[Question]
  [
Jelly has [compressed string literals](https://github.com/DennisMitchell/jellylanguage/wiki/Syntax#string-literals), using the `“...»` delimiters. The way these work is by interpreting the `...` as a base-250 integer, \$n\$, then repeatedly divmod-ing this integer until it reaches \$0 \newcommand{\d}[2]{ \left( \left\lfloor \frac {#1} {#2} \right\rfloor, #1 \text{ mod } #2 \right) }\$, building up the decompressed version as it goes by indexing into dictionaries and printable ASCII.
Jelly has 2 dictionaries, "short" and "long". "Short" contains \$20453\$ words of 5 letters or shorter. "Long" contains \$227845\$ words with 6 or more letters.
As the exact method is rather [complicated](https://github.com/DennisMitchell/jellylanguage/blob/master/jelly/interpreter.py#L1055), I'll work through how \$n\$ is decompressed:
* First, we divmod \$n\$ by \$3\$: \$n, m = \d n 3\$. We then call \$m\$ the *mode*.
* If the mode is \$0\$:
	+ Divmod \$n\$ by \$96\$, yielding \$n, c = \d n {96}\$
	+ Add the \$c\$th character in the printable ASCII range ( to `~`) to the decompressed string. If \$c\$ is \$95\$ yield a newline instead of **0x7F**
* If the mode is \$1\$:
	+ If \$n\$ is even, use the "long" dictionary and replace \$n\$ with \$\frac n 2\$
	+ If \$n\$ is odd, use the "short" dictionary and replace \$n\$ with \$\frac {n-1} 2\$
	+ Then, take the length \$l\$ of the dictionary (\$20453\$ for "short", \$227845\$ for "long"), calculate \$n, i = \d n l\$ and retrieve the \$i\$th element of the dictionary, the *word*
	+ If the decompressed string is not empty, prepend a space to the word. Finally, append the word to the decompressed string
* If the mode is \$2\$:
	+ Calculate a *flag* \$f\$ as \$n, f = \d n 3\$ and update \$n\$
	+ If \$n\$ is even, use the "long" dictionary and replace \$n\$ with \$\frac n 2\$
	+ If \$n\$ is odd, use the "short" dictionary and replace \$n\$ with \$\frac {n-1} 2\$
	+ Then, take the length \$l\$ of the dictionary (\$20453\$ for "short", \$227845\$ for "long"), calculate \$n, i = \d n l\$ and retrieve the \$i\$th element of the dictionary, the *word*
	+ If the flag doesn't equal \$1\$, swap the case of the first character of the word
	+ If the flag doesn't equal \$0\$ and the decompressed string is not empty *or* the flag equals \$0\$ and the decompressed string is empty, prepend a space to the word
	+ Finally, append the word to the decompressed string
* If \$n\$ is non-zero, go to the first step with the new value of \$n\$
---
We can work through an example, using \$n = 46886323035539\$:
* First, we divmod by \$3\$: \$n = 15628774345179, m = 2\$.
As the mode is \$2\$, we calculate \$n\$ and \$f\$ as \$n = 5209591448393, f = 0\$. \$n\$ is odd, so we're using the "short" dictionary and \$n\$ becomes \$2604795724196\$.
Calculate the index and the updated value of \$n = \left\lfloor \frac {2604795724196} {20453} \right\rfloor = 127355191\$ and \$i = 2673\$. The \$2673\$th word in the "short" dictionary is `Caird`, so we call that our word.
As \$f \ne 1\$, we swap the case of the first character of the word: `caird`. However, \$f = 0\$ and the decompressed string is empty, so we don't prepend a space. Finally, we append `caird` to the (empty) decompressed string, yielding `d = 'caird'`
As \$n = 127355191\$, which is non-zero, we go to the first step again
* Now, `d = 'caird'` and \$n = 127355191\$. Divmod by \$3\$ to get \$n = 42451730, m = 1\$.
As the mode is \$1\$ and \$n\$ is even, we're going to use the "long" dictionary this time around and \$n\$ becomes \$21225865\$
We calculate the index into the dictionary as \$n = \left\lfloor \frac {21225865} {227845} \right\rfloor = 93\$ and \$i = 36280\$. The \$36280\$th element of the "long" dictionary is `coinhering`, so we set that as our word.
