# List all possible titles for the Anno games

In the Anno video game series there are 6 games with a 7th one announced for early 2019. Their titles always feature a year in a specific pattern:

Anno 1602, Anno 1503, Anno 1701, Anno 1404, Anno 2070, Anno 2205, Anno 1800

• The digital sum is always 9.
• The years are four digits long.
• They contain at least one zero.

Within these constrains there exist 109 possible titles:

[1008,1017,1026,1035,1044,1053,1062,1071,1080,1107,1170,1206,1260,1305,1350,1404,1440,1503,1530,1602,1620,1701,1710,1800,2007,2016,2025,2034,2043,2052,2061,2070,2106,2160,2205,2250,2304,2340,2403,2430,2502,2520,2601,2610,2700,3006,3015,3024,3033,3042,3051,3060,3105,3150,3204,3240,3303,3330,3402,3420,3501,3510,3600,4005,4014,4023,4032,4041,4050,4104,4140,4203,4230,4302,4320,4401,4410,4500,5004,5013,5022,5031,5040,5103,5130,5202,5220,5301,5310,5400,6003,6012,6021,6030,6102,6120,6201,6210,6300,7002,7011,7020,7101,7110,7200,8001,8010,8100,9000]


Your objective is to list them all in any reasonable form in the fewest number of bytes.

• How flexible is the output format? Is this acceptable? Oct 15 '18 at 14:50
• @LuisMendo Yes, that's fine with me. Oct 15 '18 at 14:54
• Are lists of digits allowed? Oct 15 '18 at 18:18
• @aslum I assume you mean a lot of spaces, not just one, right? Comment markdown doesn't allow for a good representation of that. And I would assume that's allowed, given that Luis's format above is allowed. ;-) Oct 15 '18 at 20:20
• @EriktheOutgolfer I'd say no to lists of digits because they really do not look like years anymore. Oct 15 '18 at 21:50

# R, 59 51 bytes

Outputs the valid numbers as the names of a list of 201's. Why 201? Because ASCII 0 is 48, and 4*48+9 is... yeah. Saved 6 bytes by aliasing ^ to Map and another 2 by using 1:9e3 as range.

"^"=Map;x=sum^utf8ToInt^grep(0,1:9e3,,,T);x[x==201]


Try it online!

# Explanation

# Create list of sums of ASCII char values of numbers,
# with the original numbers as the names of the list
x <- Map(sum,
# Create a list from the strings where each element is the string split
# into ASCII char values
Map(utf8ToInt,
# Find all numbers between 1 and 9e3 that contain a zero
# Return the matched values as a vector of strings (6th T arg)
grep(pattern=0,x=1:9000,value=TRUE)
)
)
# Pick out elements with value 201 (i.e. 4-digits that sum to 9)
# This implicitly only picks out elements with 4 digits, since 3-digit
# sums to 9 won't have this ASCII sum, letting us use the 1:9e3 range
x[x==201]

• ah, grep, why do I never remember that it casts to character... Oct 15 '18 at 14:23

# Perl 6, 35 33 bytes

-2 bytes thanks to Jo King

{grep {.ords.sum==201&&/0/},^1e4}


Try it online!

# Python 2, 6766 64 bytes

print[y for y in range(9001)if('0'iny)*sum(map(ord,y))==201]


Try it online!

Saved:

• -1 byte, thanks to Luis felipe De jesus Munoz
• -2 bytes, thanks to Kevin Cruijssen
• 64 bytes by using the ord sum == 201 trick from other answers. Oct 16 '18 at 7:52

# Jelly, 11 bytes

9ȷṢ€æ.ẹ9ṫ19


Try it online!

### How it works

9ȷṢ€æ.ẹ9ṫ19  Main link. No arguments.

9ȷ           Set the left argument and the return value to 9000.
Ṣ€         Sort the digits of each integer in [1, ..., 9000].
æ.       Perform the dot product of each digit list and the left argument,
which gets promoted from 9000 to [9000].
Overflowing digits get summed without multiplying, so we essentially
map the digit list [a, b, c, d] to (9000a + b + c + d).
ẹ9     Find all 1-based indices of 9.
Note that 9000a + b + c + d == 9 iff a == 0 and b + c + d == 9.
ṫ19  Tail 19; discard the first 18 indices.


