Your task is to generate the nth term of the RATS sequence, where n is the input. The RATS sequence is also known as the Reverse Add Then Sort sequence. This sequence can also be found here: http://oeis.org/A004000.
test cases:
0 > 1 1 > 2 2 > 4 3 > 8 4 > 16 5 > 77 6 > 145 7 > 668 For example, the output for 5 is 77 because 16 + 61 = 77. After this the 77 is sorted.
Shortest submission wins. This is my first challenge so i hope this is not a duplicate or something.
1i"tVPU+VSU Input is a string (with single quotes) representing an integer in unary. String input is allowed by the challenge, and unary is a valid format.
1 % push number 1 to the stack i % input. Will be a string of "n" ones " % for loop: repeat n times (consumes string) t % duplicate V % convert to string P % reverse U % convert to number + % add V % convert to string S % sort U % convert to number % loop is implicitly ended % stack content is implicitly displayed Code:
$FDR+{ Explanation:
$ # Push 1 and input F # For N in range(0, input) D # Duplicate top of the stack R # Reverse top of the stack + # Add top two items { # Sort top of the stack # Implicitly print top of the stack This also works with a 0 byte program.
05AB1E by first eliminating the leading zero, and then omitting the 1, as 1E==E. Then you get just 5ABE, -2 bytes. \$\endgroup\$ uS`+vGv_GQ\1 u Q\1 reduce range(input()) on base case of "1" (string) +vG eval the string (to get a number), and add... v_G the same number, reversed first and then eval'd S` convert back to string and sort 
1ri{_sW%i+s$i}* 1 e# Push 1 as the start of the sequence. ri e# Read input and convert to integer N. { e# Run this block N times... _s e# Duplicate and convert to string. W% e# Reverse string. i+ e# Convert back to integer and add to previous value. s$ e# Convert to string and sort. i e# Convert back to integer for the next iteration. }* 
f=lambda x,n=1:x and f(x-1,int(''.join(sorted(`n+int(`n`[::-1])`))))or n Recursive function, makes use of the Python 2 shorthand for __repr__, which will break once the function reaches very large values (an L will be appended to the number's string), I'm not certain from the spec if there is a place where we can stop, but if not changing to str() only adds 6 bytes, but then it becomes slightly shorter to output as a string, at 75 bytes:
f=lambda x,n='1':x and f(x-1,''.join(sorted(str(int(n)+int(n[::-1])))))or n 1 byte saved thanks to trichoplax on this version
or in the second code block? \$\endgroup\$ Saved 1 byte thanks to @user81655
f=n=>n?+[...+[...''+(b=f(n-1))].reverse().join``+b+''].sort().join``:1 sigh JavaScript is really verbose. A lot (> 50%) of the code is just case to string + array function + join + cast to int. I've tried reduce, eval, and all sorts of stuff but this seems to be the shortest.
Try it online (All browsers work)
f=n=>n?[...+[...b=f(n-1)].reverse().join``+b+''].sort().join``:'1' if returning string allowed \$\endgroup\$ 0,1 .|-1=:0&Rr+R=o. 0,1 . § If Input is 0, unify the Output with 1 | § Else -1=:0&R § unify R with the output of this main predicate, with input = Input - 1 r+R=o. § Reverse R, add it to itself and order it, then unify with the Output. 
import Data.List g i=i:g(sort$show$read i+read(reverse i)) (g"1"!!) Usage example: (g"1"!!) 7-> "668".
It's a direct implementation of the definition: starting with "1", repeatedly append the reverse-add-sort result of the current element. The main function (g"1"!!) picks the ith element.
n->(x=1;for _=1:n x=(p=parse)(join(sort(["$(x+p(reverse("$x")))"...])))end;x) This is a lambda function that accepts an integer and returns an integer. To call it, assign it to a variable.
Ungolfed:
function f(n::Int) # Begin x at 1 x = 1 # Repeat this process n times for _ = 1:n # Add x to itself with reversed digits s = x + parse(reverse("$x")) # Refine x as this number with the digits sorted x = parse(join(sort(["$s"...]))) end # Return x after the process (will be 1 if n was 0) return x end I'm sure this can probably be golfed, as this is my first answer in Jelly/in a tacit language.
DUḌ+ðDṢḌ Performs RATS 1Ç¡ Loops D Converts integer to decimal U Reverses Ḍ Converts back to integer + Adds original and reversed ð Starts new chain D Converts back to decimal Ṣ Sorts Ḍ Back to integer again 1 Uses 1 instead of input Ḍ Uses line above ¡ For loop EDIT: Saved 1 byte, thanks to Dennis
{(1,{[~] ($_+.flip).comb.sort}...*)[$_]} # 40 ( If you want it to return an Int put a + right before [~] )
# give it a lexical name my &RATS = {…} say RATS 5; # 77 # This implementation also accepts a list of indexes # the first 10 of the sequence say RATS ^10; # (1 2 4 8 16 77 145 668 1345 6677) 
interface R{static void main(String[]a){int i,r,n=1,c=0,t=Byte.valueOf(a[0]);while(++c<=t){i=n;for(r=0;i!=0;i/=10){r=r*10+i%10;}n+=r;a[0]=n+"";char[]f=a[0].toCharArray();java.util.Arrays.sort(f);n=Integer.valueOf(new String(f));}System.out.print(n);}} interface R{ static void main(String[]args){ int input,reversed,nextValue=1,count=0,target=Byte.valueOf(args[0]); while(++count<=target){ input=nextValue; for(reversed=0;input!=0;input/=10){reversed=reversed*10+input%10;} nextValue+=reversed; args[0]=nextValue+""; char[]sortMe=args[0].toCharArray(); java.util.Arrays.sort(sortMe); nextValue=Integer.valueOf(new String(sortMe)); } System.out.print(nextValue); } } 
interface R instead of class R which is 4 bytes shorter? \$\endgroup\$ 1,`;$R≈+$S≈`n Explanation:
1,`;$R≈+$S≈`n 1 push 1 ,` `n do the following n times: ;$R≈ reverse + add $S≈ sort I can't see how I could golf it more, but I'm sure there's a way. Thanks to @LeakyNun I took the time to come down on this and golf it the proper way, I could maybe still win some bytes by using another approach.
