Look-and-say sequence

From Rosetta Code
Task
Look-and-say sequence
You are encouraged to solve this task according to the task description, using any language you may know.

The   Look and say sequence   is a recursively defined sequence of numbers studied most notably by   John Conway.


The   look-and-say sequence   is also known as the   Morris Number Sequence,   after cryptographer Robert Morris,   and the puzzle   What is the next number in the sequence 1,   11,   21,   1211,   111221?   is sometimes referred to as the Cuckoo's Egg,   from a description of Morris in Clifford Stoll's book   The Cuckoo's Egg.


Sequence Definition

  • Take a decimal number
  • Look at the number, visually grouping consecutive runs of the same digit.
  • Say the number, from left to right, group by group; as how many of that digit there are - followed by the digit grouped.
This becomes the next number of the sequence.


An example:

  • Starting with the number 1,   you have one 1 which produces 11
  • Starting with 11,   you have two 1's.   I.E.:   21
  • Starting with 21,   you have one 2, then one 1.   I.E.:   (12)(11) which becomes 1211
  • Starting with 1211,   you have one 1, one 2, then two 1's.   I.E.:   (11)(12)(21) which becomes 111221


Task

Write a program to generate successive members of the look-and-say sequence.


Related tasks


See also



8080 Assembly

<lang 8080asm>bdos: equ 5 ; CP/M calls puts: equ 9

nmemb: equ 15 ; Change this to print more or fewer members

org 100h


;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ;; Generate and output members of the sequence mvi b,nmemb ; Counter

outloop: push b ; Preserve counter across calls

mvi c,puts ; Output current member lxi d,memb call bdos ; And newline mvi c,puts lxi d,newline call bdos

lxi h,memb ; Generate next member call looksay

pop b ; Restore counter dcr b ; Done yet? jnz outloop rst 0

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; ;; Given a $-terminated string under HL, representing ;; a member of the look and say sequence, generate ;; the next one in place (ish). The memory after the ;; string is assumed to be free. looksay: push h ; Save start of string on stack mov d,h ; And in DE mov e,l mvi a,'$' ; Find end of string findend: cmp m inx h jnz findend xchg ; HL=string, DE=destination push d ; Save start of new string on stack

lookchar: mvi b,0 ; Zero counter lookloop: mov a,m ; Get current character inr b ; Compare next character inx h cmp m ; While it is the same, keep going jz lookloop

mov c,a ; Keep character mvi a,'0' ; There are B amount of these characters add b stax d ; Store the amount inx d ; And in the next location mov a,c ; Store the character stax d inx d

mvi a,'$' ; Are we done? cmp m jnz lookchar ; If not, do next character stax d ; If yes, terminate new string

;; Free up memory by copying the new string to where the old ;; string began. pop d ; Start of new string pop h ; Start of old string copyloop: ldax d ; Get char from new string mov m,a ; Store char where old string was cpi '$' ; are we done yet? inx d inx h jnz copyloop ret

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; newline: db 13,10,'$' ;; This is where the string will be stored. memb: db '1$' ; First item ; Due to how CP/M loads programs, the memory after here ; is free until we hit the stack. </lang>

Ada

<lang ada>with Ada.Text_IO, Ada.Strings.Fixed; use Ada.Text_IO, Ada.Strings, Ada.Strings.Fixed;

function "+" (S : String) return String is

  Item : constant Character := S (S'First);

begin

  for Index in S'First + 1..S'Last loop
     if Item /= S (Index) then
        return Trim (Integer'Image (Index - S'First), Both) & Item & (+(S (Index..S'Last)));
     end if;
  end loop;
  return Trim (Integer'Image (S'Length), Both) & Item;

end "+";</lang> This function can be used as follows: <lang ada>Put_Line (+"1"); Put_Line (+(+"1")); Put_Line (+(+(+"1"))); Put_Line (+(+(+(+"1")))); Put_Line (+(+(+(+(+"1"))))); Put_Line (+(+(+(+(+(+"1")))))); Put_Line (+(+(+(+(+(+(+"1"))))))); Put_Line (+(+(+(+(+(+(+(+"1")))))))); Put_Line (+(+(+(+(+(+(+(+(+"1"))))))))); Put_Line (+(+(+(+(+(+(+(+(+(+"1"))))))))));</lang>

Output:
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221

ALGOL 68

Translation of: Ada
Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386

<lang algol68>OP + = (STRING s)STRING: BEGIN

  CHAR item = s[LWB s];
  STRING out;
  FOR index FROM LWB s + 1 TO UPB s DO
     IF item /= s [index] THEN
        out := whole(index - LWB s, 0) + item + (+(s [index:UPB s]));
        GO TO return out
     FI
  OD;
  out := whole (UPB s, 0) + item;
  return out: out

END # + #;

OP + = (CHAR s)STRING:

 + STRING(s);

print ((+"1", new line)); print ((+(+"1"), new line)); print ((+(+(+"1")), new line)); print ((+(+(+(+"1"))), new line)); print ((+(+(+(+(+"1")))), new line)); print ((+(+(+(+(+(+"1"))))), new line)); print ((+(+(+(+(+(+(+"1")))))), new line)); print ((+(+(+(+(+(+(+(+"1"))))))), new line)); print ((+(+(+(+(+(+(+(+(+"1")))))))), new line)); print ((+(+(+(+(+(+(+(+(+(+"1"))))))))), new line))</lang>

Output:
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221

APL

<lang apl>

 ⎕IO←0
 d←{(1↓⍵)-¯1↓⍵}
 f←{m←(0≠d ⍵),1 ⋄ ,(d ¯1,m/⍳⍴⍵),[.5](m/⍵)}
 {(f⍣⍵) ,1}¨⍳10

</lang>

AutoHotkey

<lang autohotkey>AutoExecute:

   Gui, -MinimizeBox
   Gui, Add, Edit, w500 r20 vInput, 1
   Gui, Add, Button, x155 w100 Default, &Calculate
   Gui, Add, Button, xp+110 yp wp, E&xit
   Gui, Show,, Look-and-Say sequence

Return


ButtonCalculate:

   Gui, Submit, NoHide
   GuiControl,, Input, % LookAndSay(Input)

Return


GuiClose: ButtonExit:

   ExitApp

Return


---------------------------------------------------------------------------

LookAndSay(Input) {

---------------------------------------------------------------------------
   ; credit for this function goes to AutoHotkey forum member Laslo 
   ; http://www.autohotkey.com/forum/topic44657-161.html
   ;-----------------------------------------------------------------------
   Loop, Parse, Input          ; look at every digit
       If (A_LoopField = d)    ; I've got another one! (of the same value)
           c += 1                  ; Let's count them ...
       Else {                  ; No, this one is different!
           r .= c d                ; remember what we've got so far
           c := 1                  ; It is the first one in a row
           d := A_LoopField        ; Which one is it?
       }
   Return, r c d

}</lang>

AWK

<lang awk>function lookandsay(a) {

 s = ""
 c = 1
 p = substr(a, 1, 1)
 for(i=2; i <= length(a); i++) {
   if ( p == substr(a, i, 1) ) {
     c++
   } else {
     s = s sprintf("%d%s", c, p)
     p = substr(a, i, 1)
     c = 1
   }
 }
 s = s sprintf("%d%s", c, p)
 return s

}

BEGIN {

 b = "1"
 print b
 for(k=1; k <= 10; k++) {
   b = lookandsay(b)
   print b
 }

}</lang>

BASIC256

<lang BASIC256>

  1. look and say

dim a$(2)

i = 0 # input string index

a$[i] = "1"

print a$[i]

for n=1 to 10

 j = 1 - i  # output string index
 a$[j] = ""
 k = 1
 while (k <= length(a$[i]))
   k0 = k + 1
   while ((k0 <= length(a$[i])) and (mid(a$[i], k, 1) = mid(a$[i], k0, 1)))
     k0 = k0 + 1
   end while
   a$[j] += string(k0 - k) + mid(a$[i], k, 1)
   k = k0
 end while
 i = j
 print a$[j]

next n </lang>

BBC BASIC

<lang bbcbasic> number$ = "1"

     FOR i% = 1 TO 10
       number$ = FNlooksay(number$)
       PRINT number$
     NEXT
     END
     
     DEF FNlooksay(n$)
     LOCAL i%, j%, c$, o$
     i% = 1
     REPEAT
       c$ = MID$(n$,i%,1)
       j% = i% + 1
       WHILE MID$(n$,j%,1) = c$
         j% += 1
       ENDWHILE
       o$ += STR$(j%-i%) + c$
       i% = j%
     UNTIL i% > LEN(n$)
     = o$</lang>
Output:
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221

Bracmat

In this example we use a non-linear pattern and a negation of a pattern: the end of e sequence of equal digits is (1) the end of the string or (2) the start of a sequence starting with a different digit. <lang bracmat>( 1:?number & 0:?lines & whl

 ' ( 1+!lines:~>10:?lines
   & :?say                           { This will accumulate all that has to be said after one iteration. }
   & 0:?begin
   & ( @( !number                    { Pattern matching. The '@' indicates we're looking in a string rather than a tree structure. }
        :   ?
            (   [!begin
                %@?digit
                ?
                [?end
                ( (|(%@:~!digit) ?)  { The %@ guarantees we're testing one character - not less (%) and not more (@). The ? takes the rest. }
                & !say !end+-1*!begin !digit:?say
                & !end:?begin        { When backtracking, 'begin' advances to the begin of the next sequence, or to the end of the string. }
                )
            & ~                      { fail! This forces backtracking. Backtracking stops when all begin positions have been tried. }
            )
        )
     | out$(str$!say:?number)        { After backtracking, output string and set number to string for next iteration. }
     )
   )

);</lang>

Output:
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221

C

This program will not stop until killed or running out of memory. <lang c>#include <stdio.h>

  1. include <stdlib.h>

int main() { char *a = malloc(2), *b = 0, *x, c; int cnt, len = 1;

for (sprintf(a, "1"); (b = realloc(b, len * 2 + 1)); a = b, b = x) { puts(x = a); for (len = 0, cnt = 1; (c = *a); ) { if (c == *++a) cnt++; else if (c) { len += sprintf(b + len, "%d%c", cnt, c); cnt = 1; } } }

return 0; }</lang>

C#

<lang csharp>using System; using System.Text; using System.Linq;

class Program {

   static string lookandsay(string number)
   {
       StringBuilder result = new StringBuilder();
       char repeat = number[0];
       number = number.Substring(1, number.Length-1)+" ";
       int times = 1;
     
       foreach (char actual in number)
       {
           if (actual != repeat)
           {
               result.Append(Convert.ToString(times)+repeat);
               times = 1;
               repeat = actual;
           }
           else
           {
               times += 1;
           }
       }
       return result.ToString();
   }
   static void Main(string[] args)
   {
       string num = "1"; 
       foreach (int i in Enumerable.Range(1, 10)) {
            Console.WriteLine(num);
            num = lookandsay(num);             
       }
   }

}</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

Alternate version using Regex (C#2 syntax only): <lang csharp> using System; using System.Text.RegularExpressions;

namespace RosettaCode_Cs_LookAndSay {

   public class Program
   {
       public static int Main(string[] args)
       {
           Array.Resize<string>(ref args, 2);
           string ls = args[0] ?? "1";
           int n;
           if (!int.TryParse(args[1], out n)) n = 10;
           do {
               Console.WriteLine(ls);
               if (--n <= 0) break;
               ls = say(look(ls));
           } while(true);
           return 0;
       }
       public static string[] look(string input)
       {
           int i = -1;
           return Array.FindAll(Regex.Split(input, @"((\d)\2*)"),
               delegate(string p) { ++i; i %= 3; return i == 1; }
           );
       }
       public static string say(string[] groups)
       {
           return string.Concat(
               Array.ConvertAll<string, string>(groups,
                   delegate(string p) { return string.Concat(p.Length, p[0]); }
               )
           );
       }
   }

}</lang>

Output:
(with args
1 15):
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
1321132132111213122112311311222113111221131221
11131221131211131231121113112221121321132132211331222113112211
311311222113111231131112132112311321322112111312211312111322212311322113212221

C++

<lang cpp>#include <iostream>

  1. include <sstream>
  2. include <string>

std::string lookandsay(const std::string& s) {

   std::ostringstream r;
   for (std::size_t i = 0; i != s.length();) {
       auto new_i = s.find_first_not_of(s[i], i + 1);
       if (new_i == std::string::npos)
           new_i = s.length();
       r << new_i - i << s[i];
       i = new_i;
   }
   return r.str();

}

int main() {

   std::string laf = "1";
   std::cout << laf << '\n';
   for (int i = 0; i < 10; ++i) {
       laf = lookandsay(laf);
       std::cout << laf << '\n';
   }

}</lang>

Ceylon

<lang ceylon>shared void run() {

   function lookAndSay(Integer|String input) {
       
       variable value digits = if (is Integer input) then input.string else input;
       value builder = StringBuilder();
       
       while (exists currentChar = digits.first) {
           if (exists index = digits.firstIndexWhere((char) => char != currentChar)) {
               digits = digits[index...];
               builder.append("``index````currentChar``");
           }
           else {
               builder.append("``digits.size````currentChar``");
               break;
           }
       }
       
       return builder.string;
   }
   
   variable String|Integer result = 1;
   print(result);
   for (i in 1..14) {
       result = lookAndSay(result);
       print(result);
   }

}</lang>

Clojure

No ugly int-to-string-and-back conversions.