As `d` is non-empty, we prepend a space to our word, then append it to `d`: `d = 'caird coinhering'`
As \$n = 93\$, which is non-zero, we go to the first step again
* Now, `d = 'caird coinhering'` and \$n = 93\$. Divmod by \$3\$ to get \$n = 31, m = 0\$
As the mode is \$0\$, we calculate \$n\$ and \$c\$ as \$n = 0, c = 31\$. The \$31\$st ASCII character (alternatively, the character with ordinal \$63\$) is `?`
We add `?` to the end of `d`, resulting in `d = 'caird coinhering?'` and \$n = 0\$
As \$n = 0\$, we are done, and we return `caird coinhering?` as our decompressed string
---
Alternatively, [here](https://tio.run/##3VPLboMwEDzDV2zVA1iJoiQ0URIph6qnnNsbQpUFC7gCG9luab6e2i55uE1/oOJgez0zuzuLu6OuBU@GoZSihTdsmiOwthNSQ8FyzQSn8jiFXBT42tEKw7DAEpRSMeMaK5RkFwYF5qLtJCqFBewhisIw6GvWIIwggwnG7RRao2VQBfswu/gcTogFleP1HuaWdGHlt1nbtaUFvZA287nM1MEnkCwzuIeXmimQqN8lV6BrBB7pGh6fnw4HyGsqaa6tlpBAgWPfMG4qL4Ebxe0q9Mta@GWp2lr1u66lK8vker3YaGDx5TRrBK@mVy7PnBZJ3ZJ5WRgv8PNGlgZ57OcgJLzyw79LnUzmAKaj66ntYGRE5pu4w4@@l37fZUOr20P8L/a4Du/swOFkjl3SeTZTPe1yqjAmo1fpYpeNtJgL7VlLrIiLWkXyh9PeE5rsT/Hvn9bTGzppGoztE3xYbzbrZJnMk9Uq2RIyfAE) is a version of the `sss` function adjusted slightly.
## Task
You are to take a positive integer \$n\$ as input and output the decompressed string that \$n\$ maps to. You may also take the two dictionaries ("short" and "long") as input if you wish. The dictionaries can be [found on TIO](https://tio.run/##S0oszvj/v6CyJCM/z1hBN1lBPa0oP1chKzUnp1IhM7cgv6hEISUzuSQzPy@xqNK6oCgzr0QDIaBXnAFUoamuYAdWFQ/m6pVUlHCRb2ROfl460ESYkSAu2MTkxBI0S/7/BwA) or in the [Jelly repo](https://raw.githubusercontent.com/DennisMitchell/jellylanguage/master/jelly/dictionary.py)
You will never be given an input \$n\$ outside the native bounds of integers in your language, but your program [must *theoretically* work for arbitrarily large integers](https://codegolf.meta.stackexchange.com/questions/18196/how-can-we-clearly-define-must-work-for-theoretically-large-values)
This is [code-golf](/questions/tagged/code-golf "show questions tagged 'code-golf'"), so the shortest code in bytes wins
## Test cases
```
n			out
1			Aachen
2			aachen
3			!
47			Aah
218			aaronical
250			Abdul
745			abdominoplasties
7180			auto
8106			. aahs
364038195		Q unconfine
396478210		SoddyAberration
592528345		insulting abature
4060289999		Shool< aalborg
6079656777		CcircuityAachen
2402785962		njX abac
1192732966		flongAllel
69526673848		Kansu4Abderian
7240502929952		Eyehook% arval
7300618740196341	g noninterchangeableu mahwa
98944394632073037	/;streetwards Harz aachen
8092569718755799474	epos sporran@ Apriorism
11508828911594644653	/,e Waterbirdanelasticities
11209288138354511564	 eale Thuya&decokes		 	(note the leading space)
11683582443091721078	 Saveable! duly babbliest		(note the leading space)
13365392696538596641	tambourinists}Battu0abaca
```
Additionally, make sure your program works for \$n = 8609941742042387534416\$:
```
this
has
newlines
```
and also for \$n = 16505187\$:
```
a
b
```
And the outputs for \$1 \le n \le 100\$ (one per line): [Try it online!](https://tio.run/##NVFJcoRADLv7FWTf932bxLnnEQ10aCpdUIUHUnk9sWzmRFuSJdtIDpKizDOHKsWOgn@2iJmKBdtWNLVdQ8XC7ijLFJh2iUMu@0EpBfagUxHTPvAoY1frw1oPwMUaFlYfap3bVhBiwBEEoqZMx@hu/@CFnBNIYct0qs@hnsLwA8yCz9wIPg6cm@a3z98Ic@wCIsEsdLmhpyg2DV0hTtY2CV1rMTRhRIMvcAN63VuAI7fQpNHaF@hOL/IFf13xHnTftZXZK/AAQC0L5x83vG7FfoQnDNUkM0T97BIpRzutgy8OIkXoFR1Li5ZvxJ/mL7RamGSHwRXflWS3QfmhhEx9E7PnAdO/WYZqnv8B)
  
[Answer]
# [JavaScript (Node.js)](https://nodejs.org), ~~173 170~~ 163 bytes
Expects `([ long_dic, short_dic ])(n)`, where the dictionaries are arrays of words and **n** is a BigInt.
