# Challenge

This challenge is pretty simple: Output the total number of possible Tweets. Given the fact that $$\1{,}114{,}112\$$ Unicode characters exist, and Tweets can be up to $$\280\$$ characters long, we calculate this number like so:

$$\sum_{l=1}^{280}1{,}114{,}112^l$$

Calculated, this number becomes:

13806352273767118967309446652268353719030666987497438863889357319392349204238070937934701588382252733861344864045882375396168022346806583217340072802259387083359555134466517197703380759356965846365156977627360473773588169473731316369188960207815543356617430699549282184606847903882093344681639583363887920162736670339584230464092411246928128751923029979554646202579157612845521194257627831816510188786486205364038342997992390163377307319320795770580125829871827813635107055402218833285288428245530361070090385299410480471800097379723938710176776857154693492888113262310774952935141264656128106489715644469204223123574077687056241600407855644698999923182052787491366956979702291906138690639325270704272468782453957210692358210530379637285511395582363214078004388847219660731573151720107026864311541883029164798133330444260768597550725802772979408391647537342929574104808498797769614795400908507365237326804077132929339660443550229992211585154931729238484097424201567163596712698594520366545693861474171972272416897768579601352059483077847068448856472837079895280084051348757113747847034093121500403990822456362331568718364029903237356710927604336216180583115812490741333484187646966386074557181286407949971654502112671681454373869734610539296783900783568039838348725239932066066987739369537132033298306852573056479318562759091086979441683964087623025801925326064609304055115943692782207092740008073169526779422072953119459819275895282587278291133892079932466034984180041060600579941010427931676118356402267755461169199685582252347150686196242972686980783271232177904787906691872492956841276422272710260778534487171700075108866361285159161816004238666402986057371469198754307817645454320666214400


# Rules

• You may output a float with only zeros after the decimal point.
• Standard loopholes are forbidden.
• Trailing whitespace is allowed.
• Please explain your answer. This is not necessary, but it makes it easier for others to understand.
• Languages newer than the question are allowed. This means you could create your own language where the empty program calculates this number, but don't expect any upvotes.
• This is , so shortest code in bytes wins!
• Not that it matters for the challenge, but the number of tweets is a little less than that. Tweets consisting of only whitespace seem not to be allowed Mar 20, 2021 at 0:51
• I've edited the Mathjax slightly; using commas in the thousands place unfortunately causes a noticeable gap between the "parts" of the integer, so I've removed it. Furthermore, using $\displaystyle$ instead of $$...$$ makes it look "off" when viewing on the app/mobile site (at least, it does for me) Mar 20, 2021 at 3:54
• You cannot use 280 characters and use the full Unicode range. The 280 character limit is for non-CJK languages. Mar 20, 2021 at 15:12
• Just curious, approximately what number is that in word form? I couldn’t get Wolfie to process it Mar 21, 2021 at 1:26
• @gen-ℤreadytoperish Approximately thirteen point eight tresexagintaquingentillion, if my calculations are correct. Mar 21, 2021 at 18:40

# Emotion, 17 bytes

😍😃💁🧟🤯😃😧😍😄🤶🙆🙆😬🏃🧛😧🤠


Explanation

😍😃💁🧟 Push literal 280
🤯 Enter an iteration block over the first stack value and push the iteration element register at the begining of each loop.
😃 Push literal 1
😧 Push the sum of the second and first stack values.
😍😄🤶🙆🙆 Push literal 1114112
😬 Swap the top two stack values.
🏃🧛 Push the first stack value to the power of the second stack value.
😧 Push the sum of the second and first stack values.
🤠 Ends a control flow structure.


Try it online!