<lang clojure>(defn digits-seq

 "Returns a seq of the digits of a number (L->R)."
 [n]
 (loop [digits (), number n]
   (if (zero? number) (seq digits)
       (recur (cons (mod number 10) digits)
              (quot number 10)))))

(defn join-digits

 "Converts a digits-seq back in to a number."
 [ds]
 (reduce (fn [n d] (+ (* 10 n) d)) ds))

(defn look-and-say [n]

 (->> n digits-seq (partition-by identity)
      (mapcat (juxt count first)) join-digits))</lang>
Output:

<lang clojure>user> (take 8 (iterate look-and-say 1)) (1 11 21 1211 111221 312211 13112221 1113213211)</lang>

Common Lisp

<lang lisp>(defun compress (array &key (test 'eql) &aux (l (length array)))

 "Compresses array by returning a list of conses each of whose car is

a number of occurrences and whose cdr is the element occurring. For instance, (compress \"abb\") produces ((1 . #\a) (2 . #\b))."

 (if (zerop l) nil
   (do* ((i 1 (1+ i))
         (segments (acons 1 (aref array 0) '())))
        ((eql i l) (nreverse segments))
     (if (funcall test (aref array i) (cdar segments))
       (incf (caar segments))
       (setf segments (acons 1 (aref array i) segments))))))

(defun next-look-and-say (number)

 (reduce #'(lambda (n pair)
             (+ (* 100 n)
                (* 10 (car pair))
                (parse-integer (string (cdr pair)))))
         (compress (princ-to-string number))
         :initial-value 0))</lang>

Example use:

<lang lisp>(next-look-and-say 9887776666) ;=> 19283746</lang>

Straight character counting: <lang lisp>(defun look-and-say (s)

  (let ((out (list (char s 0) 0)))
    (loop for x across s do

(if (char= x (first out)) (incf (second out)) (setf out (list* x 1 out))))

    (format nil "~{~a~^~}" (nreverse out))))

(loop for s = "1" then (look-and-say s)

     repeat 10
     do (write-line s))</lang>

Crystal

Translation of: Ruby

The simplest one: <lang ruby>class String

 def look_and_say
   gsub(/(.)\1*/){ |s| s.size.to_s + s[0] }
 end

end

ss = '1' 12.times { puts ss; ss = ss.to_s.look_and_say }</lang>

Translation of: Ruby from Perl

<lang ruby>def lookandsay(str)

 str.gsub(/(.)\1*/) { |s| s.size.to_s + $1 }

end

num = "1" 12.times { puts num; num = lookandsay(num) }</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211

D

Short Functional Version

<lang d>import std.stdio, std.algorithm, std.range;

enum say = (in string s) pure => s.group.map!q{ text(a[1],a[0]) }.join;

void main() {

   "1".recurrence!((t, n) => t[n - 1].say).take(8).writeln;

}</lang>

Output:
["1", "11", "21", "1211", "111221", "312211", "13112221", "1113213211"]

Fast Imperative Version

Same output. <lang d>import core.stdc.stdio, std.math, std.conv, std.algorithm, std.array;

void showLookAndSay(bool showArrays)(in uint n) nothrow {

   if (n == 0) // No sequences to generate and show.
       return;
   enum Digit : char { nil = '\0', one = '1', two = '2', thr = '3' }
   // Allocate an approximate upper bound size for the array.
   static Digit* allocBuffer(in uint m) nothrow {
       immutable len = cast(size_t)(100 + 1.05 *
                                    exp(0.269 * m + 0.2686)) + 1;
       auto a = len.uninitializedArray!(Digit[]);
       printf("Allocated %d bytes.\n", a.length * Digit.sizeof);
       return a.ptr;
   }
   // Can't be expressed in the D type system:
   // a1 and a2 are immutable pointers to mutable data.
   auto a1 = allocBuffer(n % 2 ? n : n - 1);
   auto a2 = allocBuffer(n % 2 ? n - 1 : n);
   printf("\n");
   a1[0] = Digit.one;
   size_t len1 = 1;
   a1[len1] = Digit.nil;
   foreach (immutable i; 0 .. n - 1) {
       static if (showArrays)
           printf("%2u: %s\n", i + 1, a1);
       else
           printf("%2u: n. digits: %u\n", i + 1, len1);
       auto p1 = a1,
            p2 = a2;
       S0: final switch (*p1++) with (Digit) { // Initial state.
               case nil: goto END;
               case one: goto S1;
               case two: goto S2;
               case thr: goto S3;
           }
       S1: final switch (*p1++) with (Digit) {
               case nil: *p2++ = one; *p2++ = one; goto END;
               case one: goto S11;
               case two: *p2++ = one; *p2++ = one; goto S2;
               case thr: *p2++ = one; *p2++ = one; goto S3;
           }
       S2: final switch (*p1++) with (Digit) {
               case nil: *p2++ = one; *p2++ = two; goto END;
               case one: *p2++ = one; *p2++ = two; goto S1;
               case two: goto S22;
               case thr: *p2++ = one; *p2++ = two; goto S3;
           }
       S3: final switch (*p1++) with (Digit) {
               case nil: *p2++ = one; *p2++ = thr; goto END;
               case one: *p2++ = one; *p2++ = thr; goto S1;
               case two: *p2++ = one; *p2++ = thr; goto S2;
               case thr: goto S33;
           }
       S11: final switch (*p1++) with (Digit) {
               case nil: *p2++ = two; *p2++ = one; goto END;
               case one: *p2++ = thr; *p2++ = one; goto S0;
               case two: *p2++ = two; *p2++ = one; goto S2;
               case thr: *p2++ = two; *p2++ = one; goto S3;
           }
       S22: final switch (*p1++) with (Digit) {
               case nil: *p2++ = two; *p2++ = two; goto END;
               case one: *p2++ = two; *p2++ = two; goto S1;
               case two: *p2++ = thr; *p2++ = two; goto S0;
               case thr: *p2++ = two; *p2++ = two; goto S3;
           }
       S33: final switch (*p1++) with (Digit) {
               case nil: *p2++ = two; *p2++ = thr; goto END;
               case one: *p2++ = two; *p2++ = thr; goto S1;
               case two: *p2++ = two; *p2++ = thr; goto S2;
               case thr: *p2++ = thr; *p2++ = thr; goto S0;
           }
       END:
           immutable len2 = p2 - a2;
           a2[len2] = Digit.nil;
           a1.swap(a2);
           len1 = len2;
   }
   static if (showArrays)
       printf("%2u: %s\n", n, a1);
   else
       printf("%2u: n. digits: %u\n", n, len1);

}

void main(in string[] args) {

   immutable n = (args.length == 2) ? args[1].to!uint : 10;
   n.showLookAndSay!true;

}</lang>

Output:
Allocated 116 bytes.
Allocated 121 bytes.

 1: 1
 2: 11
 3: 21
 4: 1211
 5: 111221
 6: 312211
 7: 13112221
 8: 1113213211
 9: 31131211131221
10: 13211311123113112211

With: <lang d>70.showLookAndSay!false;</lang>

Output:
Allocated 158045069 bytes.
Allocated 206826462 bytes.

 1: n. digits: 1
 2: n. digits: 2
 3: n. digits: 2
 4: n. digits: 4
 5: n. digits: 6
...
60: n. digits: 12680852
61: n. digits: 16530884
62: n. digits: 21549544
63: n. digits: 28091184
64: n. digits: 36619162
65: n. digits: 47736936
66: n. digits: 62226614
67: n. digits: 81117366
68: n. digits: 105745224
69: n. digits: 137842560
70: n. digits: 179691598

Using the LDC2 compiler with n=70 the run-time is about 3.74 seconds.

Intermediate Version

This mostly imperative version is intermediate in both speed and code size: <lang d>void main(in string[] args) {

   import std.stdio, std.conv, std.algorithm, std.array, std.string;
   immutable n = (args.length == 2) ? args[1].to!uint : 10;
   if (n == 0)
       return;
   auto seq = ['1'];
   writefln("%2d: n. digits: %d", 1, seq.length);
   foreach (immutable i; 2 .. n + 1) {
       Appender!(typeof(seq)) result;
       foreach (const digit, const count; seq.representation.group) {
           result ~= "123"[count - 1];
           result ~= digit;
       }
       seq = result.data;
       writefln("%2d: n. digits: %d", i, seq.length);
   }

}</lang> The output is the same as the second version.

If you modify the first program to print only the lengths of the strings (with a .map!(s => s.length)), compiling with LDC2 the run-times of the three versions with n=55 are about 31.1, 0.10 and 0.23 seconds.

More Direct Version

Translated and modified from C code by Reddit user "skeeto": http://www.reddit.com/r/dailyprogrammer/comments/2ggy30/9152014_challenge180_easy_looknsay/

Using ideas from: http://www.njohnston.ca/2010/10/a-derivation-of-conways-degree-71-look-and-say-polynomial/

This recursive version is able to generate very large sequences in a short time without memory for the intermediate sequence (and with stack space proportional to the sequence order).

<lang d>import core.stdc.stdio, std.conv;

// On Windows this uses the printf from the Microsoft C runtime, // that doesn't handle real type and some of the C99 format // specifiers, but it's faster for bulk printing. version (LDC) version (Windows) extern(C) nothrow @nogc int printf(const char*, ...);

// http://www.njohnston.ca/2010/10/a-derivation-of-conways-degree-71-look-and-say-polynomial/ struct Sequence {

   string seq;
   uint[6] next;

}

immutable Sequence[93] sequences = [

   {"", []},
   {"1112", [63]},
   {"1112133", [64, 62]},
   {"111213322112", [65]},
   {"111213322113", [66]},
   {"1113", [68]},
   {"11131", [69]},
   {"111311222112", [84, 55]},
   {"111312", [70]},
   {"11131221", [71]},
   {"1113122112", [76]},
   {"1113122113", [77]},
   {"11131221131112", [82]},
   {"111312211312", [78]},
   {"11131221131211", [79]},
   {"111312211312113211", [80]},
   {"111312211312113221133211322112211213322112", [81, 29, 91]},
   {"111312211312113221133211322112211213322113", [81, 29, 90]},
   {"11131221131211322113322112", [81, 30]},
   {"11131221133112", [75, 29, 92]},
   {"1113122113322113111221131221", [75, 32]},
   {"11131221222112", [72]},
   {"111312212221121123222112", [73]},
   {"111312212221121123222113", [74]},
   {"11132", [83]},
   {"1113222", [86]},
   {"1113222112", [87]},
   {"1113222113", [88]},
   {"11133112", [89, 92]},
   {"12", [1]},
   {"123222112", [3]},
   {"123222113", [4]},
   {"12322211331222113112211", [2, 61, 29, 85]},
   {"13", [5]},
   {"131112", [28]},
   {"13112221133211322112211213322112", [24, 33, 61, 29, 91]},
   {"13112221133211322112211213322113", [24, 33, 61, 29, 90]},
   {"13122112", [7]},
   {"132", [8]},
   {"13211", [9]},
   {"132112", [10]},
   {"1321122112", [21]},
   {"132112211213322112", [22]},
   {"132112211213322113", [23]},
   {"132113", [11]},
   {"1321131112", [19]},
   {"13211312", [12]},
   {"1321132", [13]},
   {"13211321", [14]},
   {"132113212221", [15]},
   {"13211321222113222112", [18]},
   {"1321132122211322212221121123222112", [16]},
   {"1321132122211322212221121123222113", [17]},
   {"13211322211312113211", [20]},
   {"1321133112", [6, 61, 29, 92]},
   {"1322112", [26]},
   {"1322113", [27]},
   {"13221133112", [25, 29, 92]},
   {"1322113312211", [25, 29, 67]},
   {"132211331222113112211", [25, 29, 85]},
   {"13221133122211332", [25, 29, 68, 61, 29, 89]},
   {"22", [61]},
   {"3", [33]},
   {"3112", [40]},
   {"3112112", [41]},
   {"31121123222112", [42]},
   {"31121123222113", [43]},
   {"3112221", [38, 39]},
   {"3113", [44]},
   {"311311", [48]},
   {"31131112", [54]},
   {"3113112211", [49]},
   {"3113112211322112", [50]},
   {"3113112211322112211213322112", [51]},
   {"3113112211322112211213322113", [52]},
   {"311311222", [47, 38]},
   {"311311222112", [47, 55]},
   {"311311222113", [47, 56]},
   {"3113112221131112", [47, 57]},
   {"311311222113111221", [47, 58]},
   {"311311222113111221131221", [47, 59]},
   {"31131122211311122113222", [47, 60]},
   {"3113112221133112", [47, 33, 61, 29, 92]},
   {"311312", [45]},
   {"31132", [46]},
   {"311322113212221", [53]},
   {"311332", [38, 29, 89]},
   {"3113322112", [38, 30]},
   {"3113322113", [38, 31]},
   {"312", [34]},
   {"312211322212221121123222113", [36]},
   {"312211322212221121123222112", [35]},
   {"32112", [37]}

