Recaman's sequence

From Rosetta Code
Task
Recaman's sequence
You are encouraged to solve this task according to the task description, using any language you may know.

The Recamán's sequence generates Natural numbers.

Starting from a(0)=0, the n'th term a(n), where n>0, is the previous term minus n i.e a(n) = a(n-1) - n but only if this is both positive and has not been previousely generated.

If the conditions don't hold then a(n) = a(n-1) + n.


Task
  1. Generate and show here the first 15 members of the sequence.
  2. Find and show here, the first duplicated number in the sequence.
  3. Optionally: Find and show here, how many terms of the sequence are needed until all the integers 0..1000, inclusive, are generated.


References



11l

Translation of: Python
F recamanSucc(seen, n, r)
   ‘The successor for a given Recaman term,
    given the set of Recaman terms seen so far.’
   V back = r - n
   R I 0 > back | (back C seen) {n + r} E back

F recamanUntil(p)
   ‘All terms of the Recaman series before the
    first term for which the predicate p holds.’
   V n = 1
   V r = 0
   V rs = [r]
   V seen = Set(rs)
   V blnNew = 1B
   L !p(seen, n, r, blnNew)
      r = recamanSucc(seen, n, r)
      blnNew = r !C seen
      seen.add(r)
      rs.append(r)
      n = 1 + n
   R rs

F enumFromTo(m)
   ‘Integer enumeration from m to n.’
   R n -> @m .< 1 + n

print("First 15 Recaman:\n "recamanUntil((seen, n, r, _) -> n == 15))
print("First duplicated Recaman:\n "recamanUntil((seen, n, r, blnNew) -> !blnNew).last)
V setK = Set(enumFromTo(0)(1000))
print("Number of Recaman terms needed to generate all integers from [0..1000]:\n "(recamanUntil((seen, n, r, blnNew) -> (blnNew & r < 1001 & :setK.is_subset(seen))).len - 1))
Output:
First 15 Recaman:
 [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
First duplicated Recaman:
 42
Number of Recaman terms needed to generate all integers from [0..1000]:
 328002

ALGOL W

begin
    % calculate Recaman's sequence values                                              %

    % a hash table element - holds n, A(n) and a link to the next element with the     %
    %                        same hash value                                           %
    record AValue ( integer eN, eAn ; reference(AValue) eNext );

    % hash modulus                                                                     %
    integer HMOD;
    HMOD := 100000;

    begin
        reference(AValue) array hashTable ( 0 :: HMOD - 1 );
        integer array A ( 0 :: 14 );
        integer le1000Count, firstN, duplicateN, duplicateValue, n, An, An1, prevN, maxS;

        % adds an element to the hash table, returns true if an element with value An  %
        % was already present, false otherwise                                         %
        % if the value was already present, its eN value is returned in prevN          %
        logical procedure addAValue( integer value n, An ; integer result prevN ) ;
            begin
                integer hash;
                logical duplicate;
                reference(AValue) element;
                hash      := An rem HMOD;
                element   := hashTable( hash );
                duplicate := false;
                while element not = null and eAn(element) not = An do element := eNext(element);
                duplicate := element not = null;
                if not duplicate then hashTable( hash ) := AValue( n, An, hashTable( hash ) )
                                 else prevN := eN(element);
                duplicate
            end addAValue ;

        % initialise the hash table                                                    %
        for h := 0 until HMOD - 1 do hashTable( h ) := null;

        % calculate the values of the sequence until we have found values that         %
        % include all numbers in 1..1000                                               %
        % also store the first 15 values                                               %

