quarantine 'em on Twitter: "Not 100% certain that this is the winner but it looks suspiciously similar in construction to Rayo's number, which won a large number duel at MIT. Well done https://t.co/BtTYsQlaiW"

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quarantine 'em Retweeted fabs();

This isn't well-defined, but in most scenarios your mum is at least contained within the observable universe, so this is bounded above by about a googolhttps://twitter.com/raspofabs/status/1165781895734530051 …

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fabs(); @raspofabs

Replying to @qntm

The number of atoms making up your mum in a your mum is so fat joke.

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quarantine 'em‏ @qntm Apr 17

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quarantine 'em Retweeted Ⓒⓐⓜⓔⓡⓞⓝ Ⓓⓔⓥⓘⓝⓔ

I'm not 100% certain what the leading exclamation mark here is for, but ignoring that, there are 277 nines, so that's 10^(10^277 - 1), slightly more than a googolplexhttps://twitter.com/Cameron__Devine/status/1165850650279739392 …

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Ⓒⓐⓜⓔⓡⓞⓝ Ⓓⓔⓥⓘⓝⓔ @Cameron__Devine

Replying to @qntm

!1e9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

3 replies 0 retweets 5 likes

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quarantine 'em Retweeted Natan Melzer

In Knuth's arrow notation this is "99↑↑89"... except that "base 9^999" means we read "99" as 9 * 9^999 + 9 in decimal, and "89" as 8 * 9^999 + 9 in decimal. So that's a little less than (10↑1000)↑↑↑2. Three arrows. Sizeable...https://twitter.com/NatanMelzer/status/1166376298559758336 …

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Natan Melzer @NatanMelzer

Replying to @qntm

99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99^99 base 9^999

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quarantine 'em‏ @qntm Apr 17

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quarantine 'em Retweeted Wayne Werner

If by "And so on" we assume that the nines continue for the remainder of the Tweet, that's 276 nines. That's still three arrows, but a MUCH larger exponenthttps://twitter.com/waynejwerner/status/1165837725410504704 …

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Wayne Werner @waynejwerner

Replying to @qntm

9^^^99999999999999999999999... And so on

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quarantine 'em Retweeted halcanary

haha Knuth up-arrows go brrrhttps://twitter.com/halcanary/status/1165734655292837890 …

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halcanary @halcanary

Replying to @qntm

Using Knuth's up-arrow notation: 2↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑8

2 replies 4 retweets 39 likes

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quarantine 'em‏ @qntm Apr 17

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quarantine 'em Retweeted tef e.books, depressed birb fan

Nobody specifically entered Graham's number, but this is slightly bigger than that, and easily outclasses any expression using a small, writeable number of up-arrowshttps://twitter.com/tef_ebooks/status/1165725289705607169 …

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tef e.books, depressed birb fan @tef_ebooks

Replying to @qntm

ackerman's function fed with grahams number, then passed into ackerman's function again, graham's number of times, plus one

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quarantine 'em‏ @qntm Apr 17

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quarantine 'em Retweeted Lost Snowdrift

This is a pretty large number. Kruskal's much faster-growing TREE function is such that TREE(3) obliterates Graham's number in terms of scale. This, however, is TREE^41(420)https://twitter.com/LostSnowdrift/status/1166096988904906752 …

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Lost Snowdrift @LostSnowdrift

Replying to @qntm

TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(TREE(420)))))))))))))))))))))))))))))))))))))))))? https://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem#TREE(3) …

2 replies 1 retweet 9 likes

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quarantine 'em‏ @qntm Apr 17

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quarantine 'em Retweeted d ⧉ aan

I could be dead wrong about this but I'm fairly sure Kruskal's TREE function is computable with much less than a ~120 state Turing Machinehttps://twitter.com/ijks_w/status/1165727070305038337 …

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d ⧉ aan @ijks_w

Replying to @qntm

to get low-hanging fruit: the Nth Busy Beaver number, where N is the number of characters in this tweet - as explained in https://www.scottaaronson.com/writings/bignumbers.html …

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quarantine 'em Retweeted Linus Hamilton

This value is smaller than, but probably in the vicinity of, BB(9999) in terms of scale, therefore far surpassing BB(120) abovehttps://twitter.com/linyks/status/1165748190890483712 …

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Linus Hamilton @linyks

Replying to @qntm

9999th Busy Beaver number for alternating Turing Machines (TMs with choices made alternately by Alice, who wants to prolong execution, and Bob, who wants to cut it short).

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quarantine 'em‏ @qntm Apr 17

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quarantine 'em Retweeted pete

The Radó maximum shift function is closely related to BB in terms of growth rate but this is the first entry to start using such functions recursively instead of just passing in basic literal numbers. Excellent callhttps://twitter.com/mcallisterjp/status/1166353418002284545 …

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pete @mcallisterjp

Replying to @qntm

xₛ₍₉₉₎, such that x₀=s(99); xₙ=s^xₙ₋₁(xₙ₋₁); s=Radó maximum shift function; ^=function iteration

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quarantine 'em Retweeted Drake Thomas

Not 100% certain that this is the winner but it looks suspiciously similar in construction to Rayo's number, which won a large number duel at MIT. Well donehttps://twitter.com/MaskedTorah/status/1167006809296531456 …

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Drake Thomas @MaskedTorah

Replying to @qntm

F(n) = Max number uniquely defined by a sentence in PA in the standard model with at most n chars. F1(n) = same as F but where language of PA has the function F. F2 for F1&F, etc. Fw = PA with access to a symbol F_m(n). Then # is Fw(BB(BB(BB(BB(BB(BB(BB(BB(BB(BB(BB(99))))))))))))

3:40 PM - 17 Apr 2020

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1. quarantine 'em‏ @qntm Apr 17
  
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  Now to take a big sip of coffee and assume I made no errors whatsoever
  
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1. blah‏ @___Azu Apr 17
  
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  removing some spaces in this (a step before, say, removing some vowels while still preserving legibility) lets you at fairly minimal effort replace that 99 with 9^9^9^9
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