];

void evolve(in uint seq, in uint n) nothrow @nogc {

   if (n <= 0) {
       printf(sequences[seq].seq.ptr);
   } else {
       foreach (immutable next; sequences[seq].next) {
           if (next == 0)
               break;
           evolve(next, n - 1);
       }
   }

}

void main(in string[] args) {

   immutable uint n = (args.length != 2) ? 10 : args[1].to!uint;
   immutable base = 8;
   immutable string[base] results = ["", "1", "11", "21", "1211",
                                     "111221", "312211", "13112221"];
   if (n < base) {
       printf("%s\n", results[n].ptr);
       return;
   }
   evolve(24, n - base);
   evolve(39, n - base);
   '\n'.putchar;

}</lang>

E

<lang e>def lookAndSayNext(number :int) {

 var seen := null
 var count := 0
 var result := ""
 def put() {
   if (seen != null) {
     result += count.toString(10) + E.toString(seen)
   }
 }
 for ch in number.toString(10) {
   if (ch != seen) {
     put()
     seen := ch
     count := 0
   }
   count += 1
 }
 put()
 return __makeInt(result, 10)

}

var number := 1 for _ in 1..20 {

 println(number)
 number := lookAndSayNext(number)

}</lang>

EchoLisp

<lang scheme> (lib 'math) ;; for (number->list) = explode function (lib 'list) ;; (group)

(define (next L) (for/fold (acc null) ((g (group L))) (append acc (list (length g) (first g)))))


(define (task n starter) (for/fold (L (number->list starter)) ((i n)) (writeln (list->string L)) (next L)))

</lang>

Output:

<lang scheme> (task 10 1) 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 </lang>

Elixir

<lang elixir>defmodule LookAndSay do

 def next(n) do
   Enum.chunk_by(to_char_list(n), &(&1))
   |> Enum.map(fn cl=[h|_] -> Enum.concat(to_char_list(length cl), [h]) end)
   |> Enum.concat
   |> List.to_integer
 end
 
 def sequence_from(n) do
   Stream.iterate n, &(next/1)
 end
 
 def main([start_str|_]) do
   {start_val,_} = Integer.parse(start_str)
   IO.inspect sequence_from(start_val) |> Enum.take 9
 end
 
 def main([]) do
   main(["1"])
 end

end

LookAndSay.main(System.argv)</lang>

Output:
[1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, 31131211131221]

Regex version: <lang elixir>defmodule RC do

 def look_and_say(n) do
   Regex.replace(~r/(.)\1*/, to_string(n), fn x,y -> [to_string(String.length(x)),y] end)
   |> String.to_integer
 end

end

IO.inspect Enum.reduce(1..9, [1], fn _,acc -> [RC.look_and_say(hd(acc)) | acc] end) |> Enum.reverse</lang>

Output:
[1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, 31131211131221,
 13211311123113112211]

Erlang

<lang erlang>-module(str). -export([look_and_say/1, look_and_say/2]).

%% converts a single number look_and_say([H|T]) -> lists:reverse(look_and_say(T,H,1,"")).

%% converts and accumulates as a loop look_and_say(_, 0) -> []; look_and_say(Start, Times) when Times > 0 ->

   [Start | look_and_say(look_and_say(Start), Times-1)].

%% does the actual conversion for a number look_and_say([], Current, N, Acc) ->

   [Current, $0+N | Acc];

look_and_say([H|T], H, N, Acc) ->

   look_and_say(T, H, N+1, Acc);

look_and_say([H|T], Current, N, Acc) ->

   look_and_say(T, H, 1, [Current, $0+N | Acc]).</lang>
Output:
1> c(str).
{ok,str}
2> str:look_and_say("1").
"11"
3> str:look_and_say("111221").
"312211"
4> str:look_and_say("1",10).
["1","11","21","1211","111221","312211","13112221",
 "1113213211","31131211131221","13211311123113112211"]

ERRE

<lang> PROGRAM LOOK

PROCEDURE LOOK_AND_SAY(N$->N$)

     LOCAL I%,J%,C$,O$
     I%=1
     REPEAT
       C$=MID$(N$,I%,1)
       J%=I%+1
       WHILE MID$(N$,J%,1)=C$ DO
         J%+=1
       END WHILE
       O$+=MID$(STR$(J%-I%),2)+C$
       I%=J%
     UNTIL I%>LEN(N$)
     N$=O$

END PROCEDURE

BEGIN

     NUMBER$="1"
     FOR I%=1 TO 10 DO
       LOOK_AND_SAY(NUMBER$->NUMBER$)
       PRINT(NUMBER$)
     END FOR

END PROGRAM </lang>

11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221

F#

Library functions somehow missing in F# out of the box (but present in haskell) <lang fsharp> let rec brk p lst =

 match lst with
 | [] -> (lst, lst)
 | x::xs ->
   if p x
   then ([], lst)
   else
     let (ys, zs) = brk p xs
     (x::ys, zs)

let span p lst = brk (not << p) lst

let rec groupBy eq lst =

 match lst with
 | [] ->  []
 | x::xs ->
   let (ys,zs) = span (eq x) xs

(x::ys)::groupBy eq zs

let group lst : list<list<'a>> when 'a : equality = groupBy (=) lst </lang>

Implementation <lang fsharp> let lookAndSay =

 let describe (xs: char list) =
   List.append (List.ofSeq <| (List.length xs).ToString()) [List.head xs]
 let next xs = List.collect describe (group xs)
 let toStr xs = String (Array.ofList xs)
 Seq.map toStr <| Seq.unfold (fun xs -> Some (xs, next xs)) ['1']

let getNthLookAndSay n = Seq.nth n lookAndSay

Seq.take 10 lookAndSay </lang>

Factor

<lang factor>: (look-and-say) ( str -- )

   unclip-slice swap [ 1 ] 2dip [
       2dup = [ drop [ 1 + ] dip ] [
           [ [ number>string % ] dip , 1 ] dip
       ] if
   ] each [ number>string % ] [ , ] bi* ;
look-and-say ( str -- str' ) [ (look-and-say) ] "" make ;

"1" 10 [ dup print look-and-say ] times print</lang>

Forth

<lang forth>create buf1 256 allot create buf2 256 allot buf1 value src buf2 value dest

s" 1" src place

append-run ( digit run -- )
 dest count +
 tuck c!  1+ c!
 dest c@ 2 + dest c! ;
next-look-and-say
 0 dest c!
 src 1+ c@  [char] 0  ( digit run )
 src count bounds do
   over i c@ =
   if   1+
   else append-run  i c@ [char] 1
   then
 loop
 append-run
 src dest to src to dest ;
look-and-say ( n -- )
 0 do next-look-and-say  cr src count type loop ;

10 look-and-say</lang>

Fortran

<lang fortran>module LookAndSay

 implicit none

contains

 subroutine look_and_say(in, out)
   character(len=*), intent(in) :: in
   character(len=*), intent(out) :: out
   integer :: i, c
   character(len=1) :: x
   character(len=2) :: d
   out = ""
   c = 1
   x = in(1:1)
   do i = 2, len(trim(in))
      if ( x == in(i:i) ) then
         c = c + 1
      else
         write(d, "(I2)") c
         out = trim(out) // trim(adjustl(d)) // trim(x)
         c = 1
         x = in(i:i)
      end if
   end do
   write(d, "(I2)") c
   out = trim(out) // trim(adjustl(d)) // trim(x)
 end subroutine look_and_say

end module LookAndSay</lang>

<lang fortran>program LookAndSayTest

 use LookAndSay
 implicit none

 integer :: i
 character(len=200) :: t, r
 t = "1"
 print *,trim(t)
 call look_and_say(t, r)
 print *, trim(r)
 do i = 1, 10
    call look_and_say(r, t)
    r = t
    print *, trim(r)
 end do

end program LookAndSayTest</lang>

FreeBASIC

Translation of: BASIC256

<lang freebasic> Dim As Integer n, j, k, k0, r Dim As String X(2) Dim As Integer i = 0 ' índice de cadena de entrada X(0) = "1"

Input "Indica cuantas repeticiones: ", r Print Chr(10) & "Secuencia:"

Print X(i) For n = 1 To r-1

   j = 1 - i  ' índice de cadena de salida
   X(j) = ""
   k = 1
   While k <= Len(X(i))
       k0 = k + 1
       While ((k0 <= Len(X(i))) And (Mid(X(i), k, 1) = Mid(X(i), k0, 1)))
           k0 += 1
       Wend
       X(j) += Str(k0 - k) + Mid(X(i), k, 1)
       k = k0
   Wend
   i = j
   Print X(j)

Next n End </lang>

Output:
Indica cuantas repeticiones: 10

Secuencia:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

Frink

<lang frink> LookAndSay = "12211123" println["Starting Value: " + LookAndSay]

/* Anonymous function

  There was an issue processing a standard function with MapList. However,
  the anonymous function formatting works without issue with the MapList function.
  • /

LASStr = { |LaS|

  length[LaS@0] + length[LaS@1] + LaS@0
  /*
     The returned results from the Regex are divided between a distinct matching
     character and any following identical characters. For example, a string
     of 2222 would be returned from this function as [2,222].
     The function adds the length of both elements (1 and 3 in the example
     above) and returns that value with the matching character.
     i.e. Length of 1st element = 1, Length of 2nd element = 3, value of 1st element = 2
     1 + 3 = 4 & value 2. Returned result is "42" i.e. "Four 2s."
  */

}

// Calculate the next 10 Look and Say Sequence Values

for i = 1 to 10 {

  LookAndSayReg = LookAndSay =~ %r/(\d)(\1{0,})/g
  LookAndSay = join["",mapList[LASStr,LookAndSayReg]]
  println["$i - $LookAndSay"]

} </lang>

Output:
Starting Value: 1
1 - 11
2 - 21
3 - 1211
4 - 111221
5 - 312211
6 - 13112221
7 - 1113213211
8 - 31131211131221
9 - 13211311123113112211
10 - 11131221133112132113212221

Gambas

Code is modified from the [PureBasic] example

Click this link to run this code <lang gambas>Public Sub Main() Dim i, j, cnt As Integer Dim txt$, curr$, result$ As String

txt$ = "1211" i = 1

Print "Sequence: " & txt$ & " = ";

 Repeat
   j = 1
   result$ = ""
     Repeat
       curr$ = Mid(txt$, j, 1)
       cnt = 0
         Repeat
           Inc cnt 
           Inc j 
         Until Mid(txt$, j, 1) <> curr$
       result$ &= Str(cnt) & curr$
     Until j > Len(txt$)
   Print result$
   txt$ = result$
   Dec i 
 Until i <= 0

End</lang> Output:

Sequence: 1211 = 111221

GAP

<lang gap>LookAndSay := function(s)

 local c, r, cur, ncur, v;
 v := "123";
 r := "";
 cur := 0;
 ncur := 0;
 for c in s do
   if c = cur then
     ncur := ncur + 1;
   else
     if ncur > 0 then
       Add(r, v[ncur]);
       Add(r, cur);
     fi;
     cur := c;
     ncur := 1;
   fi;
 od;
 Add(r, v[ncur]);
 Add(r, cur);
 return r;

end;

LookAndSay("1"); # "11" LookAndSay(last); # "21" LookAndSay(last); # "1211" LookAndSay(last); # "111221" LookAndSay(last); # "312211" LookAndSay(last); # "13112221" LookAndSay(last); # "1113213211" LookAndSay(last); # "31131211131221" LookAndSay(last); # "13211311123113112211" LookAndSay(last); # "11131221133112132113212221" LookAndSay(last); # "3113112221232112111312211312113211" LookAndSay(last); # "1321132132111213122112311311222113111221131221" LookAndSay(last); # "11131221131211131231121113112221121321132132211331222113112211" LookAndSay(last); # "311311222113111231131112132112311321322112111312211312111322212311322113212221"</lang>

Go

<lang go>package main

import (

   "fmt"
   "strconv"

)

func lss(s string) (r string) {

   c := s[0]
   nc := 1
   for i := 1; i < len(s); i++ {
       d := s[i]
       if d == c {
           nc++
           continue
       }
       r += strconv.Itoa(nc) + string(c)
       c = d
       nc = 1
   }
   return r + strconv.Itoa(nc) + string(c)

}

func main() {

   s := "1"
   fmt.Println(s)
   for i := 0; i < 8; i++ {
       s = lss(s)
       fmt.Println(s)
   }

