[Question]
  [
**Terminology**
An increasing number is one where each digit is greater than or equal to all the digits to the left of it (ex. 12239)
A decreasing number is one where each digit is less than or equal to all the digits to the left of it (ex. 95531)
A bouncy number is any number that is not increasing or decreasing. Since this requires at least 3 digits, the first bouncy number is 101
**The task**
Given an integer n greater than or equal to 1, find the sum of the first n bouncy numbers
**Rules**
* This is code golf, so the answer with the shortest amount of bytes wins
* If your language has limits on integer size (ex. 2^32-1) n will be small enough that the sum will fit in the integer
* Input can be any reasonable form (stdin, file, command line parameter, integer, string, etc)
* Output can be any reasonable form (stdout, file, graphical user element that displays the number, etc)
**Test cases**
```
1 > 101
10 > 1065
44701 > 1096472981
```
  
[Answer]
# [Jelly](https://github.com/DennisMitchell/jelly), ~~10~~ 8 bytes
```
ṢeṚƬ¬µ#S
```
[Try it online!](https://tio.run/##AR0A4v9qZWxsef//4bmiZeG5msaswqzCtSNT//80NDcwMQ "Jelly – Try It Online")
### How it works
```
ṢeṚƬ¬µ#S  Main link. No arguments.
      #   Read an integer n from STDIN and call the chain to the left with argument
          k = 0, 1, 2, ... until n of them return a truthy result.
          Yield the array of successful values of k.
     µ    Monadic chain. Argument: k (integer)
Ṣ           Sort, after promoting k to its digit array.
  ṚƬ        Reverse 'til the results are no longer unique and yield unique results.
            Calling Ṛ on k promotes it to its digit array. If k = 14235, the 
            result is [14235, [5,3,2,4,1], [1,4,2,3,5]].
 e          Check if the result to the left appears in the result to the right.
    ¬       Negate the resulting Boolean.
       S  Take the sum.
```
[Answer]
# [Pyth](https://pyth.readthedocs.io), 10 bytes
```
s.f!SI#_B`
```
[Try it here!](https://pyth.herokuapp.com/?code=s.f%21SI%23_B%60&input=10&debug=0)
### How it works?
```
s.f!SI#_B` – Full program. Takes an integer Q from STDIN and outputs to STDOUT.
 .f        – Find the first Q positive integers that satisfy a certain condition.
   !SI#_B  – The condition. Returns true for bouncy numbers only.
       _B` – Cast the number to a string and bifurcate (pair) it with its reverse.
      #    – Filter-keep those...
     I     – That are invariant under...
    S      – Sorting.
           – To clarify, I (invariant) is a Pyth operator that takes two inputs, a 
             function and a value and checks whether function(value) == value, so
             this is technically not a built-in.
   !       – Logical not. The empty list gets mapped to true, other values to false.
s          – Sum.
```
[Answer]
# [K (ngn/k)](https://gitlab.com/n9n/k), 37 bytes
```
{+/x{{(a~&\a)|a~|\a:10\x}(1+)/x+1}\0}
```
[Try it online!](https://tio.run/##FchBCsMgEADAe1/RQymKlLjFJo37lChkL5aSkkLIYYPRr5t4G2Z6zJ@5lGCjajhGQfnuSO6Ud0cWtOMkQMmGFSSnU1ltvA1bXmy4Mo7/CcUY6PtDxg0X6dNlHQBBg6/Qp9pXpTGdrt@3pnv2b/DlAA "K (ngn/k) – Try It Online")
`{` `}` is a function with argument `x`
`x{` `}\0` applies the `{}` on `0` `x` times, preserving the intermediate results
`(1+)` is the successor function
`{` `}(1+)/x+1` applies the successor function starting from `x+1` until the `{}` returns true
`10\x` are the decimal digits of `x`
`a:` assign to `a`
`|\` is the max-scan (partial maxima) of `a`
`&\` analogously, is the min-scan
`a~|\a` does `a` match its max-scan?
`|` or
`a~&\a` its min-scan?