# PowerShell, 50 49 bytes

999..1e4-match0|?{([char[]]"$_"-join'+'|iex)-eq9}  Try it online! Constructs a range from 999 to 10000, then uses inline -match as a filter to pull out those entries that regex match against 0. This leaves us with 1000, 1001, 1002, etc. We then pipe that into a Where-Object clause where we take the current number as a string "$_", cast it as a char-array, -join those characters together with + and Invoke-Expression (similar to eval) to come up with their digit sum. We check whether that is -equal to 9, and if so it's passed on the pipeline. At program completion, those numbers are picked up from the pipeline and implicitly output.

# JavaScript (ES6), 78 73 bytes

Saved 2 bytes thanks to @KevinCruijssen

Returns a space-separated string.

f=(n=9e3)=>n>999?f(n-9)+(eval([...n+''].join+)&/0/.test(n)?n+' ':''):''


Try it online!

### How?

We iterate over the range $$\[1008..9000]\$$ with an increment of $$\9\$$, ignoring numbers that don't have a $$\0\$$.

All these numbers are multiples of $$\9\$$, so the sum of their digits is guaranteed to be a multiple of $$\9\$$ as well.

Because valid numbers have at least one $$\0\$$, they have no more than two $$\9\$$'s, which means that the sum of the remaining digits is at most $$\18\$$. Therefore, it's enough to test if the sum of the digits is odd.

Hence the test:

(eval([...n + ''].join+) & /0/.test(n)

• You can save a byte changing the 1008 to 999, since it doesn't contain a 0 anyway, and 999+9 = 1008. Oct 15 '18 at 15:13
• Or even 2 bytes by changing it to f=(n=9e3)=>n<1e3?'':(eval([...n+''].join+)<10&/0/.test(n)?[n,,]:'')+f(n-9) (does contain a trailing comma though, so f=(n=9e3)=>n<1e3?'':(eval([...n+''].join+)<10&/0/.test(n)?n+' ':'')+f(n-9) with space delimiter including trailing space might look prettier) Oct 15 '18 at 15:17
• @KevinCruijssen Thanks! I'm actually trying to update this for a while, but I have like 500B/s of Internet bandwidth where I am tonight. :/ Oct 15 '18 at 16:42
• I know the feeling.. Lately our internet at home is crap for some reason.. Can't download anything above 10 MB, and sometimes have to refresh videos or pages with 10+ images a few times before it completely loads.. Really annoying when I'm working from home on Mondays/Tuesdays.. >.> Tomorrow someone comes to fix it (and I'm not leaving him until it's fixed xD) Oct 15 '18 at 16:44

# JavaScript (Node.js), 89 bytes

[...Array(9e3)].map(_=>i++,i=1e3).filter(a=>(s=[...a+""]).sort()[0]<1&eval(s.join+)==9)


Try it online!

• -4 bytes thanks to @ETHproductions

# JavaScript (Node.js), 129127126124115114111110105979392 90 bytes

[...Array(9e3)].map(f=(_,i)=>eval(s=[...(i+=1e3)+""].sort().join+)-9|s[0]?0:i).filter(f)


Try it online!

### Explanation

[...Array(9e3)].map(f=(_,i)=>eval(s=[...(i+=1e3)+""].sort().join+)-9|s[0]?0:i).filter(f)
[...Array(9e3)].map(f=(_,i)=>                                                  )           // Create a 9000-length array and loop over it; store the loop body
[...(i+=1e3)+""]                                       // Add 1000 to the index and split it into an array of characters (17 -> ["1", "0", "1", "7"])
.sort()                                // Sort the array of characters in ascending order by their code points ("0" will always be first) (["1", "0", "1", "7"] -> ["0", "1", "1", "7"])
s=                       .join+                        // Join them together with "+" as the separator (["0", "1", "1", "7"] -> "0+0+2+9"); store the result
eval(                                 )-9                     // Evaluate and test if it's different than 9
s[0]                // Take the first character of the string and implicitly test if it's different than "0"
|    ?0              // If either of those tests succeeded, then the number doesn't meet challenge criteria - return a falsey value
:i            // Otherwise, return the index
.filter(f) // Filter out falsey values by reusing the loop body


First time doing code golf in JavaScript. I don't think I need to say it, but if I'm doing something wrong, please notify me in the comments below.