```
d=>F=(n,C)=>n?(B=Buffer,(g=k=>x=+[n%(n/=k=BigInt(k),k)])(3)?(q=x^2||g(3),b=B(d[g(2)][g(d[x].length)]),b[-q%2]^=32,q%3>0^C?' '+b:b):B([g(96)^95?x+32:10]))+F(n,1):''
```
[Try it online!](https://tio.run/##jZbfcqM2FMbv/RRapikwIV4hhICkJI0zu9NOrzrpTDvjdboCZMMGS44Qid1s3qAzvell@x59nn2BPkJ6SNabfzgpFxpj8dP5dDjfER/4OW9yXS3MjlSFuM6VbAy6RGIp8uOVzNEVSpEWZ22lhWPlZVUXvy60ykXTWK43KKoc5sfIrpWc2R6ym1JpY6PJcM4XToPSfSTOee2sl3PeL1amVDJAOzmyp1rN0QdR1ytUzRcAIljPVEpyvdpb6Eoa5@6P4VeXzZVrv3eHRh0bmJw5ruvuDQbT9LpI99@mjvSO3HRfHjijdNROp0J7ziw9TfeX6fZYbjnyNdyMqtn3sOyp6526E9cJ3APnLF2ekI8fZ3DjZenIKcYzh7gTGIvxcjKshZyZEh72svHO2RaZnKQB8c62gn18cnRgI3s7283c3ZEDQMLckyQ8WG4HZNfHE9fdfguqfHfXtq/3xgMEmfIl6rs8ZB3yvBTS6p1GE@@GJhtp/j/oYCP9agN4n6aR3Ki8tF6iiR/LTcq1klXOa@sZOsQb6MOsaGvr@dgRDTfFzgo1r6Ra1LwxlWisPtqP@4J3dGuU9cK@Yx@zfnqIOC@bF94YoziI/eSRfqB/RK0Es04rKayNdMJoFBMfP6GPVVGsDjOhNe/sZfXRYUJCEgf0aexKNm1twIGIZ9y0@oGCdbVghkmcwCUfxy6Vqr@BzdeZ0jOrVznDUcJCFkXRY/oor3TeVmbVZ5d1tVBMojhMGHlMyw@/dJpz65mc@35CooAkjD2mp12XO6xr0Vdwa@VJSBiLgpg@qHegf@CQNgr1KnTFZf@@I5AeYpKQBJaR9@k3KwF5O91CXJ8/scqaDjBmfhxR7CcsoOtWA/QMSfCYNELnJZczwbNatGjOywturekkTigNEsoCgmGl4HPqgX691xgthLngumjQd1z/hu7azbrOMZQLS8AsURhGSUIjKm9psVANaqC9ay6/RYfQ15Wumrn1MOchjmOoF/gBCihlYdesutieQD9zEJ5VuuBS3Bg1r27N@oUmED2O/SAOQhrCGoze0kjwWqCfynbFvy5Erk7vW/wLzQCLCWweJ34Edoniz/QxP79J1SsETWaFMp5lNQQ29/btBwFoTQiDeg26mmM3eQfa8HmmWjioqsY0VyNuTIu72uP3s8YwpMqPKMGUBJC6gFKfyY4uq@adLDkMUlzUYPMe5SzEISRcPu1M72S2uTMNJoPhVOk38A4dZ4wgnFguRG5EgSZud2ZfwpEPhzeaduev60h3D3UfBqoWw1rNHJj0bp9I78gDZP/7zx@f/voTRhvtIvvT37/bAF651/8B "JavaScript (Node.js) – Try It Online")
### Commented
```
d =>                        // outer function taking d[] = dictionaries
F = (n, C) =>               // inner function taking n and a flag C
n ?                         // if n is not equal to 0:
  ( B = Buffer,             //   B = alias for Buffer
    ( g = k =>              //   g is a helper function taking a divisor k
      x = +[                //     it computes and returns x = n modulo k
        n % (               //     and updates n to floor(n / k) afterwards
          n /=              //
            k = BigInt(k),  //     k is first converted to a BigInt
          k                 //     but x is converted back to a Number
        )                   //
      ]                     //
    )(3) ?                  //   extract the mode; if it's not 0:
      ( q =                 //     q = flag; we force q = 3 in mode 1
          x ^ 2 || g(3),    //     otherwise, it's extracted from n
        b = B(              //     b is a Buffer:
          d[g(2)]           //       select the relevant dictionary
          [g(d[x].length)]  //       and extract the relevant word from it
        ),                  //     end of Buffer
        b[-q % 2] ^= 32,    //     if q is even, toggle the case of the 1st char.