`+/` sum
[Answer]
# [Python 2](https://docs.python.org/2/), 84 bytes
```
i=s=0
n=input()
while n:b=`i`>`sorted(`i`)`[2::5]<`i`[::-1];n-=b;s+=b*i;i+=1
print s
```
**[Try it online!](https://tio.run/##DcZBCoAgEADAu6/waEmQUQRr20ckECloITZJI3q9NaeJb95P7kohTNgKRuJ4Z1WJZ6djkwwBPfnZp/PK26r@V951AMMy/XcAjVksNxhs0hhqsqTRiHgRZ5lK6fuxNR8 "Python 2 – Try It Online")** Or see a [test-suite](https://tio.run/##JY7RDoIgGEav5Sm4AxI3MVsb9vsizg1RmmwGTmjV05PW3bk439m3fuLsXZVGPxkghCQLAUrkwLr1GSlDr9kuBjupQVnVquC3aCa6M1NdJeWlv@3cSVmIvnEF6CbkoE@2sTkItG7WRRzSHka/ICzDQ0@DjCZEdPcbPgBbh6ngFT9zUfK6vpaCSZT9x4fAMSlawlFm3makx1WWvg "Python 2 – Try It Online").
[Answer]
# JavaScript (ES6), 77 bytes
```
f=(i,n=0,k,p)=>i&&([...n+''].map(x=>k|=x<p|2*(p<(p=x)))|k>2&&i--&&n)+f(i,n+1)
```
[Try it online!](https://tio.run/##bcyxDsIgEADQ3Q@53lkgbeceP2IcSC0GqXCxxjDw76ircX3Du7mX25dHkKdO@bK25hmDSjyoqITYBgA8GWNS33Vnc3eChW2sXGap0xFlRuFCRDXaCSBoDZCo99@jH6ktOe15W82Wr@hxJDr8yPCHPtbe "JavaScript (Node.js) – Try It Online")
### Commented
```
f = (                     // f = recursive function taking:
  i,                      //   i = number of bouncy numbers to find
  n = 0,                  //   n = current value
  k,                      //   k = bitmask to flag increasing/decreasing sequences
  p                       //   p = previous value while iterating over the digits
) =>                      //
  i && (                  // if there's still at least one number to find:
    [...n + '']           //   turn n into a string and split it
    .map(x =>             //   for each digit x in n:
      k |=                //     update k:
        x < p |           //       set bit #0 if x is less than the previous digit
        2 * (p < (p = x)) //       set bit #1 if x is greater than the previous digit
                          //       and update p
    )                     //   end of map()
    | k > 2               //   if both bits are set (n is bouncy):
    && i--                //     decrement i
    && n                  //     and add n to the total
  ) + f(i, n + 1)         //   add the result of a recursive call with n + 1
```
[Answer]
# Python 2, ~~110~~ ~~92~~ 89 bytes
```
n=input()
x=s=0
while n:b={-1,1}<=set(map(cmp,`x`[:-1],`x`[1:]));s+=x*b;n-=b;x+=1
print s
```
[**Try it online**](https://tio.run/##Lc1NCoMwEEDhtXOK4iqpCk4RCrFzkiJo0gGlNQaT0kjp2dMfunur77ktjIs9JLNcmPI8T5Ym6@5BSIjkqYbHON14Z5WmZ4Ulvk7kOYh5cMLMruxjf1YVdr9A1UnZ@oLiXre2It3GghDcOtmw8@mj/zVUkIV1UxzZiO9ZQsbRsAtKrzxcEwLW0DTHGt8)
This function determines if a number is bouncy:
```
lambda x:{-1,1}<=set(map(cmp,`x`[:-1],`x`[1:]))
```
[Answer]
# [Retina](https://github.com/m-ender/retina/wiki/The-Language), 93 bytes
```
K`:
"$+"{/(?=.*;(_*);\1_)(?=.*_(_*);\2;)/^{`:(#*).*
:$1#;$.($1#
)`\d
*_;
)`:(#+).*
$1:$1
\G#
```
[Try it online!](https://tio.run/##K0otycxLNPz/3zvBiktJRVupWl/D3lZPy1ojXkvTOsYwXhPMjYdwjaw19eOqE6w0lLU09bS4rFQMla1V9DSAFJdmQkwKl1a8NZABlNYGSasYAhVwxbgr//9vaAAA "Retina – Try It Online") Explanation:
```
K`:
```
Initialise `s=i=0`. (`s` is the number of `#`s before the `:`, `i` the number of `#`s after.)
```
"$+"{
...
)`
```
Repeat `n` times.
```
/(?=.*;(_*);\1_)(?=.*_(_*);\2;)/^{`
...
)`
```
Repeat while `i` is not bouncy.
```
:(#*).*
:$1#;$.($1#
```
Increment `i` and make a copy in decimal.
```
\d
*_;
```
Convert the digits of the copy to unary. The bounciness test uses the unary copy, so it only works once `i` has been incremented at least once.
```
:(#+).*
$1:$1
```
Add `i` to `s` and delete the copy of the unary digits, so that for the next pass of the inner loop the bounciness test fails and `i` gets incremented at least once.