• -3 bytes thanks to @Luis felipe De jesus Munoz

• -6 bytes thanks to @Kevin Cruijssen

• [...Array(9e3)] instead Array(9e3).fill() saves 2 bytes Oct 15 '18 at 15:03
• .map(a=>+a) instead .map(Number) saves another byte Oct 15 '18 at 15:05
• You can remove the space at (_, i) to save a byte, and s[0]+s[1]+s[2]+s[3] can be eval(s.join+) to save an additional 4 bytes. Oct 15 '18 at 15:29
• Also, I'm pretty sure the || can be | in your answer. Oct 15 '18 at 15:41
• If you use .map() only to generate the range, and keep the filtering separate, you can save 8 bytes: Try it online! Oct 15 '18 at 18:16

# Python 2, 57 bytes

n=999
exec"n+=9\nif'0'inn>int(n,11)%10>8:print n\n"*n


Try it online!

2 bytes thanks to Dennis

Uses an exec loop to counts up n in steps of 9 as 1008, 1017, ..., 9981, 9990, printing those that meet the condition.

Only multiples of 9 can have digit sum 9, but multiples of 9 in this range can also have digits sum of 18 and 27. We rule these out with the condition int(n,11)%10>8. Interpreting n in base 11, its digit sum is equal to the number modulo 10, just like in base 10 a number equals its digit sum modulo 9. The digits sum of (9, 18, 27) correspond to (9, 8, 7) modulo 10, so taking those>8 works to filter out nines.

The number containing a zero is check with string membership. '0'inn. This condition is joined with the other one with a chained inequality, using that Python 2 treats strings as greater than numbers.

• I like how heavily golfed Python seems to often have enormously long autogenerated executables... Oct 16 '18 at 9:13

# sed and grep (and seq), 7264 63 bytes

seq 9e3|sed s/\\B/+/g|bc|grep -wn 9|sed s/:9//|grep 0|grep ....

• Some of these aren't four digits long (but I'm not sure what the final grep is, so maybe I'm running it wrong?) Oct 16 '18 at 1:06
• @Sparhawk: The last grep ensures that the number is 4 digits long
– Thor
Oct 16 '18 at 15:58
• @Thor Ah right. For some reason I parsed that as an ellipsis. Oct 16 '18 at 20:58

[i|i<-show<$>[1..5^6],201==sum(fromEnum<$>i),elem '0'i]


Thanks to @Laikoni, see the comments.

import Data.Char (digitToInt)

[i | i <- show <$> [1000..9999] , sum (digitToInt <$> i) == 9
, '0' elem i
]

• Welcome to PPCG and Haskell golfing in particular! You can save a few bytes by dropping (-48+) and comparing the sum against 201 instead of 9. Incidentally this also allows you to use 1 instead of 1000 for the range. Oct 16 '18 at 12:10
• Also your previous version without main=print was fine as per this consensus on Meta. Oct 16 '18 at 12:13
• 9999 can be 5^6 instead. Oct 16 '18 at 13:38
• Ha, there's always another byte to shave! Thanks :-)
– mb21
Oct 16 '18 at 13:56

# R, 82 bytes

write((x=t(expand.grid(1:9,0:9,0:9,0:9)))[,colSums(x)==9&!apply(x,2,all)],1,4,,"")


Try it online!

Generates a matrix x of all possible 4-digit numbers, excluding leading zeros, going down columns. Then filters for column (digital) sums of 9 and containing zero, i.e., not all are nonzero. write prints down the columns, so we write to stdout with a width of 4 and a separator of "".

Outgolfed by J.Doe

• Nice answer! I came up with a different route... Oct 15 '18 at 14:17

# Japt, 20 18 bytes.

-2 bytes thanks to @Shaggy and @ETHproductions

A³òL² f_=ì)x ¥9«Z×


A³òL² f_=ì)x ¥9«Z×  Full program
A³òL²               Range [1000, 10000]
f_            Filter by :
=ì)         Convert to array
x ¥9     Sum equal to 9?
«    And
Z×  Product not 0


Try it online!

• This is actually 28 bytes. Using a literal integer instead is 22 bytes but A³ò9000 f_ìx ¥9©ZsøT gets you back down to 20. Oct 15 '18 at 17:28
• You can save 1 byte by using ì instead of s and ¬, which has to be done in the filter: f_=ì)x ¥9.... Then you can save another by checking if the product of Z is zero with «Z×: Try it online! Oct 15 '18 at 17:59

# Java 8, 128117 115 bytes

v->{int i=109,r[]=new int[i],n=i;for(;i>0;n++)if((n+"").chars().sum()==201&(n+"").contains("0"))r[--i]=n;return r;}


-11 bytes thanks to @nwellnhof.