}</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221

Groovy

<lang groovy>def lookAndSay(sequence) {

   def encoded = new StringBuilder()
   (sequence.toString() =~ /(([0-9])\2*)/).each { matcher ->
       encoded.append(matcher[1].size()).append(matcher[2])
   }
   encoded.toString()

}</lang> Test Code <lang groovy>def sequence = "1" (1..12).each {

   println sequence
   sequence = lookAndSay(sequence)

}</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211

Haskell

<lang haskell>import Control.Monad (liftM2) import Data.List (group)

-- this function is composed out of many functions; data flows from the bottom up lookAndSay :: Integer -> Integer lookAndSay = read -- convert digits to integer

          . concatMap                              -- concatenate for each run,
              (liftM2 (++) (show . length)         --    the length of it
                           (take 1))               --    and an example member
          . group                                  -- collect runs of the same digit
          . show                                   -- convert integer to digits

-- less comments lookAndSay2 :: Integer -> Integer lookAndSay2 = read . concatMap (liftM2 (++) (show . length)

                                           (take 1))
           . group . show


-- same thing with more variable names lookAndSay3 :: Integer -> Integer lookAndSay3 n = read (concatMap describe (group (show n)))

 where describe run = show (length run) ++ take 1 run

main = mapM_ print (iterate lookAndSay 1) -- display sequence until interrupted</lang>

Haxe

<lang haxe>using Std;

class Main {

static function main() { var test = "1"; for (i in 0...11) { Sys.println(test); test = lookAndSay(test); } }

static function lookAndSay(s:String) { if (s == null || s == "") return "";

var results = ""; var repeat = s.charAt(0); var amount = 1; for (i in 1...s.length) { var actual = s.charAt(i); if (actual != repeat) { results += amount.string(); results += repeat; repeat = actual; amount = 0; } amount++; } results += amount.string(); results += repeat;

return results; } }</lang>

Icon and Unicon

<lang Icon>procedure main() every 1 to 10 do

  write(n := nextlooknsayseq(\n | 1))

end

procedure nextlooknsayseq(n) #: return next element in look and say sequence n2 := "" n ? until pos(0) do {

  i := tab(any(&digits)) | fail  # or fail if not digits
  move(-1) 
  n2 ||:= *tab(many(i)) || i     # accumulate count+digit
  }

return n2 end</lang>

Output:
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221

J

Solution:

<lang j>las=: ,@((# , {.);.1~ 1 , 2 ~:/\ ])&.(10x&#.inv)@]^:(1+i.@[)</lang>

Example: <lang j> 10 las 1 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211 11131221133112132113212221</lang>

Note the result is an actual numeric sequence (cf. the textual solutions given in other languages).

Java

Translation of: C#
Works with: Java version 1.5+

<lang java5>public static String lookandsay(String number){ StringBuilder result= new StringBuilder();

char repeat= number.charAt(0); number= number.substring(1) + " "; int times= 1;

for(char actual: number.toCharArray()){ if(actual != repeat){ result.append(times + "" + repeat); times= 1; repeat= actual; }else{ times+= 1; } } return result.toString(); }</lang> Testing: <lang java5>public static void main(String[] args){ String num = "1";

for (int i=1;i<=10;i++) { System.out.println(num); num = lookandsay(num); } }</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

JavaScript

Translation of: Perl

<lang javascript>function lookandsay(str) {

   return str.replace(/(.)\1*/g, function(seq, p1){return seq.length.toString() + p1})

}

var num = "1"; for (var i = 10; i > 0; i--) {

   alert(num);
   num = lookandsay(num);

}</lang>

Without RegExp

<lang javascript>function lookSay(digits) {

   var result = ,
       chars = (digits + ' ').split(),
       lastChar = chars[0],
       times = 0;
   
   chars.forEach(function(nextChar) {
       if (nextChar === lastChar) {
           times++;
       }
       else {
           result += (times + ) + lastChar;
           lastChar = nextChar;
           times = 1;
       }
   });
   
   return result;

}

(function output(seed, iterations) {

   for (var i = 0; i < iterations; i++) {
       console.log(seed);
       seed = lookSay(seed);
   }

})("1", 10);</lang>

jq

Works with: jq version 1.4

<lang jq>def look_and_say:

 def head(c; n): if .[n:n+1] == c then head(c; n+1) else n end;
 tostring
 | if length == 0 then ""
   else head(.[0:1]; 1) as $len
     | .[0:$len] as $head
     | ($len | tostring) + $head[0:1] + (.[$len:] | look_and_say)
   end ;
  1. look and say n times

def look_and_say(n):

 if n == 0 then empty
 else look_and_say as $lns
      | $lns, ($lns|look_and_say(n-1))
 end ;</lang>

Example

1 | look_and_say(10)
Output:
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221

Julia

Works with: Julia version 1.1

<lang julia>function lookandsay(s::String)

   rst = IOBuffer()
   c = 1
   for i in 1:length(s)
       if i != length(s) && s[i] == s[i+1]
           c += 1
       else
           print(rst, c, s[i])
           c = 1
       end
   end
   String(take!(rst))

end


function lookandsayseq(n::Integer)

   rst = Vector{String}(undef, n)
   rst[1] = "1"
   for i in 2:n
       rst[i] = lookandsay(rst[i-1])
   end
   rst

end

println(lookandsayseq(10))</lang>

Output:
String["1", "11", "21", "1211", "111221", "312211", "13112221", "1113213211", "31131211131221", "13211311123113112211"]

K

<lang k> las: {x{0$,//$(#:'n),'*:'n:(&1,~=':x)_ x:0$'$x}\1}

 las 8

1 11 21 1211 111221 312211 13112221 1113213211 31131211131221</lang>

Kotlin

<lang scala>// version 1.0.6

fun lookAndSay(s: String): String {

   val sb = StringBuilder()
   var digit = s[0]
   var count = 1
   for (i in 1 until s.length) {
       if (s[i] == digit)
           count++
       else {
           sb.append("$count$digit")
           digit = s[i]
           count = 1
       }
   }
   return sb.append("$count$digit").toString()

}

fun main(args: Array<String>) {

   var las = "1"
   for (i in 1..15) {
       println(las)
       las = lookAndSay(las)
   }

}</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
1321132132111213122112311311222113111221131221
11131221131211131231121113112221121321132132211331222113112211
311311222113111231131112132112311321322112111312211312111322212311322113212221

Lasso

The Look-and-say sequence is a recursive RLE, so the solution can leverage the same method as used for RLE. <lang Lasso>define rle(str::string)::string => { local(orig = #str->values->asCopy,newi=array, newc=array, compiled=string) while(#orig->size) => { if(not #newi->size) => { #newi->insert(1) #newc->insert(#orig->first) #orig->remove(1) else if(#orig->first == #newc->last) => { #newi->get(#newi->size) += 1 else #newi->insert(1) #newc->insert(#orig->first) } #orig->remove(1) } } loop(#newi->size) => { #compiled->append(#newi->get(loop_count)+#newc->get(loop_count)) } return #compiled } define las(n::integer,run::integer) => { local(str = #n->asString) loop(#run) => { #str = rle(#str) } return #str } loop(15) => {^ las(1,loop_count) + '\r' ^}</lang>

Output:
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
1321132132111213122112311311222113111221131221
11131221131211131231121113112221121321132132211331222113112211
311311222113111231131112132112311321322112111312211312111322212311322113212221
132113213221133112132113311211131221121321131211132221123113112221131112311332111213211322211312113211

LiveCode

This function takes a string and returns the next Look-And-Say iteration of it: <lang Lua>function lookAndSay S

  put 0 into C
  put char 1 of S into lastChar
  repeat with i = 2 to length(S)
     add 1 to C
     if char i of S is lastChar then next repeat
     put C & lastChar after R
     put 0 into C
     put char i of S into lastChar
  end repeat
  return R & C + 1 & lastChar

end lookAndSay

on demoLookAndSay

  put 1 into x
  repeat 10
     put x & cr after message
     put lookAndSay(x) into x
  end repeat
  put x after message

end demoLookAndSay</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221

<lang logo>to look.and.say.loop :in :run :c :out

 if empty? :in [output (word :out :run :c)]
 if equal? first :in :c [output look.and.say.loop bf :in :run+1 :c :out]
 output look.and.say.loop bf :in 1 first :in (word :out :run :c)

end to look.and.say :in

 if empty? :in [output :in]
 output look.and.say.loop bf :in 1 first :in "||

end

show cascade 10 [print ? look.and.say ?] 1</lang>

Lua

<lang lua>--returns an iterator over the first n copies of the look-and-say sequence function lookandsayseq(n)

 local t = {1}
 return function()
   local ret = {}
   for i, v in ipairs(t) do
     if t[i-1] and v == t[i-1] then
       ret[#ret - 1] = ret[#ret - 1] + 1
     else
       ret[#ret + 1] = 1
       ret[#ret + 1] = v
     end
   end
   t = ret
   n = n - 1
   if n > 0 then return table.concat(ret) end
 end

end for i in lookandsayseq(10) do print(i) end</lang>

Alternative solution, using LPeg: <lang lua>require "lpeg" local P, C, Cf, Cc = lpeg.P, lpeg.C, lpeg.Cf, lpeg.Cc lookandsay = Cf(Cc"" * C(P"1"^1 + P"2"^1 + P"3"^1)^1, function (a, b) return a .. #b .. string.sub(b,1,1) end) t = "1" for i = 1, 10 do

 print(t)
 t = lookandsay:match(t)

end</lang>

M4

Using regular expressions:

Translation of: Perl

<lang M4>divert(-1) define(`for',

  `ifelse($#,0,``$0,
  `ifelse(eval($2<=$3),1,
  `pushdef(`$1',$2)$4`'popdef(`$1')$0(`$1',incr($2),$3,`$4')')')')

define(`las',

  `patsubst(`$1',`\(\(.\)\2*\)',`len(\1)`'\2')')


define(`v',1) divert for(`x',1,10,

  `v

define(`v',las(v))')dnl v</lang>

Mathematica

Custom Functions: <lang Mathematica>RunLengthEncode[x_List]:=(Through[{First,Length}[#]]&)/@Split[x]

LookAndSay[n_,d_:1]:=NestList[Flatten[Reverse/@RunLengthEncode[#]]&,{d},n-1]</lang>

If second argument is omitted the sequence is started with 1. Second argument is supposed to be a digits from 0 to 9. If however a larger number is supplied it will be seen as 1 number, not multiple digits. However if one wants to start with a 2 or more digit number, one could reverse the sequence to go back to a single digit start. First example will create the first 13 numbers of the sequence starting with 1, the next example starts with 7:

<lang Mathematica>FromDigits /@ LookAndSay[13] // Column FromDigits /@ LookAndSay[13, 7] // Column</lang>

gives back:

1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
1321132132111213122112311311222113111221131221

7
17
1117
3117
132117
1113122117
311311222117
13211321322117
1113122113121113222117
31131122211311123113322117
132113213221133112132123222117
11131221131211132221232112111312111213322117
31131122211311123113321112131221123113111231121123222117 

Maxima

<lang maxima>collect(a) := block(

  [n: length(a), b: [ ], x: a[1], m: 1],
  for i from 2 thru n do
     (if a[i] = x then m: m + 1 else (b: endcons([x, m], b), x: a[i], m: 1)),
  b: endcons([x, m], b)

)$

look_and_say(s) := apply(sconcat, map(lambda([p], sconcat(string(p[2]), p[1])), collect(charlist(s))))$

block([s: "1"], for i from 1 thru 10 do (disp(s), s: look_and_say(s))); /* "1"

  "11"
  "21"
  "1211"
  "111221"
  "312211"
  "13112221"
  "1113213211"
  "31131211131221"
  "13211311123113112211" */</lang>

MAXScript

<lang maxscript>fn lookAndSay num = (

   local result = ""
   num += " "
   local current = num[1]
   local numReps = 1
   for digit in 2 to num.count do
   (
       if num[digit] != current then
       (
           result += (numReps as string) + current
           numReps = 1
           current = num[digit]
       )
       else
       (
           numReps += 1
       )
   )
   result

)

local num = "1"

for i in 1 to 10 do (

   print num
   num = lookAndSay num

)</lang>

Metafont

<lang metafont>vardef lookandsay(expr s) = string r; r := ""; if string s:

 i := 0;
 forever: exitif not (i < length(s));
   c := i+1;
   forever: exitif ( (substring(c,c+1) of s) <> (substring(i,i+1) of s) );
     c := c + 1;
   endfor
   r := r & decimal (c-i) & substring(i,i+1) of s;
   i := c;
 endfor

fi r enddef;

string p; p := "1"; for el := 1 upto 10:

 message p;
 p := lookandsay(p);

endfor

end</lang>

MiniScript

<lang MiniScript>// Look and Say Sequence repeats = function(digit, string) count = 0 for c in string if c != digit then break count = count + 1 end for return str(count) end function

numbers = "1" print numbers for i in range(1,10) // warning, loop size > 15 gets long numbers very quickly number = "" position = 0 while position < numbers.len repeatCount = repeats(numbers[position], numbers[position:]) number = number + repeatCount + numbers[position] position = position + repeatCount.val end while print number numbers = number end for</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221