```
\G#
```
Convert `s` to decimal.
121 byte version calculates in decimal, so might work for larger values of `n`:
```
K`0:0
"$+"{/(?=.*;(_*);\1_)(?=.*_(_*);\2;)/^{`:(\d+).*
:$.($1*__);$.($1*__)
)+`;\d
;$&*_;
)`\d+:(\d+).*
$.(*_$1*):$1
:.*
```
[Try it online!](https://tio.run/##PYsxCsMwDEV3HSOIYsngWBklQg7QK5jIhXTokqF0y@Ed00C2/z7vfd@/z/6S1p41a4YB43CMYZkTW3AmK@L0R79wMhrXo2ooW6TEoJgCCruT3QsoVisbGD7YDah29w66xd49UhTQfrQm@QQ "Retina – Try It Online") Explanation:
```
K`0:0
```
Initialise `s=i=0`.
```
"$+"{
...
)`
```
Repeat `n` times.
```
/(?=.*;(_*);\1_)(?=.*_(_*);\2;)/^{`
...
)
```
Repeat while `i` is not bouncy.
```
:(\d+).*
:$.($1*__);$.($1*__)
```
Increment `i` and make a copy.
```
+`;\d
;$&*_;
```
Convert the digits of the copy to unary. The bounciness test uses the unary copy, so it only works once `i` has been incremented at least once.
```
\d+:(\d+).*
$.(*_$1*):$1
```
Add `i` to `s` and delete the copy of the unary digits, so that for the next pass of the inner loop the bounciness test fails and `i` gets incremented at least once.
```
:.*
```
Delete `i`.
[Answer]
# [05AB1E](https://github.com/Adriandmen/05AB1E), 12 bytes
```
µN{Â‚Nå_iNO¼
```
[Try it online!](https://tio.run/##MzBNTDJM/f//0Fa/6sNNjxpm@R1eGp/p539oz///JibmBoYA "05AB1E – Try It Online")
**Explanation**
```
µ              # loop over increasing N until counter equals input
     Nå_i      # if N is not in
 N{Â‚          # the pair of N sorted and N sorted and reversed
         NO    # sum N with the rest of the stack
           ¼   # and increment the counter
```
[Answer]
# Java 8, ~~114~~ 112 bytes
```
n->{int i=0,s=0;for(;n>0;++i)s+=(""+i).matches("0*1*2*3*4*5*6*7*8*9*|9*8*7*6*5*4*3*2*1*0*")?0:--n*0+i;return s;}
```
Uses a regular expression to check if the number is increasing or decreasing. Try it online [here](https://tio.run/##TVDBcoIwEL33K3Y4wSJMqFprU@xM7z15dDxEjBoLC5MEO47l22lAtO4hL3nZffP2HcVJRGUl6bj9bqt6k6sMslwYA19CEVyewJUiK/VOZBI@y5qy80APX2Dqwu@QAt7zTX8OYsYK6@BUqi0UTtJfWq1ov1qD0HsTPEgN2psrpNBStLh0uiplI5Myviu1z2nBeBiqwISp73nuEhfCZgdpfI9hgs84xglO8QVn@Ipz/J07mLnn1NFj950gQy/4YG9RRMhCxbW0tSYwvGlvTvjjes4p1cVGauM8kfwZuAskI0jYCCaTGUug@Z/pbPa2XT/jDt5vAnEuaW8PHLoF7v1dLc/GyiIuaxtXLh6bk3/NIe7CHcZXah3cI27aPw).
Ungolfed:
```
n -> { // lambda
    int i = 0, // the next number to check for bounciness
        s = 0; // the sum of all bouncy numbers so far
    for(; n > 0; ++i) // iterate until we have summed n bouncy numbers, check a new number each iteration
        s += ("" + i) // convert to a String
             .matches("0*1*2*3*4*5*6*7*8*9*" // if it's not an increasing  number ...
             + "|9*8*7*6*5*4*3*2*1*0*") ? 0 // ... and it's not a decreasing number ...
             : --n*0 // ... we have found another bouncy number ...
               + i; // ... add it to the total.
    return s; // return the sum
}
```
[Answer]
# Python 2, 250 bytes
```
n=input()
a=0
b=0
s=0
while a<n:
    v=str(b)
    h=len(v)
    g=[int(v[f])-int(v[0]) for f in range(1,h) if v[f]>=v[f-1]]
    d=[int(v[f])-int(v[0]) for f in range(1,h) if v[f]<=v[f-1]]
    if len(g)!=h-1 and len(d)!=h-1:
       a+=1
       s+=b
    b+=1
print s
```
[Answer]
# [Nibbles](http://golfscript.com/nibbles), 9 bytes
```
+<$|,~~?`.`<;`p$\
```
(it would be 8.5 bytes if filesystems could store half bytes)
Similar to the Jelly answer.