• This is disgusting. I love it +1 :D Mar 20, 2021 at 3:55
– user100690
Mar 20, 2021 at 7:24
• @Nordine Lotfi It's 17 bytes. I have hyperlinked the codepage consisting of 256 emoji in my answer. Valid Emotion programs contain only those characters. Each emoji represents one byte. Mar 22, 2021 at 6:14
• @NordineLotfi Unicode is one code page to rule them all. However Quantum 64's answer is using a different code page where the required emojis only take up 1 byte. I'd suggest reading Joel's post about Unicode and code pages. Mar 22, 2021 at 9:52
• @NordineLotfi: Your console seems to be using UTF-8, in which the string is indeed 68 bytes (17 characters of 4 bytes each). But Emotion uses its own character encoding that represents its emojis in a single byte (at the expense of being unable to represent any other characters). Mar 22, 2021 at 16:52

div(x^281-x)$x-1 x=8^5*34  Try it online! ## Why does it work? 1. $$\8^5\cdot 34=1114112\$$, but 1 byte shorter. 2. $$\\sum_{l=1}^n x^l=\frac{x^{n+1}-x}{x-1}\$$. # Boring answer, still 25 bytes sum$map(1114112^)[1..280]

• If anyone is curious about the mathematical proof for (2), look here. Mar 20, 2021 at 15:13

# Ruby, 23 bytes

-1 byte thanks to @Level River St.

x=17<<16
p x**281/~-x-1


Try it online!

### Ruby, 24 bytes

An interesting iterative solution. It uses the fact that a + a*a + a*a*a... is equal to a*(1+a*(1+a*(1+...))).

p eval'+1114112*x=1'*280


Try it online!

• For the 2nd answer, x=34<<15 or x=17<<16 is a byte shorter than x=1114112. Mar 20, 2021 at 10:13
• @Level River St Pretty neat! And it looks like to save a byte in many of the other answers here too. Mar 20, 2021 at 10:21

# Wolfram Language (Mathematica), 21 bytes

Tr[1114112^Range@280]


Try it online!

-1 byte from @GregMartin

# C (gcc), 183181167162 161 bytes

-1 byte ceilingcat

c;d;char s[1695],t[1695];x=281;y=1694;main(z){for(t[y-1]=1;--x;)for(z=y;z--;c/=10,d/=10)s[z]=(d+=s[z]+(t[z]=(c+=t[z]*17<<16)%10))%10;for(;y--;s[y]+=48);puts(s);}


Try it online!

# JavaScript (Node.js), 33 bytes

f=(n=280n)=>n&&1114112n**n--+f(n)


Try it online!

## 29 bytes

Porting @Delfad0r's solution saves 4 bytes:

_=>((x=1114112n)**281n-x)/~-x


Try it online!

• _=>(x=1114112n)**281n/~-x-1n
– l4m2
Mar 20, 2021 at 0:46

# MATL, 14 bytes

16W17*X$280:^s  Try it online! ### Explanation 16 % Push 16 W % 2 raised to that 17 % Push 17 * % Multiply. Gives 1114112 X$     % Convert to symbolic (to achieve arbitrary precision)
280    % Push 280
:      % Range. Gives [1 2 ... 280]
^      % Power, element-wise
s      % Sum. Implicit display