Try it online.

Explanation:

v->{                              // Method with empty unused parameter & int-array return
int i=109,                      //  Index-integer, starting at 109
r[]=new int[i],             //  Result-array of size 109
n=i;                        //  Number integer, starting at 109
for(;i>0;                      //  Loop as long as i is not 0 yet:
n++)                       //    After every iteration, increase n by 1
if((n+"").chars().sum()==201 //   If the sum of the unicode values of n is 201,
//   this means there are four digits, with digit-sum = 9
&(n+"").contains("0"))    //   and n contains a 0:
r[--i                      //    Decrease i by 1 first
]=n;                  //    And put n in the array at index i
return r;}                      //  Return the array as result

• What about chars().sum()==201? Oct 15 '18 at 15:33
• @nwellnhof Ah, of course. Thanks! Oct 15 '18 at 15:37

# R, 85 bytes

(just competing for the best abuse of R square brackets ... :P )

[=for;i[a<-0:9,j[a,k[a,w[a,if(sum(s<-c(i,j,k,w))==9&any(!s)&i)write(s,1,s='')]]]]


Try it online!

• Holy for loops, Batman!
– BLT
Oct 18 '18 at 17:43

# 05AB1E, 151312 10 bytes

₄4°ŸεW°ö9Q


-2 bytes thanks to @Emigna
-3 bytes thanks to @Grimy

Try it online.

Explanation:

₄4°Ÿ        # Create a list in the range [1000,10000]
ʒ       # Filter this list by:
W      #  Get the smallest digit in the number (without popping the number itself)
°     #  Take 10 to the power this digit
ö    #  Convert the number from this base to an integer (in base-10)
9Q  #  Check if it's equal to 9

• If the smallest digit is $$\d=0\$$ it will become $$\1\$$ with the $$\10^d\$$ (°). And the number in base-1 converted to an integer in base-10 (ö) would act like a sum of digits.
• If the smallest digit is $$\d=1\$$ it will become $$\10\$$ with the $$\10^d\$$ (°). And the number in base-10 converted to an integer in base-10 (ö) will of course remain the same.
• If the smallest digit is $$\d=2\$$ it will become $$\100\$$ with the $$\10^d\$$ (°). And the number in base-100 convert to an integer in base-10 (ö) would act like a join with 0 in this case (i.e. 2345 becomes 2030405).
• If the smallest digit is $$\d=3\$$ it will become $$\1000\$$ with the $$\10^d\$$ (°). And the number in base-100 convert to an integer in base-10 (ö) would act like a join with 00 in this case (i.e. 3456 becomes 3004005006).
• ... etc. Smallest digits $$\d=[4,9]\$$ would act the same as $$\d=2\$$ and $$\d=3\$$ above, with $$\d-1\$$ amount of 0s in the 'join'.

If the smallest digit is $$\>0\$$ with the given range $$\[1000,10000]\$$, the resulting number after °ö would then be within the range $$\[1111,9000000009000000009000000009]\$$, so can never be equal to $$\9\$$. If the result is equal to $$\9\$$ (9Q) it would mean the smallest digit is $$\d=0\$$, resulting in a base-1 with °ö; and the sum of the digits was $$\9\$$.

• ₄4°Ÿʒ0å}ʒSO9Q. Splitting filters are usually shorter Oct 15 '18 at 15:58
• @Emigna Ah, I was looking for a shorter way for the range, but completely forgot about 4°. Thanks. And you're indeed right that multiple loose filters (at the end) are shorter. Will also add it to one of my tip answers. Thanks for both bytes! Oct 15 '18 at 16:12
• And my other 13-byter (inspired by the ord sum == 201 trick) is 4°Lʒ0å}ʒÇOт·-. Leaving this here, maybe someone can golf it further Oct 16 '18 at 5:06
• ₄4°ŸʒD0åôO9Q. Using a single filter is usually shorter. May 7 '19 at 12:52
• Nevermind, here's a 10: ₄4°ŸʒW°ö9Q May 7 '19 at 13:21