Nim

<lang nim>proc NextInLookAndSaySequence (current: string): string =

 assert(len(current) > 0)
 Result = ""
 var ch = current[0]
 var count = 1
 for i in countup(1, len(current)-1):
   if current[i] != ch:
     Result &= $count & ch
     ch = current[i]
     count = 1
   else:
     count += 1
 Result &= $count & ch

proc LookAndSay (n = 10) =

 var next = "1"
 for i in countup(1, n):
   next = NextInLookAndSaySequence(next)
   echo next
 

LookAndSay() </lang>

Objective-C

<lang objc>#import <Foundation/Foundation.h>

-(NSString*)lookAndSay:(NSString *)word{

   if (!word) {
       return nil;
   }
   NSMutableString *result = [NSMutableString new];
   
   char repeat = [word characterAtIndex:0];
   int times = 1;
   word = [NSString stringWithFormat:@"%@ ",[word substringFromIndex:1] ];
   
   for (NSInteger index = 0; index < word.length; index++) {
       char actual = [word characterAtIndex:index];
       if (actual != repeat) {
           [result appendFormat:@"%d%c", times, repeat];
           times = 1;
           repeat = actual;
       } else {
           times ++;
       }
   }
   return [result copy];

}

- (void)applicationDidFinishLaunching:(NSNotification *)aNotification {

   NSString *num = @"1";
   for (int i=1;i<=10;i++) {
       NSLog(@"%@", num);
       num = [self lookAndSay:num];
   }

} </lang>

OCaml

Functional

This function computes a see-and-say sequence from the previous one: <lang ocaml>let rec seeAndSay = function

 | [], nys -> List.rev nys
 | x::xs, [] -> seeAndSay(xs, [x; 1])
 | x::xs, y::n::nys when x=y -> seeAndSay(xs, y::1+n::nys)
 | x::xs, nys -> seeAndSay(xs, x::1::nys)</lang>

It can be used like this: <lang ocaml>> let gen n =

   let xs = Array.create n [1] in
   for i=1 to n-1 do
     xs.(i) <- seeAndSay(xs.(i-1), [])
   done;
   xs;;

val gen : int -> int list array = <fun>

> gen 10;; - : int list array =

 [|[1]; [1; 1]; [2; 1]; [1; 2; 1; 1]; [1; 1; 1; 2; 2; 1]; [3; 1; 2; 2; 1; 1];
   [1; 3; 1; 1; 2; 2; 2; 1]; [1; 1; 1; 3; 2; 1; 3; 2; 1; 1];
   [3; 1; 1; 3; 1; 2; 1; 1; 1; 3; 1; 2; 2; 1];
   [1; 3; 2; 1; 1; 3; 1; 1; 1; 2; 3; 1; 1; 3; 1; 1; 2; 2; 1; 1]|]</lang>

With regular expressions in the Str library

<lang ocaml>#load "str.cma";;

let lookandsay =

 Str.global_substitute (Str.regexp "\\(.\\)\\1*")
                       (fun s -> string_of_int (String.length (Str.matched_string s)) ^
                                 Str.matched_group 1 s)

let () =

 let num = ref "1" in
 print_endline !num;
 for i = 1 to 10 do
   num := lookandsay !num;
   print_endline !num;
 done</lang>

With regular expressions in the Pcre library

<lang ocaml>open Pcre

let lookandsay str =

 let rex = regexp "(.)\\1*" in
 let subs = exec_all ~rex str in
 let ar = Array.map (fun sub -> get_substring sub 0) subs in
 let ar = Array.map (fun s -> String.length s, s.[0]) ar in
 let ar = Array.map (fun (n,c) -> (string_of_int n) ^ (String.make 1 c)) ar in
 let res = String.concat "" (Array.to_list ar) in
 (res)

let () =

 let num = ref(string_of_int 1) in
 for i = 1 to 10 do
   num := lookandsay !num;
   print_endline !num;
 done</lang>

run this example with 'ocaml -I +pcre pcre.cma script.ml'

Imperative

<lang ocaml>(* see http://oeis.org/A005150 *)

let look_and_say s = let n = String.length s and buf = Buffer.create 0 and prev = ref s.[0] and count = ref 0 in let append () = Buffer.add_char buf (char_of_int (48 + !count));

               Buffer.add_char buf !prev in

String.iter (fun c ->

  if c = !prev then incr count else
  begin
     append ();
     prev := c;
     count := 1
  end

) s; append (); Buffer.contents buf;;

(* what about length of successive strings ? *) let iter f a n = let rec aux r n v = if n = 0

                   then List.rev(r::v)
                   else aux (f r) (n - 1) (r::v) in

aux a n [];;

let las = iter look_and_say "1";;

(* the first sixty terms *)

List.map (String.length) (las 59);; (*

  [1; 2; 2; 4; 6; 6; 8; 10; 14; 20; 26; 34; 46; 62; 78; 102; 134; 176; 226;
   302; 408; 528; 678; 904; 1182; 1540; 2012; 2606; 3410; 4462; 5808; 7586;
   9898; 12884; 16774; 21890; 28528; 37158; 48410; 63138; 82350; 107312;
   139984; 182376; 237746; 310036; 403966; 526646; 686646; 894810; 1166642;
   1520986; 1982710; 2584304; 3369156; 4391702; 5724486; 7462860; 9727930;
   12680852]
  • )

(* see http://oeis.org/A005341 *)</lang>

Oforth

<lang Oforth>import: mapping

lookAndSay ( n -- )
  [ 1 ] #[ dup .cr group map( [#size, #first] ) expand ] times( n ) ;</lang>
Output:
for n = 10
[1]
[1, 1]
[2, 1]
[1, 2, 1, 1]
[1, 1, 1, 2, 2, 1]
[3, 1, 2, 2, 1, 1]
[1, 3, 1, 1, 2, 2, 2, 1]
[1, 1, 1, 3, 2, 1, 3, 2, 1, 1]
[3, 1, 1, 3, 1, 2, 1, 1, 1, 3, 1, 2, 2, 1]
[1, 3, 2, 1, 1, 3, 1, 1, 1, 2, 3, 1, 1, 3, 1, 1, 2, 2, 1, 1]

Oz

<lang oz>declare

 %% e.g. "21" -> "1211"
 fun {LookAndSayString S}
    for DigitGroup in {Group S} append:Add do
       {Add {Int.toString {Length DigitGroup}}}
       {Add [DigitGroup.1]}
    end
 end
 %% lazy sequence of integers starting with N
 fun {LookAndSay N}
    fun lazy {Loop S}
       {String.toInt S}|{Loop {LookAndSayString S}}
    end
 in
    {Loop {Int.toString N}}
 end
 %% like Haskell's "group"
 fun {Group Xs}
    case Xs of nil then nil
    [] X|Xr then

Ys Zs

       {List.takeDropWhile Xr fun {$ W} W==X end ?Ys ?Zs}
    in
       (X|Ys) | {Group Zs}
    end
 end

in

 {ForAll {List.take {LookAndSay 1} 10} Show}</lang>

PARI/GP

<lang parigp>step(n)={

 my(v=eval(Vec(Str(n))),cur=v[1],ct=1,out="");
 v=concat(v,99);
 for(i=2,#v,
   if(v[i]==cur,
     ct++
   ,
     out=Str(out,ct,cur);
     cur=v[i];
     ct=1
   )
 );
 eval(out)

}; n=1;for(i=1,20,print(n);n=step(n))</lang>

Pascal

Works with: Free_Pascal
Library: SysUtils

<lang pascal>program LookAndSayDemo(input, output);

uses

 SysUtils;

function LookAndSay (s: string): string;

 var
   item: char;
   index: integer;
   count: integer;
 begin
   LookAndSay := ;
   item := s[1];
   count := 1;
   for index:= 2 to length(s) do
     if item = s[index] then
       inc(count)
     else
     begin

LookAndSay := LookAndSay + intTostr(count) + item;

       item := s[index];

count := 1;

     end;
     LookAndSay := LookAndSay + intTostr(count) + item;
 end;

var

 number: string;

begin

 writeln('Press RETURN to continue and ^C to stop.');
 number := '1';
 while not eof(input) do
 begin
  write(number);
  readln;
  number := LookAndSay(number);
 end;

end.</lang>

Output:
% ./LookAndSay 
Press RETURN to continue and ^C to stop.

1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
1321132132111213122112311311222113111221131221
11131221131211131231121113112221121321132132211331222113112211^C

Even faster imperative Version Improvement: setlength of result,no inttoStr and using pChar But the Code Alignment is very important.

Works with: Free_Pascal
Library: SysUtils

<lang pascal> program LookAndSayDemo(input, output); {$IFDEF FPC}

 {$Mode Delphi}  // using result
 {$optimization ON}

// i3-4330 3.5 Ghz // {$CodeAlign proc=16,loop=8} //2,6 secs

 {$CodeAlign proc=16,loop=1}  //1,6 secs so much faster ???

{$ENDIF}

uses

 SysUtils;

const

 cntChar : array[1..9] of char =
          ('1','2','3','4','5','6','7','8','9');

function LookAndSay2 (const s: string): string; //using pChar for result var

 source,
 destin : pChar;
 len,
 idxFrom,
 idxTo :  integer;
 cnt: integer;
 item: char;

begin

 idxFrom := length(s);
 source := @s[1];
 //adjust length of result
 len := round(length(s)* 1.306+10);
 setlength(result,len);
 destin := @result[1];
 dec(destin);
 idxto := 1;
 item := source^;
 inc(source);
 cnt := 1;
 for idxFrom := idxFrom downto 2 do
 begin
   if item <> source^ then
   begin
     destin[idxTo]  := cntChar[cnt];
     destin[idxTo+1]:= item;
     item := source^;
     cnt := 1;
     inc(idxto,2);
   end
   else
     inc(cnt);
   inc(source);
 end;
 destin[idxTo] := cntChar[cnt];
 destin[idxTo+1]:= item;
 setlength(result,idxto+1);

end;

var

 number: string;
 l1,l2,
 i : integer;

begin

 number := '1';
 writeln(number);
 writeln(1:4,length(number):16,1/1:10:6);
 For i := 2 to 70 do
 begin
   l1 := length(number);
   number := LookAndSay2(number);
   l2 := length(number);
   IF i <10 then
     writeln(number);
   writeln(i:4,length(number):16,l2/l1:10:6);
 end;

end.</lang>

Output:
1
   1               1  1.000000
11
   2               2  2.000000
21
   3               2  1.000000
1211
   4               4  2.000000
111221
   5               6  1.500000
312211
   6               6  1.000000
13112221
   7               8  1.333333
1113213211
   8              10  1.250000
31131211131221
   9              14  1.400000
  10              20  1.428571
  11              26  1.300000
  12              34  1.307692
  13              46  1.352941
  14              62  1.347826
  15              78  1.258065
  16             102  1.307692
........
  67        81117366  1.303580
  68       105745224  1.303608
  69       137842560  1.303535
  70       179691598  1.303600

real	0m1.639s
user	0m1.593s
sys	0m0.043s

Perl

<lang perl>sub lookandsay {

 my $str = shift;
 $str =~ s/((.)\2*)/length($1) . $2/ge;
 return $str;

}

my $num = "1"; foreach (1..10) {

 print "$num\n";
 $num = lookandsay($num);

}</lang>

Using string as a cyclic buffer: <lang perl>for (local $_ = "1\n"; s/((.)\2*)//s;) { print $1; $_ .= ($1 ne "\n" and length($1)).$2 }</lang>

Phix

<lang Phix>function lookandsay(string s) string res = "" integer p = s[1], c = 1

   for i=2 to length(s) do
       if p=s[i] then
           c += 1
       else
           res &= sprintf("%d%s",{c,p})
           p = s[i]
           c = 1
       end if
   end for
   res &= sprintf("%d%s",{c,p})
   return res

end function

string s = "1" ?s for i=1 to 10 do

   s = lookandsay(s)
   ?s

end for</lang>

Output:
"1"
"11"
"21"
"1211"
"111221"
"312211"
"13112221"
"1113213211"
"31131211131221"
"13211311123113112211"
"11131221133112132113212221"

PHP

<lang php><?php

function lookAndSay($str) {

return preg_replace_callback('#(.)\1*#', function($matches) {

return strlen($matches[0]).$matches[1]; }, $str); }

$num = "1";

foreach(range(1,10) as $i) {

echo $num."
"; $num = lookAndSay($num); }

?></lang>

PicoLisp

<lang PicoLisp>(de las (Lst)

  (make
     (while Lst
        (let (N 1  C)
           (while (= (setq C (pop 'Lst)) (car Lst))
              (inc 'N) )
           (link N C) ) ) ) )</lang>