• This seems to map directly to a 10-byter in 05AB1E (and the currently shortest known 05AB1E answer here is 10 bytes): žH17*Ƶ´LmO - 16WžH, 17*17*, X$‎, 280Ƶ´, :L, ^m, sO Mar 20, 2021 at 1:03 # J, 20 bytes 1#.1114112x^1+i.@280  Try it online! Wasn't able to improve the straightforward answer. • Does using the equation (from the Haskell answer) instead of a sum help? – user Mar 21, 2021 at 23:57 • It did play around with it and wasn't able to make it help (even though you save on the numeric representation by 1 or 2 bytes you lose that or more in the rearrangement), but ofc that doesn't mean it's impossible... Mar 22, 2021 at 0:01 # Factor, 38 27 bytes -11 bytes thanks to Bubbler! 1114112 280 [1,b] n^v sum .  Try it online! • Using vector functions is much shorter. Mar 23, 2021 at 23:20 • @Bubbler Thank you! I always forget to use them... Mar 24, 2021 at 4:47 # Vyxalsr, 9 bytes ⁺∑ƛ»∆#∆»e  Try it Online! ⁺∑ # push 280 ƛ # Map foreach (1...280) »∆#∆» # Base-255 compressed 1112114 e # Exponentiation  The s flag sums the ToS at the end of execution. The r flag reverses the order in which functions take their arguments, so you have $$\ 1112114^n\$$ instead of $$\ n^{1112114} \$$. # Vyxals, 9 bytes »∆#∆»⁺∑ɾe  Try it Online! Thanks to Underslash for this • 10 bytes but it doesn't use -r. May 23, 2021 at 0:29 • @Underslash Thanks! May 23, 2021 at 0:30 # Jelly, 11 10 bytes ⁴‘æ«*Ɱ280S  Try it online! -1 byte thanks to Arnauld! ## How it works ⁴‘æ«*Ɱ280S - Main link. No arguments ⁴ - Yield 16 and set the argument to 16 ‘ - Increment to 17 æ« - Implicitly using the left argument, bitshift left, yielding 1114112 280 - Yield 280 Ɱ - Over each integer 1 ≤ i ≤ 280: * - Yield 1114112 to the power i S - Sum  • Alternate 11 bytes. Sadly just as long. Mar 20, 2021 at 0:18 • Mar 21, 2021 at 2:43 • @UnrelatedString Those two are the same, and don't produce the correct output Mar 21, 2021 at 12:13 • @Arnauld Yep, but you can remove the second ⁴ as the æ« doesn't hook to the *, thanks! Mar 21, 2021 at 12:14 • I had a second one but got the same link twice! They do both produce the same wrong output, of course. Don't know how I didn't catch that... Mar 21, 2021 at 13:00 # dc, 19 B14112dd281^r-r1-/p  Uses @Delfad0r's sum trick. Also in dc, B = 11, and B14112 = 1114112. Try it online! # 05AB1E, 10 bytes •HYH•Ƶ´LmO  Try it online! Likely Makonede's answer. Thanks to Kevin's tip on compressing integers in 05AB1E. Make sure you upvote it! ## How it works •HYH•Ƶ´LmO - Program •HYH• - Compressed integer: 1114112 Ƶ´ - Compressed integer: 280 L - Range [1, 2, ..., 280] m - Raise 1114112 to the power of each O - Sum  • That was one of the three solutions I found. Others: ŽΓH5BƵ´LmO, ŽHH₁*Ƶ´LmO Mar 19, 2021 at 23:26 • Another one: žH17*Ƶ´LmO Mar 20, 2021 at 1:00 # Pyth, 14 bytes(?) s^LhC\􏿿S280 Try it here! Pyth supposedly uses an SBCS, but I can't find any details about its actual codepage. The reason I have my doubts about the byte count is that the code contains a character with codepoint 1114111, which certainly isn't normally represented in a single byte (and I VERY much doubt is in Pyth's code page). The code is 11 characters long, and Pyth reports the mystery character as taking 4 bytes. However, the interpreter reports a length of 12, and the number 1114111 takes 3 bytes. In short, 11, 12, 13, and 14 are all possible byte counts for this program. I've chosen 14, as its the most likely (since, again, Pyth says the character is 4 bytes). • This should be 14 bytes as according to Pyth the character you have used is 4 bytes. Mar 20, 2021 at 4:26 # Bash, 27 bytes seq -f1114112^%g -s+ 280|bc  Try it online! # PowerShell, 44 bytes 1..280|%{$k+="[bigint]1114112*"*$_+1|iex};$k


Try it online!