        A( 0 ) := An1 := n := 0;
        le1000Count := 0;
        maxS := firstN := duplicateN := duplicateValue := -1;
        while le1000Count < 1000 do begin
            logical le0, duplicate;
            n  := n + 1;
            An := An1 - n;
            le0 := ( An <= 0 );
            if le0 then An := An1 + n;
            prevN := -1;
            duplicate := addAValue( n, An, prevN );
            if duplicate and not le0 then begin
                An := An1 + n;
                duplicate := addAValue( n, An, prevN )
            end if_duplicate_and_not_le0 ;
            if duplicate then begin
                % the value was already present %
                if firstN < 0 then begin   % have the first duplicate                  %
                    firstN         := n;
                    duplicateN     := prevN;
                    duplicateValue := An;
                end if_firstN_lt_0
                end
            else if An <= 1000 then le1000Count := le1000Count + 1;;
            if n < 15 then A( n ) := An;
            if An > maxS then maxS := An;
            An1 := An
        end while_le1000Count_lt_1000 ;

        % show the first 15 values of the sequence                                     %
        write( "A( 0 .. 14 ): " );
        for n := 0 until 14 do writeon( i_w := 1, A( n ) );
        % positions of the first duplicate                                             %
        write( i_w := 1
               , s_w := 0
               , "First duplicates: "
               , duplicateN
               , " "
               , firstN
               , " ("
               , duplicateValue
               , ")"
               );
        % number of elements required to include the first 1000 integers               %
        write( i_w := 1, "first element to include all 1..1000: ", n );
        write( i_w := 1, "max sequence value encountered: ", maxS )
    end

end.
Output:
A( 0 .. 14 ): 0  1  3  6  2  7  13  20  12  21  11  22  10  23  9
First duplicates: 20 24 (42)
first element to include all 1..1000: 328002
max sequence value encountered: 1942300

APL

Works with: Dyalog APL
recaman{⎕IO0
    genNext{
        R[N-1]-N
        (R<0)(R):⍵,[N-1]+N
        ,[N-1]-N
    }
    'First 15: '
    reca(genNext14),0
    'First repetition: '
    ⊃⌽recagenNext{≢∪}reca
    'Length of sequence containing [0..1000]:'
    recagenNext{(1001).∊⊂}reca
}
Output:
First 15: 
0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
First repetition: 
42
Length of sequence containing [0..1000]:
328003

AppleScript

The third of these tasks probably stretches Applescript a bit beyond the point of its usefulness – it takes about 1 minute to find the result, and even that requires the use of NSMutableSet, from the Apple Foundation classes.

use AppleScript version "2.4"
use framework "Foundation"
use scripting additions

on run
  
  -- FIRST FIFTEEN RECAMANs ------------------------------------------------------
  
  script term15
    on |λ|(i)
      15 = (i as integer)
    end |λ|
  end script
  set strFirst15 to unwords(snd(recamanUpto(true, term15)))
  
  set strFirstMsg to "First 15 Recamans:" & linefeed
  display notification strFirstMsg & strFirst15
  delay 2
  
  -- FIRST DUPLICATE RECAMAN ----------------------------------------------------
  
  script firstDuplicate
    on |λ|(_, seen, rs)
      setSize(seen) as integer is not (length of (rs as list))
    end |λ|
  end script
  set strDuplicate to (item -1 of snd(recamanUpto(true, firstDuplicate))) as integer as string
  
  set strDupMsg to "First duplicated Recaman:" & linefeed
  display notification strDupMsg & strDuplicate
  delay 2
  
  -- NUMBER OF RECAMAN TERMS NEEDED TO GET ALL OF [0..1000]
  -- (takes about a minute, depending on system)
  
  set setK to setFromList(enumFromTo(0, 1000))
  script supersetK
    on |λ|(i, setR)
      setK's isSubsetOfSet:(setR)
    end |λ|
  end script
  
  display notification "Superset size result will take c. 1 min to find ..."
  set dteStart to current date
  
  set strSetSize to (fst(recamanUpto(false, supersetK)) - 1) as string
  
  set dteEnd to current date
  
  set strSetSizeMsg to "Number of Recaman terms needed to generate" & ¬
    linefeed & "all integers from [0..1000]:" & linefeed
  set strElapsed to "(Last result took c. " & (dteEnd - dteStart) & " seconds to find)"
  display notification strSetSizeMsg & linefeed & strSetSize
  
