Count all possible Tweets

Question

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.
• If possible, please link to an online interpreter (e.g. TIO) to run your program on.
• 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 code-golf, so shortest code in bytes wins!

Answer 1

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.

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Answer 2

Haskell, 25 bytes

div(x^281-x)$x-1
x=8^5*34

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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]

Answer 3

Ruby, 23 bytes

-1 byte thanks to @Level River St.

Based on the formula presented in @Delfad0r's answer.

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

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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

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Answer 4

Wolfram Language (Mathematica), 21 bytes

Tr[1114112^Range@280]

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-1 byte from @GregMartin

Answer 5

C (gcc), 183 181 167 162 161 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);}

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Answer 6

JavaScript (Node.js), 33 bytes

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

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29 bytes

Porting @Delfad0r's solution saves 4 bytes:

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

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Answer 7

MATL, 14 bytes

16W17*X$280:^s

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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

Answer 8

J, 20 bytes

1#.1114112x^1+i.@280

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Wasn't able to improve the straightforward answer.

Answer 9

Factor, 38 27 bytes

-11 bytes thanks to Bubbler!

1114112 280 [1,b] n^v sum .

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Answer 10

Vyxal sr, 9 bytes

⁺∑ƛ»∆#∆»e

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⁺∑        # 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} \$.

Vyxal s, 9 bytes

»∆#∆»⁺∑ɾe

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Thanks to Underslash for this

Answer 11

Jelly, 11 10 bytes

⁴‘æ«*Ɱ280S

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-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

Answer 12

dc, 19

B14112dd281^r-r1-/p

Uses @Delfad0r's sum trick.

Also in dc, B = 11, and B14112 = 1114112.

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Answer 13

05AB1E, 10 bytes

•HYH•Ƶ´LmO

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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

Answer 14

Pyth, 14 bytes(?)

s^LhC\􏿿S280

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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).

Answer 15

Bash, 27 bytes

seq -f1114112^%g -s+ 280|bc

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Answer 16

PowerShell, 44 bytes

1..280|%{$k+="[bigint]1114112*"*$_+1|iex};$k

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-18 bytes thanks to @ZaelinGoodman

Powershell v7, 37 bytes, thanks to @mazzy

1..280|%{$k+="1114112n*"*$_+1|iex};$k

Answer 17

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;}

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Credits

• fix increasing byte count by 1, and 1 byte reduction by iota.

Answer 18

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.

Answer 19

Icon, 59 49 bytes

• 10 bytes thanks to mik!

procedure main();write(1114112^281/1114111-1);end

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Original solution, 59 bytes:

procedure main()
t:=0
t+:=1114112^(1to 280)&\z
write(t)
end

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Answer 20

Husk, 14 bytes

ṁ`^1114112ḣ280

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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

Answer 21

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}oiisiiioddddoddoisiioiiioddddodddosiioiooiiodddddoddooddoiisiiiodddddoisoddddoddoddooiisodddddoioddsodo{d}ioiiso{d}iiosiiioddoiiioioddddodddoisdo{d}iiosioiiiododddddosiiioio{d}iioiso{d}iooiioosdoio{d}iioddoiisiiioiiooddddddodosoiioo{d}oiiioiodsododdddodddoiisdo{d}ooiisodddodoiisiio{d}oiisiio{d}ooiisioiioiioodddddodddodoiodoiisoddosiiioiioddddddoddoisiioiodo{d}iooiisdodddddoiioioddo{d}oiioosiioiooddoodoiio{d}iooisiioiiio{d}ioiisoodddoiiodddooiiio{d}iioosiodddoioioiiio{d}ioisio{d}oiisiioiioddo{d}ioiisodddoddddosoddoisoddodddddosiioiioddoiiiodo{d}oiisiiioiodddddodosiiio{d}ioioiododoisiiiooiio{d}oiisoiiioiodoiio{d}oiisiiooiiio{d}ioiisdododddddosodso{d}iioisoddddoioiioddddodddoiosiiiododdoddooosiiiodddddosiiio{d}iodoiiosiio{d}oiisiiiooiodddoddoioodsdodo{d}ioisiiio{d}ioisiiio{d}oooiisiiioddoddddodoiiisdooddoodddoiiio{d}ioioisdodddoddddoisioddso{d}ioisiio{d}ioiisdo{d}ioisiio{d}ooiisoddoiosdoddoooddo{d}oiioisododdo{d}oiisioiioddddosddo{d}ioisoiioiiio{d}ioiisodododdodododddoiisiiioio{d}ioisiiiododdoiodoiododododdoiisiioooddddodoisoo{d}o

Also just does the digits.

Answer 22

Charcoal, 15 bytes

ＩΣＸ⍘!<JOγ…¹¦²⁸¹

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    !<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

Answer 23

Canvas, 12 bytes

－„｛“ＦＲ］„；＾］∑

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Answer 24

Perl 5, 33 bytes

say+(($q=1114112)**281-$q)/($q-1)

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Answer 25

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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Answer 26

cQuents, 14 bytes

#280;1114112^$

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Explanation

#280            n = 280
    ;           output sum of first n terms
     1114112^$  each term is 1114112 ^ current index

Answer 27

Perl 5 -Mbigint, 26 bytes

say 1114112**281/1114111-1

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Answer 28

Perl 6, 30 bytes

say 1114112**281 div 1114111-1

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Answer 29

F# (.NET Core), 60 bytes

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

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Answer 30

PowerShell, 39 36 bytes

Based on the formula presented in @Delfad0r's answer.

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

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