-18 bytes thanks to @ZaelinGoodman

# Powershell v7, 37 bytes, thanks to @mazzy

1..280|%{$k+="1114112n*"*$_+1|iex};$k  • You don't need to declare K; it will adopt whatever type is being added to it with += if it is not yet initialized. You also don't need to concatenate the type accelerator with the number, so you can do "[bigint]1114112"*$.... Finally, the addition operator uses the type of whatever is on the left hand, so you do not need to make 1 a string to concatenate it. those changes all together get you down to 44 bytes Try it online! Mar 20, 2021 at 22:12
• with PS7 you can to use n-suffix also. 1..280|%{$k+="1114112n*"*$_+1|iex};$k Mar 21, 2021 at 5:06 # Java (JDK), 101 bytes v->{var x=java.math.BigInteger.ZERO;for(int i=280;i>0;)x=x.add(x.valueOf(17<<16).pow(i--));return x;}  Try it online! ## Credits • fix increasing byte count by 1, and 1 byte reduction by iota. • I believe x should start as BigInteger.ZERO. 101 bytes Mar 21, 2021 at 3:09 • @iota Oh yes, I started this believing this would be a multiplication somehow and the ONE stayed there... Mar 21, 2021 at 9:34 # Scala, 47 44 bytes Corrected answer and saved 3 bytes thanks to @mik! ()=>{val x=17<<16:BigInt;x.pow(281)/(x-1)-1}  Try it in Scastie! Not particularly short, but it uses the simplification of the sum to avoid all the exponentiation + summation, unlike the other answers. • Your answer is 1 too big. Correct and shorter: ()=>{val x=17<<16:BigInt;x.pow(281)/(x-1)-1}. – mik Mar 22, 2021 at 13:09 • @mik Oops, didn't see that. And thanks for the golf! – user Mar 22, 2021 at 13:12 • @mik Welp, that was dumb. Thanks again :) – user Mar 22, 2021 at 13:40 # Icon, 59 49 bytes • 10 bytes thanks to mik! procedure main();write(1114112^281/1114111-1);end  Try it online! ## Original solution, 59 bytes: procedure main() t:=0 t+:=1114112^(1to 280)&\z write(t) end  Try it online! • Much shorter: procedure main() write(1114112^281/1114111-1) end – mik Mar 22, 2021 at 12:55 • @mik Thanks! Yes, this approach in much shorter. Mar 22, 2021 at 12:59 # Husk, 14 bytes ṁ^1114112ḣ280  Try it online! I found out that if you run a Husk program that doesn't take an argument, and include an argument, it errors. Here's an example. ## Explanation  ḣ280 # Range from 1 to 280 ṁ # Map over range and sum... $$$$^1114112 # ...114112^n  • yep, adding arguments when there is no input causes type inference problems. That is normal. Mar 26, 2021 at 4:31 • @Razetime I found out because I put the result of ḣ280 in the argument to test it, and instead of removing the argument after I was done, I only cleared it. Then I closed the arguments menu and went back to testing it. Mar 28, 2021 at 10:32 # Deadfish~, 8398 bytes 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 Disclaimer: This crashes TIO. calculates then outputs the number. # Deadfish / Deadfish~, 6536 bytes 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 Just does the digits. # Deadfish~, 5965 bytes iiosdo{d}oiisiiodddoiiodddoosiiioddddosddodoio{d}iooiisdoiodddoioddddodddoiiisodddddooiioododddoosiioiiodddddoiioddosddo{d}ioiiso{d}oiiiodddoiisiioooiiiodododddodsoddodddodosdooddodddosdooioddddddoiioiioddddoddoiisoddddddoso{d}iioioiodso{d}iioddoiisoddoiosdo{d}oiisiiio{d}oiiisoddddddosddoiioddddddoioiiio{d}oioisioiiioodddddosdo{d}iioosiodddosiiioddddoosdoddo{d}ioiioioodsdoddoddo{d}oiisoioiiioo{d}iioiosddoddoddosodddo{d}ioisiioiio{d}oiiooioioiioiio{d}oiisiiodoiiiodddddododoisiiioddddoio{d}ooiisiiiodddddoisdo{d}oiioosioddsoddddddosdodo{d}oiiisdodddddooiioddsoddddoooddddoiioiooiioododdddoisiiio{d}ioiisoddoo{d}oiiioosdo{d}oiisiiioddoddsoddddddoiioioiiiodddodoiiioddddoiiodddoiiiododdddoisioioiiioddoododdddosiiioddddoiiio{d}oiisoiiioddddosddooddddoiioiiioo{d}ioisiioiiiodddddoiiioddddosddoddddoddoiioddoisiiodddoiiioiiio{d}ioiisdooiodddo{d}oiioddoiisiiioio{d}ioisioododooiioioo{d}ioisiiiodddododddoiisiioiiiooddddododso{d}iioisdo{d}iiodoiisdoddddoiio{d}oiisiioiioddddoiiioiio{d}oiiiosdoo{d}iioddoiiisoddddddooiooiioiio{d}ioisiiodddosoddddoiiiodddddooiiiodddosdoodoiio{d}iioddoioisiioddddosiiioddddoiiiooio{d}oiiioosoddddoiiioddddoddoiodddoiisoiioddo{d}oiiiso{d}iiosodddooiosoiioiiio{d}iioisdo{d}ioioisdododdoddddoiiso{d}iioiodddoiioisooddoiioddddoodoiioddoiioddddoddoiiosioiioiio{d}ioisioiiodo{d}ioioisdoddddoioodddodooiisodddddoddosioiiododdddosiiioiodddddoddoiisdo{d}ioisiiododdddodoioiisdoodoioddoddodsdoddoddddoddoiisioddoiiioddo{d}oiiiosdodddddoioddoisooddoiioo{d}iioioso{d}oioisiiodddoosddooddddodddoiisiiioddddoddoiisoddddddododdoiisiiioiioddddoiioo{d}oiisioiiio{d}oioiosioiiio{d}iioisododo{d}ioiisdo{d}iiosiiioio{d}ioiioiiiodddoiioddddodoiisiiio{d}oiisioodo{d}oiioodoiisdoodddddoodoisdodddodddoisdooddddoddoisdo{d}iiosoiooddodddoiiioiiio{d}iodoiisiiio{d}ooiiiso{d}oiiiosdodddodddoisoodddddodddodoiisodsdo{d}oiisoiiio{d}ioiisdo{d}oooiiisoddoddddosddoiioddodddddoiosoddddddosdodo{d}iodoioisiiiodoioodoiiodddoiio{d}ioisiodoiioiiioddddddoiodso{d}iioisdooo{d}iooiiodosiioddddoioddodoiisiiioodddodsoddddodddoisoddddddoiioddddoisodddoiosiioddoiiodoio{d}ioioisdo{d}iodoiisiioddodsdoioddo{d}ioisioioddoooiioiiio{d}iioddoiisoddooioddoioioiioiiodddo{d}oiisiiioodoiiodo{d}oiisioioddddosodddoisiio{d}ooiiso{d}oiisiiioiodddooioddooiioiiiodoioooo{d}iioioddoiisdo{d}iioddoiisiodddosiiioiododddodso{d}ioiioiiiooiiioddddoioiiioddoiioddo{d}oiiooiso{d}ioiiso{d}oiisiio{d}ioiiosdoddoiiio{d}oiisiiodddosoddddddodosiodddosiiio{d}oiisiiio{d}oiisoddosiiiodddddosoiioiiodoio{d}iiosoioddosoddddoiiodddddododoiisiioiiio{d}iioioiioiiio{d}iiododoiisioddodddoiiiosddoiiodddodddosddodddddoisdodddooddddooiiosoddddooiiio{d}iioioiiiodddododoiso{d}oiisiiioio{d}ooiisodosdoooddddoiiiodddddodoiisodddoo{d}oiisiiioddddoddoisioiioddddoddoisioddddoisiiiodddddoddoiodoiisiiio{d}oiiosiioiioddoddododdooisiododddoiisdoodddddodddoiioiso{d}ioisiioddoiiioiiodo{d}ioiioooodddoiisoooddosiio{d}oiisiiiodoiiodddoddsoddoddoo{d}oiisiiiodddddosioiiio{d}oiiosiiioodddddoisoddoiiodddddo{d}oiiisdodddddoso{d}ioisiioddoiiioddoddosddoddddoioddoiso{d}iioisoddddoiiodddodddodoiisodsdo{d}oiiisdoddddodsododoiioddoodoiiiodddo{d}ioisoiiioiioddddodo{d}ooiiiso{d}oiiisdodddo{d}oiisiiioddddoiiiododddoiosddoddddodosiioiio{d}oiiso{d}oiisiiioo{d}ioiiodoiso{d}iioisoddddddoosodddoo{d}oiisoodoiioo{d}oiiooisooo{d}iioodooisioiiiodddoddddoisiododsoddddddoddoisiiiodddddoiso{d}iioiosdoddddodsdoddddo{d}oiiisoddodddoddosoddoddoioisioioio{d}ioisiiodddoiioddsodddoio{d}ioiosiioiiiododddoddsodddddoiodddoddoiiioiiioododoioioiiioddddddosdoddo{d}ioisoiiiodddodddoisiiio{d}ioiisoddodddddoosiiiodddddosodddoisiioiioioddoodoiiodddoiioiiodddo{d}oioiioiiodddoddoiisioddsodddddodsdodddddodddoiisiiiooioddddoiiio{d}oiisiioiioddddoodsdoodddoioddoiiiodddddoisdodddddosddo{d}oiisiiioiiodoioddddodddoisdo{d}ooiiisdoddddo{d}oiisioddddoiioiodsdododdoiio{d}iooiiosddodddoiiioioddddoiiio{d}oiiioio{d}oiiisoddddddoddoiodoisio{d}ooiiso{d}oiiiosoo{d}oiiisdo{d}iioosoioiodddoiiioddddoiooddoisioioiiodo{d}ioiisdodddddoiiioddo{d}oiioisoo{d}oiiiodoiosddoddddoiioioio{d}iodoiiiso{d}iiosiiiodo{d}oiisodooiiioddddodoisiio{d}ioiisdo{d}oiisioiiiodddddoddooisioiiio{d}ioiosodso{d}oiisiiiodddodddoiioooiodsdoddddodddoiisdodododdoiioiiiodddoodddosdoddo{d}oiisiiiodddoiooiio{d}ioiisdo{d}ioioisdoddoddo{d}oiisiiioiiodddddodsooddo{d}ioisiiododoio{d}oiiodooiosiioio{d}ioisiioiio{d}ioisoiododosddoddddosdoddoiiioddoddddoioiio{d}iodoiisioddoso{d}iioisodddoioiodddddoso{d}ooiisiiioiodddddoiioioiio{d}oiiiosododddddosdodddddoiodsdododddddosiodddoiosooddddddododdoiisiioo{d}oiisiiooiiiododooddddosoddddddoiiioiiioddddoddosddo{d}ioiiododdoiiioodoisododddddodddoiisiioiiodddodddosioiioddddodddoiisioioddoiiioiioddddddoddoiisdodddoioddddosiiioddoddso{d}oiiiso{d}iodoiiisdoddoiiioddoiiodddddoodddoisiioiiodddddosodddoddo{d}oiiisdodododdddoiodddoiiosioiiio{d}oioiiso{d}iiosioddodosiio{d}oiisiioddoiio{d}oiiisoddddddodddoiiso{d}oiisiooddddooisioddsodddddodoiiioiiio{d}iiosiiioio{d}iiooddoiisiiio{d}oiiiso{d}iiosiiiodddo{d}oooiiisdo{d}oiisiiioddddoddoisiioiiioddddodddosiioiooiiodddddoddooddoiisiiiodddddoisoddddo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 Also just does the digits. # Charcoal, 15 bytes ＩΣＸ⍘!<JOγ…¹¦²⁸¹  Try it online! Link is to verbose version of code. Explanation:  !<JO Literal string of printable ASCII ⍘ γ Convert from base 95 Ｘ Vectorised raise to power …¹¦²⁸¹ Range from 1 to 280 Σ Take the sum Ｉ Cast to string Implicitly print  # Canvas, 12 bytes －„｛“ＦＲ］„；＾］∑  Try it here! # Perl 5, 33 bytes say+(($q=1114112)**281-$q)/($q-1)