  -- CLEARED REFERENCE TO NSMUTABLESET ------------------------------------- 
  set setK to missing value
  
  -- REPORT ----------------------------------------------------------------
  unlines({strFirstMsg & strFirst15, "", ¬
    strDupMsg & strDuplicate, "", ¬
    strSetSizeMsg & strSetSize, "", ¬
    strElapsed})
end run

-- nextR :: Set Int -> Int -> Int
on nextR(seen, i, n)
  set bk to n - i
  if 0 > bk or setMember(bk, seen) then
    n + i
  else
    bk
  end if
end nextR

-- recamanUpto :: Bool -> (Int -> Set Int > [Int] -> Bool) -> (Int, [Int])
on recamanUpto(bln, p)
  script recaman
    property mp : mReturn(p)'s |λ|
    on |λ|()
      set i to 1
      set r to 0
      set rs to {r}
      set seen to setFromList(rs)
      repeat while not mp(i, seen, rs)
        set r to nextR(seen, i, r)
        setInsert(r, seen)
        if bln then set end of rs to r
        set i to i + 1
      end repeat
      set seen to missing value -- clear pointer to NSMutableSet
      {i, rs}
    end |λ|
  end script
  recaman's |λ|()
end recamanUpto

-- GENERIC FUNCTIONS -------------------------------------------------------

-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
  if m  n then
    set lst to {}
    repeat with i from m to n
      set end of lst to i
    end repeat
    return lst
  else
    return {}
  end if
end enumFromTo

-- fst :: (a, b) -> a
on fst(tpl)
  if class of tpl is record then
    |1| of tpl
  else
    item 1 of tpl
  end if
end fst

-- intercalateS :: String -> [String] -> String
on intercalateS(sep, xs)
  set {dlm, my text item delimiters} to {my text item delimiters, sep}
  set s to xs as text
  set my text item delimiters to dlm
  return s
end intercalateS

-- Lift 2nd class handler function into 1st class script wrapper 
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
  if class of f is script then
    f
  else
    script
      property |λ| : f
    end script
  end if
end mReturn

-- NB All names of NSMutableSets should be set to *missing value*
-- before the script exits.
-- ( scpt files containing residual ObjC pointer values can not be saved)
-- setFromList :: Ord a => [a] -> Set a
on setFromList(xs)
  set ca to current application
  ca's NSMutableSet's ¬
    setWithArray:(ca's NSArray's arrayWithArray:(xs))
end setFromList

-- setMember :: Ord a => a -> Set a -> Bool
on setMember(x, objcSet)
  missing value is not (objcSet's member:(x))
end setMember

-- setInsert :: Ord a => a -> Set a -> Set a
on setInsert(x, objcSet)
  objcSet's addObject:(x)
  objcSet
end setInsert

-- setSize :: Set a -> Int
on setSize(objcSet)
  objcSet's |count|() as integer
end setSize

-- snd :: (a, b) -> b
on snd(tpl)
  if class of tpl is record then
    |2| of tpl
  else
    item 2 of tpl
  end if
end snd

-- unlines :: [String] -> String
on unlines(xs)
  set {dlm, my text item delimiters} to ¬
    {my text item delimiters, linefeed}
  set str to xs as text
  set my text item delimiters to dlm
  str
end unlines

-- unwords :: [String] -> String
on unwords(xs)
  intercalateS(space, xs)
end unwords
Output:
First 15 Recamans:
0 1 3 6 2 7 13 20 12 21 11 22 10 23 9

First duplicated Recaman:
42

Number of Recaman terms needed to generate
all integers from [0..1000]:
328002

(Last result took c. 40 seconds to find)