Usage: <lang PicoLisp>: (las (1)) -> (1 1)

(las @)

-> (2 1)

(las @)

-> (1 2 1 1)

(las @)

-> (1 1 1 2 2 1)

(las @)

-> (3 1 2 2 1 1)

(las @)

-> (1 3 1 1 2 2 2 1)

(las @)

-> (1 1 1 3 2 1 3 2 1 1)

(las @)

-> (3 1 1 3 1 2 1 1 1 3 1 2 2 1)</lang>

PowerBASIC

This uses the RLEncode function from the PowerBASIC Run-length encoding entry. <lang powerbasic>FUNCTION RLEncode (i AS STRING) AS STRING

   DIM tmp1 AS STRING, tmp2 AS STRING, outP AS STRING
   DIM Loop0 AS LONG, count AS LONG
   FOR Loop0 = 1 TO LEN(i)
       tmp1 = MID$(i, Loop0, 1)
       IF tmp1 <> tmp2 THEN
           IF count > 1 THEN
               outP = outP & TRIM$(STR$(count)) & tmp2
               tmp2 = tmp1
               count = 1
           ELSEIF 0 = count THEN
               tmp2 = tmp1
               count = 1
           ELSE
               outP = outP & "1" & tmp2
               tmp2 = tmp1
           END IF
       ELSE
           INCR count
       END IF
   NEXT
   outP = outP & TRIM$(STR$(count)) & tmp2
   FUNCTION = outP

END FUNCTION

FUNCTION lookAndSay(BYVAL count AS LONG) AS STRING

   DIM iii AS STRING, tmp AS STRING
   IF count > 1 THEN
       iii = lookAndSay(count - 1)
   ELSEIF count < 2 THEN
       iii = "1"
   END IF
   tmp = RLEncode(iii)
   lookAndSay = tmp

END FUNCTION

FUNCTION PBMAIN () AS LONG

   DIM v AS LONG
   v = VAL(INPUTBOX$("Enter a number."))
   MSGBOX lookAndSay(v)

END FUNCTION</lang>

PowerShell

<lang powershell>function Get-LookAndSay ($n = 1) {

   $re = [regex] '(.)\1*'
   $ret = ""
   foreach ($m in $re.Matches($n)) {
       $ret += [string] $m.Length + $m.Value[0]
   }
   return $ret

}

function Get-MultipleLookAndSay ($n) {

   if ($n -eq 0) {
       return @()
   } else {
       $a = 1
       $a
       for ($i = 1; $i -lt $n; $i++) {
           $a = Get-LookAndSay $a
           $a
       }
   }

}</lang>

Output:
PS> Get-MultipleLookAndSay 8
1
11
21
1211
111221
312211
13112221
1113213211

Prolog

Works with SWI-Prolog.

<lang Prolog>look_and_say(L) :- maplist(write, L), nl, encode(L, L1), look_and_say(L1).

% This code is almost identical to the code of "run-length-encoding" encode(In, Out) :- packList(In, R1), append(R1,Out).


% use of library clpfd allows packList(?In, ?Out) to works % in both ways In --> Out and In <-- Out.

- use_module(library(clpfd)).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % % ?- packList([a,a,a,b,c,c,c,d,d,e], L). % L = [[3,a],[1,b],[3,c],[2,d],[1,e]] . % ?- packList(R, [[3,a],[1,b],[3,c],[2,d],[1,e]]). % R = [a,a,a,b,c,c,c,d,d,e] . % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% packList([],[]).

packList([X],1,X) :- !.


packList([X|Rest],[XRun|Packed]):-

   run(X,Rest, XRun,RRest),
   packList(RRest,Packed).


run(Var,[],[1,Var],[]).

run(Var,[Var|LRest],[N1, Var],RRest):-

   N #> 0,
   N1 #= N + 1,
   run(Var,LRest,[N, Var],RRest).


run(Var,[Other|RRest], [1,Var],[Other|RRest]):-

   dif(Var,Other).

</lang>

Output:
 ?- look_and_say([1]).
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
..........................

Pure

<lang pure>using system;

// Remove the trailing "L" from the string representation of bigints. __show__ x::bigint = init (str x);

say x = val $ strcat $ map (sprintf "%d%s") $ look $ chars $ str x with

 look [] = [];
 look xs@(x:_) = (#takewhile (==x) xs,x) : look (dropwhile (==x) xs);

end;

iteraten 5 say 1; // [1,11,21,1211,111221]

// This prints the entire sequence, press Ctrl-C to abort. do (puts.str) (iterate say 1);</lang>

PureBasic

<lang PureBasic>If OpenConsole()

 Define i, j, cnt, txt$, curr$, result$
 Print("Enter start sequence: "): txt$=Input()
 Print("How many repetitions: "): i=Val(Input())
 ;
 PrintN(#CRLF$+"Sequence:"+#CRLF$+txt$)
 Repeat
   j=1
   result$=""
   Repeat
     curr$=Mid(txt$,j,1)
     cnt=0
     Repeat
       cnt+1
       j+1
     Until Mid(txt$,j,1)<>curr$
     result$+Str(cnt)+curr$
   Until j>Len(txt$)    
   PrintN(result$)
   txt$=result$
   i-1
 Until i<=0
 ;
 PrintN(#CRLF$+"Press ENTER to exit."): Input()
 CloseConsole()

EndIf</lang>

Output:
 Enter start sequence: 1
 How many repetitions: 7
 
 Sequence:
 1
 11
 21
 1211
 111221
 312211
 13112221
 1113213211

Python

Translation of: C sharp – C#

<lang python>def lookandsay(number):

   result = ""
   repeat = number[0]
   number = number[1:]+" "
   times = 1
   for actual in number:
       if actual != repeat:
           result += str(times)+repeat
           times = 1
           repeat = actual
       else:
           times += 1
   return result

num = "1"

for i in range(10):

   print num
   num = lookandsay(num)</lang>

Functional

Works with: Python version 2.4+

<lang python>>>> from itertools import groupby >>> def lookandsay(number): return .join( str(len(list(g))) + k for k,g in groupby(number) )

>>> numberstring='1' >>> for i in range(10): print numberstring numberstring = lookandsay(numberstring)</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

As a generator
<lang python>>>> from itertools import groupby, islice >>> >>> def lookandsay(number='1'): while True: yield number number = .join( str(len(list(g))) + k for k,g in groupby(number) )


>>> print('\n'.join(islice(lookandsay(), 10))) 1 11 21 1211 111221 312211 13112221 1113213211 31131211131221 13211311123113112211</lang>

Using regular expressions

Translation of: Perl

<lang python>import re

def lookandsay(str):

   return re.sub(r'(.)\1*', lambda m: str(len(m.group(0))) + m.group(1), str)

num = "1" for i in range(10):

   print num
   num = lookandsay(num)</lang>

Q

<lang q>las:{{raze string[count@'x],'@'[;0]x:where[differ x]_x}\[x;1#"1"]} las 8</lang>

Output:
,"1"
"11"
"21"
"1211"
"111221"
"312211"
"13112221"
"1113213211"
"31131211131221"

R

Returning the value as an integer limits how long the sequence can get, so the option for integer or character return values are provided. <lang R>look.and.say <- function(x, return.an.int=FALSE) {

  #convert number to character vector
  xstr <- unlist(strsplit(as.character(x), ""))
  #get run length encoding   
  rlex <- rle(xstr)
  #form new string
  odds <- as.character(rlex$lengths)
  evens <- rlex$values
  newstr <- as.vector(rbind(odds, evens))
  #collapse to scalar
  newstr <- paste(newstr, collapse="")
  #convert to number, if desired
  if(return.an.int) as.integer(newstr) else newstr

}</lang> Example usage: <lang R>x <- 1 for(i in 1:10) {

  x <- look.and.say(x)
  print(x)

}</lang>

Racket

<lang Racket>

  1. lang racket

(define (encode str)

 (regexp-replace* #px"(.)\\1*" str (lambda (m c) (~a (string-length m) c))))

(define (look-and-say-sequence n)

 (reverse (for/fold ([r '("1")]) ([n n]) (cons (encode (car r)) r))))

(for-each displayln (look-and-say-sequence 10)) </lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221

Raku

(formerly Perl 6)

Works with: rakudo version 2018.03

In Raku it is natural to avoid explicit loops; rather we use the sequence operator to define a lazy infinite sequence. We'll print the first 15 values here.

<lang perl6>.say for ('1', *.subst(/(.)$0*/, { .chars ~ .[0] }, :g) ... *)[^15];</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
1321132132111213122112311311222113111221131221
11131221131211131231121113112221121321132132211331222113112211
311311222113111231131112132112311321322112111312211312111322212311322113212221

REXX

Programming note:   this version works with any string   (a null is assumed, which causes   1   to be used).

If a negative number is specified (the number of iterations to be used for the calculations), only the length of
the number (or character string) is shown.

simple version

<lang rexx>/*REXX program displays the sequence (and/or lengths) for the look and say series.*/ parse arg N ! . /*obtain optional arguments from the CL*/ if N== | N=="," then N=20 /*Not specified? Then use the deault. */ if !== | !=="," then !=1 /* " " " " " " */

    do j=1  for abs(N)                          /*repeat a number of times to show NUMS*/
    if j\==1  then != $lookAndSay(!)            /*invoke function to calculate next #. */
    if N<0    then say 'length['j"]:" length(!) /*Also,  display the sequence's length.*/
              else say '['j"]:"      !          /*display the number to the terminal.  */
    end   /*j*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ $lookAndSay: procedure; parse arg x,,$ /*obtain the (passed) argument {X}. */ fin = '0'x /*use unique character to end scanning.*/ x=x || fin /*append the FIN character to string.*/

            do k=1  by 0                        /*now,  process the given sequence.    */
               y=  substr(x, k, 1)              /*pick off one character to examine.   */
            if y== fin  then return $           /*if we're at the end, then we're done.*/
            _= verify(x, y, , k)  - k           /*see how many characters we have of Y.*/
            $= $ || _ || y                      /*build the  "look and say"  sequence. */
            k= k  + _                           /*now, point to the next character.    */
            end   /*k*/</lang>
output   when using the default input values of:     20   1
[1]: 1
[2]: 11
[3]: 21
[4]: 1211
[5]: 111221
[6]: 312211
[7]: 13112221
[8]: 1113213211
[9]: 31131211131221
[10]: 13211311123113112211
[11]: 11131221133112132113212221
[12]: 3113112221232112111312211312113211
[13]: 1321132132111213122112311311222113111221131221
[14]: 11131221131211131231121113112221121321132132211331222113112211
[15]: 311311222113111231131112132112311321322112111312211312111322212311322113212221
[16]: 132113213221133112132113311211131221121321131211132221123113112221131112311332111213211322211312113211
[17]: 11131221131211132221232112111312212321123113112221121113122113111231133221121321132132211331121321231231121113122113322113111221131221
[18]: 31131122211311123113321112131221123113112211121312211213211321322112311311222113311213212322211211131221131211132221232112111312111213111213211231131122212322211331222113112211
[19]: 1321132132211331121321231231121113112221121321132122311211131122211211131221131211132221121321132132212321121113121112133221123113112221131112311332111213122112311311123112111331121113122112132113213211121332212311322113212221
[20]: 11131221131211132221232112111312111213111213211231132132211211131221131211221321123113213221123113112221131112311332211211131221131211132211121312211231131112311211232221121321132132211331121321231231121113112221121321133112132112312321123113112221121113122113121113123112112322111213211322211312113211
output   when using the input values of:     17   ggg
[1]: ggg
[2]: 3g
[3]: 131g
[4]: 1113111g
[5]: 3113311g
[6]: 132123211g
[7]: 11131211121312211g
[8]: 31131112311211131122211g
[9]: 132113311213211231132132211g
[10]: 11131221232112111312211213211312111322211g
[11]: 3113112211121312211231131122211211131221131112311332211g
[12]: 1321132122311211131122211213211321322112311311222113311213212322211g
[13]: 1113122113121122132112311321322112111312211312111322211213211321322123211211131211121332211g
[14]: 31131122211311122122111312211213211312111322211231131122211311123113322112111312211312111322111213122112311311123112112322211g
[15]: 132113213221133122112231131122211211131221131112311332211213211321322113311213212322211231131122211311123113223112111311222112132113311213211221121332211g
[16]: 11131221131211132221231122212213211321322112311311222113311213212322211211131221131211132221232112111312111213322112132113213221133112132113221321123113213221121113122123211211131221222112112322211g
[17]: 31131122211311123113321112132132112211131221131211132221121321132132212321121113121112133221123113112221131112311332111213122112311311123112112322211211131221131211132221232112111312211322111312211213211312111322211231131122111213122112311311221132211221121332211g
output   when using the input value of:     -60
length[1]: 1
length[2]: 2
length[3]: 2
length[4]: 4
length[5]: 6
length[6]: 6
length[7]: 8
length[8]: 10
length[9]: 14
length[10]: 20
length[11]: 26
length[12]: 34
length[13]: 46
length[14]: 62
length[15]: 78
length[16]: 102
length[17]: 134
length[18]: 176
length[19]: 226
length[20]: 302
length[21]: 408
length[22]: 528
length[23]: 678
length[24]: 904
length[25]: 1182
length[26]: 1540
length[27]: 2012
length[28]: 2606
length[29]: 3410
length[30]: 4462
length[31]: 5808
length[32]: 7586
length[33]: 9898
length[34]: 12884
length[35]: 16774
length[36]: 21890
length[37]: 28528
length[38]: 37158
length[39]: 48410
length[40]: 63138
length[41]: 82350
length[42]: 107312
length[43]: 139984
length[44]: 182376
length[45]: 237746
length[46]: 310036
length[47]: 403966
length[48]: 526646
length[49]: 686646
length[50]: 894810
length[51]: 1166642
length[52]: 1520986
length[53]: 1982710
length[54]: 2584304
length[55]: 3369156
length[56]: 4391702
length[57]: 5724486
length[58]: 7462860
length[59]: 9727930
length[60]: 12680852

faster version

This version appends the generated parts of the sequence, and after it gets to a certain size (chunkSize),
it appends the sequence generated (so far) to the primary sequence, and starts with a null sequence.
This avoids appending a small character string to a growing larger and larger character string. <lang rexx>/*REXX program displays the sequence (and/or lengths) for the look and say series.*/ parse arg N ! . /*obtain optional arguments from the CL*/ if N== | N=="," then N=20 /*Not specified? Then use the default.*/ if !== | !=="," then !=1 /* " " " " " " */