        q % 3 > 0 ^ C ?     //     if (q mod 3 != 0) XOR C is true:
          ' ' + b           //       append a space and the word
        :                   //     else:
          b                 //       just append the word
      )                     //
    :                       //   else (mode 0):
      B([                   //     use a Buffer to generate a character:
        g(96) ^ 95 ? x + 32 //       extract the character ID and add 32
                   : 10     //       or force a line-feed if it's 95
      ])                    //     end of Buffer
  )                         //
  + F(n, 1)                 //   append the result of a recursive call
:                           // else (n = 0):
  ''                        //   stop the recursion
```
[Answer]
# [Jelly](https://github.com/DennisMitchell/jelly), 11 [bytes](https://github.com/DennisMitchell/jelly/iki/Code-page)
Obligatory Jelly answer.
```
ḃØ⁵ịØJ⁾“»jV
```
**[Try it online!](https://tio.run/##ATgAx/9qZWxsef//4biDw5jigbXhu4vDmErigb7igJzCu2pW////ODYwOTk0MTc0MjA0MjM4NzUzNDQxNg "Jelly – Try It Online")**
### How?
```
ḃØ⁵ịØJ⁾“»jV - Link: integer, n
ḃ           - convert (n) to bijective base:
 Ø⁵         -   250
   ị        - index into:
    ØJ      -   Jelly's code-page
      ⁾“»   - list of characters = "“»"
         j  - join -> "“...»"
          V - evaluate as Jelly code
```
[Answer]
# [Python 3](https://docs.python.org/3/), 245 bytes
```
def f(n,s,l,i=0):
	t=""
	while n:
		if n%3:
			if n%3>1:f=n//3%3;n//=9
			else:f=-1;n//=3
			d=[l,s][n%2];w=d[n//2%len(d)];t+=((t!="")^(f>0))*" "+[w[0].swapcase(),w[0]][f%2]+w[1:];n//=2*len(d)
		else:t+=chr((n//3%96+22)%117+10);n//=288
	return t
```
[Try it online!](https://tio.run/##VVLdruM0EL6un8InUlGyLT3@SRz7HLrigFZC4gotEkhNkZzEabxNnchxthRpb3kAHpEXOUzSgxC5yXh@vm/mmxluoe0df32tTYOb2G3Hbbe1e5I8oVXYRxFaXVvbGezgvbINdms@W2/me/rU7N3jI1/zZ/jt1Rwy3WjA/TVdXHx21ftDtx2PB7dmx@frvj5AhK074@I6OT6HzT6OwwOwJb/FzXuSJO8iHG0O1wM57sarHio9mjjZzu/joQGMzfVAn44LPnt3h0FvxABWtT6Ol66U2DCWrCnNN5Qk93wp0cqbMHmHw2vj@wv@ZLruhu1l6H3Ata2C7Z32N4Tuo//sJwMz6y0u8R5bN0whTnbj0NkQR0URAbVeAiHWuzF4O8QJ@M7ga2Ko@g9xN7ZA8T9P17sTJA9@Lo/M74Opgqkj4Nri6NQHsM5g4b///AtHSwd7fE5eKYhavOiqNQ6x2dZ3m8/2A0rze7xFjMp72PfOVrpDLCNLrKynDuVptkTLur9Y1w@dHoM1I8qpXLL0FHokKRHzY4e1bkfERUq4pAoqi5/w5KreNdYZxJVIc8koFBYf@7q@vZTGez0PijLFMib5zFZYN05dsO6EdalhDQalRBAmFXxzadv33TdA1ZW9PyFBciUykecwUfF9ZX012XD7d/KUsFxmSoAEhfv064xYIUoVyzlTAroumlngl64zHRIqY0LkXKYgSfGjhj5SkMF4qx3KASsjTDEFWRD@cDPQyHmNtf8MquWcEEFlnhKqBE/pqjhhB4q6YHzVancyuuzMhC@6vWqkpEpTrlLBGYFKnq@Kx2e4DGPCVft6xD9o/wd@W5kkII5QIHmeZblSaZ6uCjP0Ix7hIL123@IXOI/e2/ECs2VEShALDMBPU5FxAN8a/IuGVkrra@3MssbKLquklAGBlJRLnqUZ1AnAx0bPl91ON/1Vbar@vGQKSJEMWieK5rDJXELmR/15Ge4Bw8XccKnLsgPggCjnwK6YgAXxeQti1iXoS9lPcM52DOOX73QIE5nXov8B "Python 3 – Try It Online")
Pretty poorly optimized; working on a better approach.
]