-1) n. But it's also roughly on that order. So certainly (256 up up 257) arrows beats 10^100.
23. BULLION = Graham's number
-- Now we've left arrows behind and have to count nested levels of arrows. Graham's number, famously, has 64 nested levels of arrows. In chained arrow notation it's about 3->3->64->2.
24. PYGMALION = 2->3->3->3
-- Becomes 2->3->(2->3->2->3)->2. Many, many nested levels of arrows. How many? Well, 2->3->2->3 = 2->3->8->2 = 2->3->(2->3->(...->(2->3))), which itself is 7 nested levels of arrows.
25. MEDALLION = 3->2->3->3
-- Becomes 3->2->(3->2->2->3)->2. Hmm. How many nested levels here? 3->2->2->3 = 3->2->9->2 = 3->2->(3->2->(...->(3->2))). This is 8 nested levels of arrows, so it ends up (much) larger than the above.
-- On the Googology wiki we've skipped over many orders of magnitude of fun with explicit functions. Now we're getting into the realm of numbers we know exist, but can define very little else about. Loader's number, a small C program that diagonalizes a calculus of constructions, is just skirting the top edge of computability. We still have an algorithm that produces it, but no way of conceptualizing just how incredibly far it goes...
27. BRAZILIAN = A(BB(99), BB(99))
-- ...but now we get to the Busy Beaver functions. You might wonder whether 100 Turing Machine states can compare with 512 bytes of C, but that's missing the point. loader.c has a concise proof that it terminates. Busy Beaver machines have no such restriction; likely the "best" machines can't be proven to stop without simulation. This qualitative advantage makes the BB function much, much more powerful.
28. ECTHELION = BB(100)
-- Similar to how changing the base of a tower of powers doesn't matter much, the Ackerman function does basically nothing compared to adding one more state to the BB function. I suspect this may be impossible to actually PROVE, but it's intuitively obvious; Busy Beavers grow far faster than any of the nice explicit functions above, including Ackerman's.
29. EMIL LEON = BIG FOOT + 1
-- You'd probably just look at the hierarchy on the Googology Wiki, see that this one's at the top, and stop there. But let's delve into the reasons.
How can unbounded computation possibly get beaten? Well, note that if you consider the actual BB(100) machine, which exists and has a brief description, you can definitely prove that it halts (though not AT ALL briefly). To look at it another way, it is possible to computationally produce lower bounds to BB(100) that eventually achieve the correct value.
Appropriately, EMIL LEON Post's work helps pave the way forward. He worked on questions of computability and provability. Rayo's number, from the Googology Wiki, ends up beating BB by considering numbers "defined" by first-order set theory. Thanks to Tarski's undefinability theorem, there is no way to prove which statements in first-order set theory validly "define" a number and which don't, using first-order set theory itself (which is easily rich enough to describe Turing machines and all the mathematics we've used so far). We're veering dangerously away from mathematics and into philosophy, but if you believe in a Platonic ideal of "truth", many of these describable numbers do "exist" even though we can never know or verify them. So, Rayo's number, in some Platonic ideal, dominates all the numbers we can hope to describe with normal non-self-referential mathematics.
"Thanks, Emil Leon!"
What's next? Tarski's undefinability theorem means Rayo's number itself isn't definable in first-order set theory, but it IS definable in a logic that is "one level" up. Now, BIG FOOT moves forward by taking that one meta step to MANY meta steps (where "many" is described using ordinal numbers and is itself rather hard to grasp). Yeesh.
At the time of writing, BIG FOOT is the qualitatively largest number mentioned on the Googology wiki. So of course I took it and added 1. Take that, Googologists!!!
Step #3 is to Index. As you can see on the bottom of the page, you need the lexicographic rank of the original permutation. It's convenient that 29! is close to 29 digits. :)
Plug "lexicographic rank of (1,3,10,15,25,13,23,19,24, 7,14,6,5,8,18,20,9,28,2,26,4,16,11,12,17,27,29,22,21)" into Wolfram Alpha, and you'll get the 29 desired indices:
13895194436626511361327337458

 1 1 T TRILLIAN = 2 up^(10^100) 2 2 3 H APHELION = Number of permutations of these numbers 3 8 A REPTILIAN = Number of particles in the known Universe 4 9 N RIGELLIAN = Largest named number in "The Sand Reckoner" 5 5 K TRISKELION = 20^270354175698445357 6 1 S SICILIAN = 13^315760124882724518 7 9 A AUSTRALIAN = Largest named number in the Avatamsaka Sutra 8 4 M VERMILION = A(4, 5) 9 4 I PAVILION = 2 up up 8 10 3 L GILLIAN = 43^45^47^49^41^3 11 6 L CHAMELEON = 68^75^96^92^39^3 12 6 I STALLION = 2^38^64^20^57^13 13 2 O COTILLION = 3^2^2^4^35^15 14 6 N JULIAN = ((((10^100)!)!)!)! 15 5 A HYLIAN = 6 inside square 16 1 N NAPOLEON = A(5, 1) 17 1 S SMULLYAN = (10^100)^...^7^6^5^4^3 18 3 W ORWELLIAN = 2->(10^100)->2 19 6 E PERIHELION = 3^4^5^6^7^...^(10^100) 20 1 R REBELLION = A(10^100, 10^100) 21 3 L FILLION = 3 up^(10^100) 3 22 2 O CORELLIAN = Moser's number 23 7 N BULLION = Graham's number 24 3 G PYGMALION = 2->3->3->3 25 3 D MEDALLION = 3->2->3->3 26 7 O SCALLION = Loader's number 27 4 Z BRAZILIAN = A(BB(99), BB(99)) 28 5 E ECTHELION = BB(100) 29 8 N EMIL LEON = BIG FOOT + 1

"THANKS A MILLION! ANSWER LONG DOZEN"

Step #4 is to Solve. The answer is LONG DOZEN.