                                                /* [↑]  !:   starting char for the seq.*/
    do j=1  for abs(N)                          /*repeat a number of times to show NUMS*/
    if j\==1  then != $lookAndSay(!)            /*invoke function to calculate next #. */
    if N<0    then say 'length['j"]:" length(!) /*Also,  display the sequence's length.*/
              else say '['j"]:"      !          /*display the number to the terminal.  */
     end   /*j*/

exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ $lookAndSay: procedure; parse arg x,,$ ! /*obtain the (passed) argument {X}. */

            chSize= 1000                        /*define a sensible chunk size.        */
            fin = '0'x                          /*use unique character to end scanning.*/
            x=x || fin                          /*append the  FIN  character to string.*/
              do k=1  by 0                      /*now,  process the given sequence.    */
                 y=  substr(x, k, 1)            /*pick off one character to examine.   */
              if y== fin  then return $         /*if we're at the end, then we're done.*/
              _= verify(x, y, , k)  - k         /*see how many characters we have of Y.*/
              $= $ || _ || y                    /*build the  "look and say"  sequence. */
              k= k  + _                         /*now, point to the next character.    */
              if length($)<chSize  then iterate /*Less than chunkSize?  Then keep going*/
              != ! || $                         /*append   $   to the  !  string.      */
              $=                                /*now,  start   $   from scratch.      */
              chSize= chSize + 100              /*bump the  chunkSize (length) counter.*/
              end   /*k*/
           return ! || $                        /*return the ! string plus the $ string*/</lang>
output   is identical to the 1st REXX version   (the simple version).


Ring

<lang ring> number = "1" for nr = 1 to 10

   number = lookSay(number)
   see number + nl

next

func lookSay n

    i = 0 j = 0 c="" o=""
    i = 1
    while i <= len(n)
          c = substr(n,i,1)
          j = i + 1
          while substr(n,j,1) = c
                j += 1
          end
          o += string(j-i) + c
          i = j
     end
     return o 

</lang>

Ruby

The simplest one: <lang ruby> class String

 def look_and_say
   gsub(/(.)\1*/){|s| s.size.to_s + s[0]}
 end

end

ss = '1' 12.times {puts ss; ss = ss.look_and_say} </lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
Translation of: Perl

<lang ruby>def lookandsay(str)

 str.gsub(/(.)\1*/) {$&.length.to_s + $1}

end

num = "1" 10.times do

 puts num
 num = lookandsay(num)

end</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

Using Enumerable#chunk <lang ruby>def lookandsay(str)

 str.chars.chunk{|c| c}.map{|c,x| [x.size, c]}.join

end

puts num = "1" 9.times do

 puts num = lookandsay(num)

end</lang> The output is the same above.

Without regular expression:

<lang ruby># Adding clusterization (http://apidock.com/rails/Enumerable/group_by) module Enumerable

 # clumps adjacent elements together
 # >> [2,2,2,3,3,4,2,2,1].cluster
 # => [[2, 2, 2], [3, 3], [4], [2, 2], [1]]
 def cluster
   cluster = []
   each do |element|
     if cluster.last && cluster.last.last == element
       cluster.last << element
     else
       cluster << [element]
     end
   end
   cluster
 end

end</lang>

Using Array#cluster defined above:

<lang ruby>def print_sequence(input_sequence, seq=10)

 return unless seq > 0
 puts input_sequence.join
 result_array = input_sequence.cluster.map do |cluster|
   [cluster.count, cluster.first]
 end
 print_sequence(result_array.flatten, seq-1)

end

print_sequence([1])</lang> The output is the same above.

Rust

<lang rust>fn next_sequence(in_seq: &[i8]) -> Vec<i8> {

   assert!(!in_seq.is_empty());
   let mut result = Vec::new();
   let mut current_number = in_seq[0];
   let mut current_runlength = 1;
   for i in &in_seq[1..] {
       if current_number == *i {
           current_runlength += 1;
       } else {
           result.push(current_runlength);
           result.push(current_number);
           current_runlength = 1;
           current_number = *i;
       }
   }
   result.push(current_runlength);
   result.push(current_number);
   result

}

fn main() {

   let mut seq = vec![1];
   for i in 0..10 {
       println!("Sequence {}: {:?}", i, seq);
       seq = next_sequence(&seq);
   }

}</lang>

Output:
Sequence 0: [1]
Sequence 1: [1, 1]
Sequence 2: [2, 1]
Sequence 3: [1, 2, 1, 1]
Sequence 4: [1, 1, 1, 2, 2, 1]
Sequence 5: [3, 1, 2, 2, 1, 1]
Sequence 6: [1, 3, 1, 1, 2, 2, 2, 1]
Sequence 7: [1, 1, 1, 3, 2, 1, 3, 2, 1, 1]
Sequence 8: [3, 1, 1, 3, 1, 2, 1, 1, 1, 3, 1, 2, 2, 1]
Sequence 9: [1, 3, 2, 1, 1, 3, 1, 1, 1, 2, 3, 1, 1, 3, 1, 1, 2, 2, 1, 1]

Scala

Recursive

<lang Scala>import scala.annotation.tailrec

object LookAndSay extends App {

 loop(10, "1")
 @tailrec
 private def loop(n: Int, num: String): Unit = {
   println(num)
   if (n <= 0) () else loop(n - 1, lookandsay(num))
 }
 private def lookandsay(number: String): String = {
   val result = new StringBuilder
   @tailrec
   def loop(numberString: String, repeat: Char, times: Int): String =
     if (numberString.isEmpty) result.toString()
     else if (numberString.head != repeat) {
       result.append(times).append(repeat)
       loop(numberString.tail, numberString.head, 1)
     } else loop(numberString.tail, numberString.head, times + 1)
   loop(number.tail + " ", number.head, 1)
 }

}</lang>

Output:
See it running in your browser by (JavaScript, non JVM) or by Scastie (JVM).

using Iterator

Library: Scala
<lang scala>def lookAndSay(seed: BigInt) = {
 val s = seed.toString
 ( 1 until s.size).foldLeft((1, s(0), new StringBuilder)) {
   case ((len, c, sb), index) if c != s(index) => sb.append(len); sb.append(c); (1, s(index), sb)
   case ((len, c, sb), _) => (len + 1, c, sb)
 } match {
   case (len, c, sb) => sb.append(len); sb.append(c); BigInt(sb.toString)
 }

}

def lookAndSayIterator(seed: BigInt) = Iterator.iterate(seed)(lookAndSay)</lang>

using Stream

<lang Scala>object Main extends App {

 def lookAndSay(previous: List[BigInt]): Stream[List[BigInt]] = {
   def next(num: List[BigInt]): List[BigInt] = num match {
     case Nil => Nil
     case head :: Nil => 1 :: head :: Nil
     case head :: tail =>
       val size = (num takeWhile (_ == head)).size
       List(BigInt(size), head) ::: next(num.drop(size))
   }
   val x = next(previous)
   x #:: lookAndSay(x)
 }
 (lookAndSay(1 :: Nil) take 10).foreach(s => println(s.mkString("")))

}</lang>

Seed7

<lang seed7>$ include "seed7_05.s7i";

const func string: lookAndSay (in integer: level, in string: stri) is func

 result
   var string: lookAndSay is "";
 local
   var integer: index is 2;
 begin
   if level = 1 then
     if stri <> "" then
       while index <= length(stri) and stri[index] = stri[1] do
         incr(index);
       end while;
       lookAndSay := str(pred(index)) & stri[1 len 1] & lookAndSay(level, stri[index ..]);
     end if;
   else
     lookAndSay := lookAndSay(1, lookAndSay(pred(level), stri));
   end if;
 end func;

const proc: main is func

 local
   var integer: level is 0;
 begin
   for level range 1 to 14 do
     writeln(lookAndSay(level, "1"));
   end for;
 end func;</lang>
Output:
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
1321132132111213122112311311222113111221131221
11131221131211131231121113112221121321132132211331222113112211
311311222113111231131112132112311321322112111312211312111322212311322113212221

Sidef

Translation of: Perl

<lang ruby>func lookandsay(str) {

   str.gsub(/((.)\2*)/, {|a,b| a.len.to_s + b });

}

var num = "1"; {

 say num;
 num = lookandsay(num);

} * 10;</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

Smalltalk

Works with: GNU Smalltalk

<lang smalltalk>String extend [

 lookAndSay [ |anElement nextElement counter coll newColl|
    coll := (self asOrderedCollection).
    newColl := OrderedCollection new.
    counter := 0.
    anElement := (coll first).
    [ coll size > 0 ]
    whileTrue: [
       nextElement := coll removeFirst.

( anElement == nextElement ) ifTrue: [

          counter := counter + 1.
       ] ifFalse: [

newColl add: (counter displayString). newColl add: (anElement asString). anElement := nextElement. counter := 1.

       ]
    ].
    newColl add: (counter displayString).
    newColl add: (anElement asString).
    ^(newColl join)
 ]

].

|r| r := '1'. 10 timesRepeat: [

 r displayNl.
 r := r lookAndSay.

]</lang>

Works with: Pharo

<lang smalltalk>String compile:

 'lookAndSay |anElement nextElement counter coll newColl|
    coll := (self asOrderedCollection).
    newColl := OrderedCollection new.
    counter := 0.
    anElement := (coll first).
    [ coll size > 0 ]
    whileTrue: [
       nextElement := coll removeFirst.

( anElement == nextElement ) ifTrue: [

          counter := counter + 1.
       ] ifFalse: [

newColl add: (counter displayString). newColl add: (anElement asString). anElement := nextElement. counter := 1.

       ]
    ].
    newColl add: (counter displayString).
    newColl add: (anElement asString).
    ^(' join: newColl)'
 classified: 'toys'.

result := OrderedCollection new. r := '1'. result add: r. result addAll: ((1 to: 10) collect: [ :i |

 r := r lookAndSay.

]). result. </lang>

Output:

   an OrderedCollection('1' '11' '21' '1211' '111221' '312211' '13112221' '1113213211' '31131211131221' '13211311123113112211' '11131221133112132113212221')

SNOBOL4

Works with: Macro Spitbol
Works with: Snobol4+
Works with: CSnobol

The look-and-say sequence is an iterative run-length string encoding. So looksay( ) is just a wrapper around the Run-length Encoding task. This is by far the easiest solution.