Try it online!

# Python 3.8, 44 31 30 bytes

Saved 13 bytes thanks to Makonede!!!
Saved a byte thanks to ovs!!!

print((x:=17<<16)**281//~-x-1)


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• @Makonede Nice one - thanks! :D Mar 20, 2021 at 9:52
• You can save a byte with 17<<16 == 1114112
– ovs
Mar 20, 2021 at 10:02
• @ovs Ah, most insightful - thanks! :D Mar 20, 2021 at 10:07
• This has the same length, but is simpler: print(1114112**281//1114111-1).
– mik
Mar 22, 2021 at 11:59
• @mik Interesting, but I'll stick with what I've got thank you. Mar 22, 2021 at 13:35

# cQuents, 14 bytes

#280;1114112^$ Try it online! ## Explanation #280 n = 280 ; output sum of first n terms 1114112^$  each term is 1114112 ^ current index


# Perl 5-Mbigint, 26 bytes

say 1114112**281/1114111-1


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# Perl 6, 30 bytes

say 1114112**281 div 1114111-1


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# F# (.NET Core), 60 bytes

printf"%A"(Seq.sumBy(fun i->bigint.Pow(1114112I,i)){1..280})


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# PowerShell, 39 36 bytes

[bigint]::Pow(1114112,281)/1114111-1