k=0z=table for i=0,io.read()do t={}(""..k+(""..k):reverse()):gsub("%d",function(d)t[#t+1]=d end)z.sort(t)k=k<1 and 1or tonumber(z.concat(t,"")) end print(k) k=0 z=table -- z is a pointer on the table named table -- it allows me to use its functions -- while saving 4 bytes/use for i=0,io.read() -- Iterate n times for the nth element do t={} (""..a+(""..a):reverse()) -- we add k with its "reversed" value -- and convert the whole thing to a string :gsub(".",function(d) -- for each character in it, use an anonymous fucntion t[#t+1]=d end) -- which insert them in the array t z.sort(t) a=a<1 and 1 or -- if i==0, k=1 tonumber(z.concat(t,"")) -- else we concat t in a string and convert it to number end print(k) 
a=a<1 and 1or for instance). \$\endgroup\$ ;1{↔;?+o}ⁱ⁽ ;1{↔;?+o}ⁱ⁽ { }ⁱ Repeatedly apply the following, 1 starting at 1, ; ⁽ a number of times equal to the input: ↔ reverse, ;?+ add the original input, o then sort the resulting number I'm not quite clear on what this does with zero digits, but the question doesn't state any particular handling, and they probably don't appear in the sequence anyway.
f=lambda n,a=1:n<1or int(''.join(sorted(`f(n-1)+f(n-1,-1)`))[::a]) 
n=>eval("r=1;while(n--)r=+[...+[...r+''].reverse().join``+r+''].sort().join``") 82 bytes without eval:
n=>[...Array(n)].reduce(r=>+[...+[...r+''].reverse().join``+r+''].sort().join``,1) All those conversions are painful.
@edc65 I actually saved 4 bytes by switching from map to reduce this time... no doubt you'll prove me wrong again though.
for is shorter: n=>eval("for(r=1;n--)r=+[...+[...r+''].reverse().join``+r+''].sort().join``") \$\endgroup\$ n=0, even after I've fixed the syntax errors. \$\endgroup\$ Input as Integer, result is printed to the screen.
def f(n): t=1 for i in range(n):t=int("".join(sorted(str(int(str(t)[::-1])+t)))) print t This could be a lot shorter with some recursion magic I guess, but I cant wrap my head around it yet. Gonna have a fresh look later and hopefully improve this one.
def f(n): v='1' for _ in v*n:v=''.join(sorted(str(int(v)+int(v[::-1])))) print v 
Nest[FromDigits@Sort@IntegerDigits[#+IntegerReverse@#]&,1,#]& Quite simple.
$r=[1];$i++<$argn;sort($v),$r[]=join($v))$v=str_split(bcadd(strrev($e=end($r)),$e));echo$r[$argn]; n <= 39
for($r=[1];$i++<$argn;sort($v),$r[]=join($v))$v=str_split(strrev($e=end($r))+$e);echo$r[$argn]; 
int f(int n){int a=n>0?f(n-1):0,x=10,b[]=new int[x],c=a,d=0;for(;c>0;c/=x)d=d*x+c%x;for(a+=d;a>0;a/=x)b[a%x]++;for(;a<x;c=b[a++]-->0?c*x+--a:c);return n>0?c:1;} Not the longest entry! \o/
=1:sZ+\rZ 1-indexed.
=1 first term is 1 : sequence mode each term equals s sort( ) Z previous + plus \r reverse( ) Z previous 
1?(ms 1?(ms 1? # Push 1 followed by the input ( # for n in range(0, input): m # push tos + tos[::-1] (palindromize on integers doesn't make it a palindrome) +s # add and sort the result It's a shame how in order to remain competitive, one has to port other answers. Why do I say this? Because I have a nifty 21 byte function that accomplishes the same task:
@f:1|0=[1|n⨪@f;:ʼn+]s; Example usage:
6 @f; , # Single number 0 11 r °f; z # Provide a test case unit kinda display 
ʼn do? \$\endgroup\$ m work instead of :ʼn ? \$\endgroup\$ a + reversed(a) not a, reversed(a) \$\endgroup\$ 1?(ms, it says TODO is not defined \$\endgroup\$ $ofs='' for($c="1";$args[0]--){$c="$("$(+$c+"$($c[20..0])")"|% t*y|sort)"}$c PowerShell converts strings to long by default. The maximum length would always be 19 chars.
Cool challenge! -2 bytes thanks to Mazzy!
for is shorter in the Powershell. Try it online! \$\endgroup\$ c(r:String):NNI==reduce(+,[(ord(r.i)-48)*10^(#r-i) for i in 1..#r]);f(n:INT):NNI==(n<1=>1;v:=f(n-1);v:=v+c(reverse(v::String));c(sort(v::String))) test and results [RATS sequence]
(3) -> [[i, f(i)] for i in 0..20] (3) [[0,1], [1,2], [2,4], [3,8], [4,16], [5,77], [6,145], [7,668], [8,1345], [9,6677], [10,13444], [11,55778], [12,133345], [13,666677], [14,1333444], [15,5567777], [16,12333445], [17,66666677], [18,133333444], [19,556667777], [20,1233334444]]