<lang SNOBOL4>* # Encode RLE

       define('rle(str)c,n') :(rle_end)

rle str len(1) . c :f(return)

       str span(c) @n =
       rle = rle n c :(rle)

rle_end

  • # First m members of sequence with seed n
       define('looksay(n,m)') :(looksay_end)

looksay output = n; m = gt(m,1) m - 1 :f(return)

       n = rle(n) :(looksay)

looksay_end

  • Test and display
       looksay(1,10)

end</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

SQL

<lang sql>DROP VIEW delta; CREATE VIEW delta AS

   SELECT sequence1.v AS x,
          (sequence1.v<>sequence2.v)*sequence1.c AS v,
          sequence1.c AS c
     FROM sequence AS sequence1,
          sequence AS sequence2
    WHERE sequence1.c = sequence2.c+1;

DROP VIEW rle0; CREATE VIEW rle0 AS

   SELECT delta2.x AS x,
          SUM(delta2.v) AS v,
          delta2.c AS c
     FROM delta AS delta1,
          delta as delta2
    WHERE delta1.c >= delta2.c
 GROUP BY delta1.c;

DROP VIEW rle1; CREATE VIEW rle1 AS

   SELECT sum(x)/x AS a,
          x AS b,
          c AS c
     FROM rle0
 GROUP BY v;

DROP VIEW rle2; CREATE VIEW rle2 AS

   SELECT a as v, 1 as o, 2*c+0 as c FROM rle1 UNION
   SELECT b as v, 1 as o, 2*c+1 as c FROM rle1;

DROP VIEW normed; CREATE VIEW normed AS

   SELECT r1.v as v, SUM(r2.o) as c
     FROM rle2 AS r1,
          rle2 AS r2
    WHERE r1.c >= r2.c
 GROUP BY r1.c;

DROP TABLE rle; CREATE TABLE rle(v int, c int); INSERT INTO rle SELECT * FROM normed ORDER BY c;

DELETE FROM sequence; INSERT INTO sequence VALUES(-1,0); INSERT INTO sequence SELECT * FROM rle;</lang>

Usage:

% sqlite3 
SQLite version 3.4.0
Enter ".help" for instructions
sqlite> CREATE TABLE sequence(v int, c int);
sqlite> INSERT INTO sequence VALUES(-1,0);
sqlite> INSERT INTO sequence VALUES(1,1);
sqlite> SELECT * FROM sequence;
-1|0
1|1
sqlite> .read look.sql
sqlite> SELECT * FROM sequence;
-1|0
1|1
1|2
sqlite> .read look.sql
sqlite> SELECT * FROM sequence;
-1|0
2|1
1|2
sqlite> .read look.sql
sqlite> SELECT * FROM sequence;
-1|0
1|1
2|2
1|3
1|4
sqlite> .read look.sql
sqlite> SELECT * FROM sequence;
-1|0
1|1
1|2
1|3
2|4
2|5
1|6

SQL PL

Works with: Db2 LUW
version 9.7 or higher.

With SQL PL: <lang sql pl> SET SERVEROUTPUT ON @

BEGIN

DECLARE NMBR VARCHAR(100) DEFAULT '1';
DECLARE J SMALLINT DEFAULT 1;
CALL DBMS_OUTPUT.PUT_LINE(NMBR);
WHILE (J < 10) DO
 BEGIN
 DECLARE I SMALLINT;
 DECLARE SIZE SMALLINT;
 DECLARE ACTUAL CHAR(1);
 DECLARE REPEAT CHAR(1);
 DECLARE RESULT VARCHAR(100);
 DECLARE TIMES SMALLINT;
 SET REPEAT = SUBSTR(NMBR, 1, 1);
 SET NMBR = SUBSTR(NMBR, 2) || ' ';
 SET TIMES = 1;
 SET I = 1;
 SET SIZE = LENGTH(NMBR);
 WHILE (I <= SIZE) DO
  SET ACTUAL = SUBSTR(NMBR, I, 1);
  IF (ACTUAL <> REPEAT) THEN
   SET RESULT = COALESCE(RESULT, ) || TIMES ||  || REPEAT;
   SET TIMES = 1;
   SET REPEAT = ACTUAL;
  ELSE
   SET TIMES = TIMES + 1;
  END IF;
  SET I = I + 1;
 END WHILE;
 CALL DBMS_OUTPUT.PUT_LINE(RESULT);
 SET NMBR = RESULT;
 END ;
 SET J = J + 1;
END WHILE;

END @ </lang> Output:

db2 => BEGIN
...
db2 (cont.) => END @
DB20000I  The SQL command completed successfully.

1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

Swift

Translation of: Rust

<lang swift>func lookAndSay(_ seq: [Int]) -> [Int] {

 var result = [Int]()
 var cur = seq[0]
 var curRunLength = 1
 for i in seq.dropFirst() {
   if cur == i {
     curRunLength += 1
   } else {
     result.append(curRunLength)
     result.append(cur)
     curRunLength = 1
     cur = i
   }
 }
 result.append(curRunLength)
 result.append(cur)
 return result

}

var seq = [1]

for i in 0..<10 {

 print("Seq \(i): \(seq)")
 seq = lookAndSay(seq)

}</lang>

Output:
Seq 0: [1]
Seq 1: [1, 1]
Seq 2: [2, 1]
Seq 3: [1, 2, 1, 1]
Seq 4: [1, 1, 1, 2, 2, 1]
Seq 5: [3, 1, 2, 2, 1, 1]
Seq 6: [1, 3, 1, 1, 2, 2, 2, 1]
Seq 7: [1, 1, 1, 3, 2, 1, 3, 2, 1, 1]
Seq 8: [3, 1, 1, 3, 1, 2, 1, 1, 1, 3, 1, 2, 2, 1]
Seq 9: [1, 3, 2, 1, 1, 3, 1, 1, 1, 2, 3, 1, 1, 3, 1, 1, 2, 2, 1, 1]

Tcl

<lang tcl>proc lookandsay n {

   set new ""
   while {[string length $n] > 0} {
       set char [string index $n 0]
       for {set count 1} {[string index $n $count] eq $char} {incr count} {}
       append new $count $char
       set n [string range $n $count end]
   }
   interp alias {} next_lookandsay {} lookandsay $new
   return $new

}

puts 1  ;# ==> 1 puts [lookandsay 1]  ;# ==> 11 puts [next_lookandsay] ;# ==> 21 puts [next_lookandsay] ;# ==> 1211 puts [next_lookandsay] ;# ==> 111221 puts [next_lookandsay] ;# ==> 312211</lang>

Alternatively, with coroutines:

Works with: Tcl version 8.6

<lang tcl>proc seq_lookandsay {n {coroName next_lookandsay}} {

   coroutine $coroName apply {n {
       for {} {[yield $n] ne "stop"} {set n $new} {
           set new ""
           foreach subseq [regexp -all -inline {0+|1+|2+|3+|4+|5+|6+|7+|8+|9+} $n] {
               append new [string length $subseq] [string index $subseq 0]
           }
       }
   }} $n

}

puts [seq_lookandsay 1] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay] puts [next_lookandsay]</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

TUSCRIPT

<lang tuscript> $$ MODE TUSCRIPT,{} num=1,say=""

LOOP look
 digits=STRINGS (num," ? ")
 digitgrouped=ACCUMULATE (digits,howmany)
  LOOP/CLEAR  h=howmany,digit=digitgrouped
   say=JOIN (say,"",h,digit)
  ENDLOOP
 PRINT say
 num=VALUE(say),say=""
 IF (look==14) EXIT
ENDLOOP

</lang>

Output:
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
1321132132111213122112311311222113111221131221
11131221131211131231121113112221121321132132211331222113112211
311311222113111231131112132112311321322112111312211312111322212311322113212221  

UNIX Shell

Works with: bash

<lang bash>lookandsay() {

   local num=$1 char seq i
   for ((i=0; i<=${#num}; i++)); do
       char=${num:i:1}
       if [[ $char == ${seq:0:1} ]]; then
           seq+=$char
       else
           -n $seq  && printf "%d%s" ${#seq} ${seq:0:1}
           seq=$char
       fi
   done

}

for ((num=1, i=1; i<=10; i++)); do

   echo $num
   num=$( lookandsay $num )

done</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

Ursala

The look_and_say function returns the first n results by iterating the function that maps a given sequence to its successor. <lang Ursala>#import std

  1. import nat

look_and_say "n" = ~&H\'1' next"n" rlc~&E; *= ^lhPrT\~&hNC %nP+ length

  1. show+

main = look_and_say 10</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211

VBA

<lang VBA> Public Sub LookAndSay(Optional Niter As Integer = 10) 'generate "Niter" members of the look-and-say sequence '(argument is optional; default is 10)

Dim s As String 'look-and-say number Dim news As String 'next number in sequence Dim curdigit As String 'current digit in s Dim newdigit As String 'next digit in s Dim curlength As Integer 'length of current run Dim p As Integer 'position in s Dim L As Integer 'length of s

On Error GoTo Oops 'to catch overflow, i.e. number too long

'start with "1" s = "1" For i = 1 To Niter

 'initialise
 L = Len(s)
 p = 1
 curdigit = Left$(s, 1)
 curlength = 1
 news = ""
 For p = 2 To L
   'check next digit in s
   newdigit = Mid$(s, p, 1)
   If curdigit = newdigit Then 'extend current run
     curlength = curlength + 1
   Else ' "output" run and start new run
     news = news & CStr(curlength) & curdigit
     curdigit = newdigit
     curlength = 1
   End If
 Next p
 ' "output" last run
 news = news & CStr(curlength) & curdigit
 Debug.Print news
 s = news

Next i Exit Sub

Oops:

 Debug.Print
 If Err.Number = 6 Then 'overflow
   Debug.Print "Oops - number too long!"
 Else
   Debug.Print "Error: "; Err.Number, Err.Description
 End If

End Sub </lang>

Output:
LookAndSay 7
11
21
1211
111221
312211
13112221
1113213211

(Note: overflow occurs at 38th iteration!)

VBScript

Implementation

<lang vb>function looksay( n ) dim i dim accum dim res dim c res = vbnullstring do if n = vbnullstring then exit do accum = 0 c = left( n,1 ) do while left( n, 1 ) = c accum = accum + 1 n = mid(n,2) loop if accum > 0 then res = res & accum & c end if loop looksay = res end function</lang>

Invocation

<lang vb>dim m m = 1 for i = 0 to 13 m = looksay(m) wscript.echo m next</lang>

Output:
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221
3113112221232112111312211312113211
1321132132111213122112311311222113111221131221
11131221131211131231121113112221121321132132211331222113112211
311311222113111231131112132112311321322112111312211312111322212311322113212221

Vedit macro language

This implementation generates look-and-say sequence starting from the sequence on cursor line in edit buffer. Each new sequence is inserted as a new line. 10 sequences are created in this example.

<lang vedit>Repeat(10) {

 BOL
 Reg_Empty(20)
 While (!At_EOL) {
   Match("(.)\1*", REGEXP+ADVANCE)
   Num_Str(Chars_Matched, 20, LEFT+APPEND)
   Reg_Copy_Block(20, CP-1, CP, APPEND)
 }
 Ins_Newline Reg_Ins(20)

}</lang>

Output:
1
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221

Yabasic

<lang Yabasic> dim X$(2) i = 0 // índice de cadena de entrada X$(i) = "1"

input "Indica cuantas repeticiones: " r print "\nSecuencia:"

print X$(i) for n = 1 to r-1

   j = 1 - i  // índice de cadena de salida
   X$(j) = ""
   k = 1
   while k <= len(X$(i))
       k0 = k + 1
       while ((k0 <= len(X$(i))) and (mid$(X$(i), k, 1) = mid$(X$(i), k0, 1)))
           k0 = k0 + 1
       wend
       X$(j) = X$(j) + str$(k0 - k) + mid$(X$(i), k, 1)
       k = k0
   wend
   i = j
   print X$(j)

next n print </lang>

Output:
La salida es similar a la de FreeBASIC, mostrada arriba.

Yorick

<lang yorick>func looksay(input) {

   // Special case: one digit
   if(strlen(input) == 1)
       return "1" + input;
   // Convert string into an array of digits
   digits = strchar(input)(:-1);
   // Find indices where each run starts
   w = where(digits(dif));
   start = numberof(w) ? grow(1, w+1) : [1];
   // Find length of each run
   len = grow(start, numberof(digits)+1)(dif);
   // Find digits for each run
   run = digits(start);
   // Construct output array
   result = array(string, numberof(start)*2);
   // Fill in lengths
   result(1::2) = swrite(format="%d", len);
   // Fill in digits; first must add trailing nulls to coerce single string
   // into an array of strings.
   run = transpose([run, array(char(0), numberof(run))])(*);
   result(2::2) = strchar(run);
   // Merge string array into single string
   return result(sum);

}

val = "1"; do {

  write, val;
  val = looksay(val);

} while(strlen(val) < 80);</lang>

Output:
 1
 11
 21
 1211
 111221
 312211
 13112221
 1113213211
 31131211131221
 13211311123113112211
 11131221133112132113212221
 3113112221232112111312211312113211
 1321132132111213122112311311222113111221131221
 11131221131211131231121113112221121321132132211331222113112211
 311311222113111231131112132112311321322112111312211312111322212311322113212221

zkl

Treating the task as a string manipulation problem.

Translation of: Scala

<lang zkl>fcn lookAndSay(seed){ // numeric String --> numeric String

  len,c:=[1..seed.len()-1].reduce(fcn([(len,c)]lc,index,s,sb){
     if(c!=s[index]) { sb.write(len); sb.write(c); lc.clear(1,s[index]) }
     else lc.clear(len+1,c);
  },L(1,seed[0]), seed,sb:=Sink(String));
  sb.write(len); sb.write(c);
  sb.close();

}</lang>

Output:
(0).reduce(10,fcn(seed,_){ lookAndSay(seed).println() },"1");
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112132113212221