# Pip, 18 bytes

{0Na&$+a=9}FIm,t*m  Use an ouput-format flag such as -p to get readable output. Try it online! {0Na&$+a=9}FIm,t*m
m,t*m  Range from 1000 to 10*1000
{         }FI       Filter on this function:
0Na                 There is at least one 0 in the argument
&                and
$+a The sum of the argument =9 equals 9  # Wolfram Language (Mathematica), 56 55 bytes Select[9!!~Range~9999,Tr@#==Times@@#+9&@*IntegerDigits]  Try it online! We test the range from 9!! = 945 to 9999, since there are no results between 945 and 999. Maybe there's a shorter way to write a number between 9000 and 10007, as well. Tr@#==Times@@#+9& applied to {a,b,c,d} tests if a+b+c+d == a*b*c*d+9, which ends up being equivalent to The Anno Condition. • In retrospect, 9!! isn't any shorter than 999 or something, but it beats 1000. Oct 16 '18 at 0:37 • What is 9!! ? In guessing it isnt related to factorials. Oct 16 '18 at 6:49 • @RobertFraser Double factorial: 9*7*5*3*1. Oct 16 '18 at 14:16 # Ruby, 46 42 41 bytes ?9.upto(?9*4){|x|x.sum==201&&x[?0]&&p(x)}  Try it online! ### How it works: • Iterate on strings ranging from '9' to '9999' • Check that sum of ASCII values is 201 • Check if string contains a zero (without regex, a regex would be 1 byte longer) (Thanks Laikoni for -2 bytes) • 9*3 can be just 9, because checking against 201 already requires 4 digit numbers. Oct 16 '18 at 0:01 # Octave, 49 bytes 6 bytes saved using a more convenient output format as suggested by J.Doe. Thanks to @Laikoni for a correction. y=dec2base(x=1e3:9999,10)'-48;x(sum(y)==9>all(y))  Try it online! • I don't know anything about Octave, but it looks like you can leave the disp off... Oct 15 '18 at 14:48 • @J.Doe OP has confirmed that that output format is acceptable. Thanks for the suggestion! Oct 15 '18 at 14:57 # Dart, 103 100 96 bytes f()=>List.generate(9001,(i)=>'$i').where((i)=>i.contains('0')&&i.runes.fold(0,(p,e)=>p+e)==201);


• -3 bytes by setting the value in the array to string, making the conversion once and not twice
• -4 bytes by using runes instead of codeUnits
• Pretty self-explanatory. generates a list of 9001 (0-9000) cells with the cell's index as value, filters the ones containing a 0 then the one having an ASCII sum of 201 (The result if all the ASCII characters sum to 9). These conditions implictly include that the year is 4 digits long because using 2 ASCII numbers (and the 0), you cannot reach 201.

• Welcome to PPCG. :) Oct 16 '18 at 13:36
• Thanks ! Been lurking for a while, can finally participate Oct 16 '18 at 13:43

# Bash (with seq, grep), 39 bytes

seq 0 9 1e4|egrep '([0-4].*){3}'|grep 0


Try it online!

• @xobzoo suggested seq 0 9 1e4|awk '/([0-4].*){3}/&&/0/' to save two bytes. Oct 21 '18 at 3:03

# K (ngn/k), 22 bytes

55_&(|/~a)&9=+/a:!4#10


Try it online!

• Nice! 55_&9=+/y*|/'~y:!4#10 for 21?
– mkst
Oct 17 '18 at 23:21
• @streetster thanks :) the ' in |/' looks wrong. the result includes 1116, 1125, 1134, etc which are not supposed to be there
– ngn
Oct 18 '18 at 6:43

# APL (Dyalog Unicode), 23 bytes

55↓⍸(×⌿<9=+⌿)10⊥⍣¯1⍳9e3


Try it online!

• that use of < is clever. you can make it even shorter with ⎕io←0 and ( )10⊥⍣¯1⍳9e3 -> ( )¨,⍳4⍴10
– ngn
Nov 1 '18 at 16:14

## PHP, 69, 87 bytes 74 bytes

for($i=999;$i<9001;$i++){echo((array_sum(str_split($i))==9&strpos($i,"0")!=0)?$i:" ");} for($i=999;$i++<1e4;)echo!strpos($i,48)|array_sum(str_split($i))-9?" ":$i; Note this puts a space for every "failed" number, leading to some kind of funky spacing. This can be changed to comma separation, but will add another 4 characters: ?$i.",":""

Got bigger because I wasn't checking for 0. Derp. Shortened by 13 by Titus!