```
+                # sum of
 <               # first n elements of
  $              # first input integer
   |  ~          # filter.not
    ,~           # range from 1 to infinity
       ?         # index of (begin of filter body)
        `.       # iterate while uniq
        `<       # sort
          ;`p$   # int to str of filter arg, also save value
              \  # reverse
                 # implicit $ (arg of iterate while uniq)
                 # implicit $ (saved value from `p)
```
[Answer]
# [Stax](https://github.com/tomtheisen/stax), 14 [bytes](https://github.com/tomtheisen/stax/blob/master/docs/packed.md#packed-stax)
```
Ö²àDæ▲☼Ty≡°lû↕
```
[Run and debug it](https://staxlang.xyz/#p=99fd8544911e0f5479f0f86c9612&i=1%0A10%0A44701&a=1&m=2)
[Answer]
# [Vyxal](https://github.com/Vyxal/Vyxal), 7 bytes
```
fÞȮ¬)ȯ∑
```
[Try it Online!](https://vyxal.pythonanywhere.com/#WyIiLCIiLCJmw57IrsKsKciv4oiRIiwiIiwiNSJd)
```
      ∑ # Sum of...
     ȯ  # First n integers...
----)   # Where...
f       # Digits
   ¬    # Are not
 ÞȮ     # Sorted
```
[Answer]
# [Red](http://www.red-lang.org), 108 bytes
```
func[n][i: s: 0 until[t: form i
if(t > u: sort copy t)and(t < reverse u)[s: s + i n: n - 1]i: i + 1
0 = n]s]
```
[Try it online!](https://tio.run/##PYzBCgIhFEX3fsVdNkSgMBBI9RFtxUWMCg/qOTyfQV9vrmZ3OJdzJafxzClEU/wonbfAMZBH87DorPQO6lGqfECGyknxQJ9zFcVW9x90eXGa@gbJ3ywtoy9h1g1nENiDcYGL85KmccbiDo4tjl2IFWWqg@yB63q1bvwB "Red – Try It Online")
## More readable:
```
f: func [ n ] [
    i: s: 0
    until [
       t: form i  
       if ( t > u: sort copy t ) and ( t < reverse u ) [
            s: s + i
            n: n - 1
       ]
       i: i + 1
       0 = n
    ]
    s
]
```
A good opportunity to use `form` - `form i` is 5 bytes shorter than `to-string i`
[Answer]
# [MATL](https://github.com/lmendo/MATL), ~~31~~ 30 bytes
```
l9`tVdZS&*0<a?wQtG>?.]y]QT]xvs
```
[Try it online!](https://tio.run/##y00syfn/P8cyoSQsJSpYTcvAJtG@PLDE3c5eL7YyNjAktqKs@P9/QwMA "MATL – Try It Online")
```
l       % Initialize counter to 1
9       % First value to check for being bouncy
`       % Start do-while loop
  tV    % duplicate the current number and convert to string
        % (let's use the iteration where the number is 1411)
        % stack: [1, 1411, '1411']
  d     % compute differences between the characters
        %  For non-bouncy numbers, this would be either all 
        %  positive values and 0, or all negative values and 0
        % stack: [1, 1411, [3 -3 0]]
  ZS    % keep only the signs of those differences
        % stack: [1, 1411, [1 -1 0]]
  &*    % multiply that array by its own transpose, so that
        %  each value is multiplied by every other value
        %  for non-bouncy numbers, all products will be >=0
        %  since it would have had only all -1s or all 1 (other than 0s)
        % stack: [1, 1411, [1 -1  0
                            -1  1 0
                            0  0  0]]
  0<a?  % if any value in the matrix is less than 0
    wQ    % switch the counter to top, increment it
          % stack: [1411, 2]
    tG>?  % duplicate it, check if it's gotten greater than the input limit
      .]    % if so, break out of loop
  y]    % else, duplicate bouncy number from inside stack,
        %  keeping a copy to be used for summing later
        % stack: [1411, 2, 1411]
  QT    % increment number, push True to continue loop
]     % loop end marker
x     % if we've broken out of loop, remove counter from stack
v     % concatenate the bouncy numbers we've collected in stack and 
s     % sum them
```