• I don't really know PHP, but does this code ensure that each year contains a zero? Oct 16 '18 at 0:13
• This code does not check for zero in the number. Oct 16 '18 at 11:29
• 13 bytes shorter: for($i=999;$i++<1e4;)echo!strpos($i,48)|array_sum(str_split($i))-9?" ":$i; Oct 18 '18 at 15:29 • Here´s another byte: ?"$i,":"" er ... now the other way round: ?"":"$i," Oct 18 '18 at 21:06 • Actually @Titus that adds a couple bytes. We don't need quotes around$i unless we're including a string w/ it. Oct 22 '18 at 12:40

# APL(Dyalog), 33 29 bytes

1e3+⍸(0∘∊∧9=+/)¨⍎¨∘⍕¨1e3+⍳9e3


Try it online!

Oct 16 '18 at 5:56

# Scala (766361 56 bytes)

for(n<-0 to 9000;t=n+""if t.sum==201&t.min<49)println(t)


Try it online

• Thanks to Laikoni for the suggestions
• Two more bytes shed after applying Jo King's comment
• Welcome to PPCG! Do you have an idea what needs to be added to the header or footer section to get this code to run on TIO? Try it online! Oct 18 '18 at 21:18
• @Laikoni, didn't know I could run Scala in TIO. Fixed it. Thanks for the comment. Oct 18 '18 at 22:40
• It looks like t.sum==201 works instead of t.map(_.asDigit).sum==9. Oct 18 '18 at 23:00
• You may find our tips for golfing in Scala interesting. E.g. it looks like s"$n" can be n+"" and s"$t " can be t+" ". Oct 20 '18 at 10:33
• Since you're using the sum is 201 trick, the range doesn't need to start at 999
– Jo King
Oct 24 '18 at 11:16

# Tcl, 77 bytes

time {if [incr i]>1e3&[regexp 0 $i]&9==[join [split$i ""] +] {puts \$i}} 9999


Try it online!

# Japt, 16 bytes

Returns an array of digit arrays.

L²õì l4 k_×ª9aZx


Test it

## Explanation

L                    :100
²                   :Squared
õ                  :Range [1,L²]
ì                 :Convert each to a digit array
l4              :Filter elements of length 4
k_           :Remove each Z that returns truthy (not 0)
×          :  When reduced by multiplication
ª         :  OR
Zx     :  When reduced by addition
9a       :   And subtracted from 9

• OP has ruled that digit arrays aren't valid output unfortunately :o(
– Sok
Oct 16 '18 at 8:12

# APL(NARS), 45 chars, 90 bytes

f←{⍵×⍳(0∊x)∧9=+/x←⍎¨⍕⍵}⋄f¨1e3..5e3⋄f¨5e3..9e3


test afther some formatting:

1008  1017  1026  1035  1044  1053  1062  1071  1080  1107  1170  1206  1260
1305  1350  1404  1440  1503  1530  1602  1620  1701  1710  1800  2007  2016
2025  2034  2043  2052  2061  2070  2106  2160  2205  2250  2304  2340
2403  2430  2502  2520  2601  2610  2700  3006  3015  3024  3033  3042  3051
3060  3105  3150  3204  3240  3303  3330  3402  3420  3501  3510  3600
4005  4014  4023  4032  4041  4050  4104  4140  4203  4230  4302  4320  4401
4410  4500
5004  5013  5022  5031  5040  5103  5130  5202  5220  5301  5310  5400  6003
6012  6021  6030  6102  6120  6201  6210  6300  7002  7011  7020  7101  7110
7200  8001  8010  8100  9000



possible alternative

r←f;i;x
r←⍬⋄i←1e3⋄→B
A: r←r,i
B: i+←1⋄→A×⍳(0∊x)∧9=+/x←⍎¨⍕i⋄→B×⍳i≤9e3


# Jelly, 13 bytes

ȷ4µṢÄm3Ḍ)ẹ9ṫ4


Try it online!

### How?

ȷ4µṢÄm3Ḍ)ẹ9ṫ4 - Link: no arguments
ȷ4            - literal 10^4 = 10000
µ     )     - for each in range (1,2,3,...,10000): e.g. 3042       or  5211
Ṣ          -   sort (given an integer makes digits)    [0,2,3,4]      [1,1,2,5]
Ä         -   cumulative addition                     [0,2,5,9]      [1,2,4,9]
m3       -   modulo 3 slice (1st,4th,7th...)         [0,9]          [1,9]
Ḍ      -   convert from decimal digits             9              19
ẹ9   - 1-based indices equal to nine             [9,99,999,1008,1017,...,8100,9000]
ṫ4 - tail from the 4th index                   [1008,1017,...,8100,9000]