[Answer]
# [R](https://www.r-project.org/), 96 bytes
```
function(n){while(n){y=diff(T%/%10^(0:log10(T))%%10);g=any(y<0)&any(y>0);F=F+T*g;n=n-g;T=T+1};F}
```
[Try it online!](https://tio.run/##HchBCsIwEEDRvfeIzFiKEywUjOMyJ8haEDUxUKZQLBKkZ49JNp/HX7Ln7Fd5fOIsIPj7vuP0qkj8jN6DU0el6QZ0nuagCRyiKgNN4LskSBfCfcO1PMu2c4dghKUPxrHr9Gbslj1o3JVQ7al1GEZqc6SK/Ac "R – Try It Online")
**Explanation :**
```
while(n){                         # while n > 0
        T%/%10^(0:log10(T))%%10   # split T into digits(T==TRUE==1 at the 1st loop)
y=diff(                         ) # compute the diff of digits i.e. digits[i] - digits[i+1]
g=any(y<0)&any(y>0)               # if at least one of the diff is < 0 and 
                                  # at least one is > 0 then T is "bouncy"
F=F+T*g                           # if bouncy increment F (F==FALSE==0 at the 1st loop)
n=n-g                             # decrement n by 1 if bouncy
T=T+1}                            # increment T by 1 and loop
F}                                # return F
```
[Answer]
# [Ruby](https://www.ruby-lang.org/), 76 bytes
```
->n,s=i=0{[d=(i+=1).digits,d.reverse].index(d.sort)||(s+=i;n-=1);n<1?s:redo}
```
[Try it online!](https://tio.run/##NcpBDoIwEEDRvado4qYEaFpTQcTRg5BuzBQzm2I6aDTi2SuQ@JcvPz6u79RDKs@hYCDQnw5BUg4mU0g3GrlAFf3TR/ZOUUD/kqh4iGM2TZJzoDaU89yGk7nwMXocvuku@s44sbQVRpvNCtr9odqvYm2t522RprL1rjmY9AM "Ruby – Try It Online")
[Answer]
# C (gcc), 104 bytes
```
f(b,o,u,n,c,y){for(o=u=0;b;u+=y?0:o+0*--b,++o)for(n=o,y=3;n/10;)c=n%10,n/=10,y&=(c-=n%10)<0?:c?2:y;b=u;}
```
Try it online [here](https://tio.run/##PY3BasMwDIbPyVOIQoc1K8TpCoOoIg/S9TB7zfCh9kjngxfy7Jkzyi4Sn/5Pkms@nVvXUVmKlCiQo4zzGCcVJYlhy0lLHkwftXluGktaR9ziIJGyvHBoO8PoJOw7Q6GVUvOTKNf8TfBkht4Nhz6zlcTL6sM33N59UAhzXW0U0s1ep/v5AgIzdATlAhyPr6aDhetq@7VpvsSGSzvB3f9c46gei9g@uFjIoLXHuqq@poKj2u0/3sKO4N8@@wsi18v6Cw).
Ungolfed:
```
f(n, // function: return type and type of arguments defaults to int;
     // abusing extra arguments to declare variables
  i,   // number currently being checked for bounciness
  s,   // sum of the bouncy numbers
  j,   // copy of i to be used for checking bounciness in a loop
  p,   // used for investigating the last digit of j
  b) { // whether i is not bouncy; uses the two least significant bits to indicate increasing/decreasing
    for(i = s = 0; // check numbers from zero up; initial sum is zero
        n; // continue until we have n bouncy numbers
        s += b ? 0 // not bouncy, no change to the sum
        : i + 0* --n, // bouncy, add it to the sum and one less bouncy number to go
        ++i) // either way, move to the next number
        for(j = i, b = 3; j/10; ) // make a copy of the current number, and truncate it from the right until there is just one digit left
        // bounciness starts as 0b11, meaning both increasing and decreasing; a value of 0 means bouncy
            p = j % 10, // get the last digit
            j /= 10, // truncate one digit from the right
            b &= // adjust bounciness:
                 (p -= j % 10) // compare current digit to the next
                 < 0 ? // not an increasing number, clear second to least significant bit
                 : p ? 2 // not a decreasing number, clear least significant bit
                 : b; // keep it the same
    n = s; // gcc shortcut for return s
}
```
[Answer]
## Ruby (123 106 bytes)
```
o=0
x=[?0]
(z=(x=(x.join.to_i+1).to_s.chars).sort)!=x&&z!=x.reverse ? (n-=1;o+=x.join.to_i):1while n>0
p o
```
Looks pretty ugly to me. Bounciness is defined in this block `x.sort!=x&&x.sort!=x.reverse`
]