Arturo

Translation of: Python
recamanSucc: function [seen, n, r].memoize[
    back: r - n
    (or? 0 > back contains? seen back)? -> n + r
                                        -> back
]

recamanUntil: function [p][
    n: new 1
    r: 0
    rs: new @[r]
    seen: rs
    blnNew: true
    while [not? do p][
        r: recamanSucc seen n r
        blnNew: not? in? r seen
        seen: unique seen ++ r
        'rs ++ r
        inc 'n
    ]
    return rs
]

print "First 15 Recaman numbers:"
print recamanUntil [n = 15]

print ""
print "First duplicate Recaman number:"
print last recamanUntil [not? blnNew]
Output:
First 15 Recaman numbers:
0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 

First duplicate Recaman number:
42

AWK

# syntax: GAWK -f RECAMANS_SEQUENCE.AWK
# converted from Microsoft Small Basic
BEGIN {
    found_dup = 0
    n = -1
    do {
      n++
      ap = a[n-1] + n
      if (a[n-1] <= n) {
        a[n] = ap
        b[ap] = 1
      }
      else {
        am = a[n-1] - n
        if (b[am] == 1) {
          a[n] = ap
          b[ap] = 1
        }
        else {
          a[n] = am
          b[am] = 1
        }
      }
      if (n <= 14) {
        terms = sprintf("%s%s ",terms,a[n])
        if (n == 14) {
          printf("first %d terms: %s\n",n+1,terms)
        }
      }
      if (!found_dup) {
        if (dup[a[n]] == 1) {
          printf("first duplicated term: a[%d]=%d\n",n,a[n])
          found_dup = 1
        }
        dup[a[n]] = 1
      }
      if (a[n] <= 1000) {
        arr[a[n]] = ""
      }
    } while (n <= 15 || !found_dup || length(arr) < 1001)
    printf("terms needed to generate integers 0 - 1000: %d\n",n)
    exit(0)
}
Output:
first 15 terms: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
first duplicated term: a[24]=42
terms needed to generate integers 0 - 1000: 328002

BASIC

10 DEFINT A-Z: DIM A(100)
20 PRINT "First 15 terms:"
30 FOR N=0 TO 14: GOSUB 100: PRINT A(N);: NEXT
35 PRINT
40 PRINT "First repeated term:"
50 GOSUB 100
55 FOR M=0 TO N-1: IF A(M)=A(N) THEN 70 ELSE NEXT
60 N=N+1: GOTO 50
70 PRINT "A(";N;") =";A(N)
80 END
100 IF N=0 THEN A(0)=0: RETURN
110 X = A(N-1)-N: IF X<0 THEN 160
120 FOR M=0 TO N-1
130 IF A(M)=X THEN 160
140 NEXT
150 A(N)=X: RETURN
160 A(N)=A(N-1)+N: RETURN
Output:
First 15 terms:
 0  1  3  6  2  7  13  20  12  21  11  22  10  23  9
First repeated term:
A( 24 ) = 42

Applesoft BASIC

Translation of: BASIC
Works with: Chipmunk Basic
Works with: QBasic
10 DIM A(100)
20 PRINT "First 15 terms:"
30 FOR N=0 TO 14: GOSUB 100: PRINT A(N); " ";: NEXT
35 PRINT
40 PRINT "First repeated term:"
50 GOSUB 100
55 FOR M=0 TO N-1
56 IF A(M)=A(N) THEN 70 
57 NEXT
60 N=N+1: GOTO 50
70 PRINT "A(";N;") = ";A(N)
80 END
100 IF N=0 THEN A(0)=0: RETURN
110 X = A(N-1)-N: IF X<0 THEN 160
120 FOR M=0 TO N-1
130 IF A(M)=X THEN 160
140 NEXT
150 A(N)=X: RETURN
160 A(N)=A(N-1)+N: RETURN

Chipmunk Basic

Works with: Chipmunk Basic version 3.6.4

The BASIC solution works without any changes.

GW-BASIC

Works with: PC-BASIC version any
Works with: BASICA

The BASIC solution works without any changes.

Minimal BASIC

Translation of: BASIC
Works with: Quite BASIC
10 DIM A(100)
20 PRINT "FIRST 15 TERMS:"
30 FOR N=0 TO 14
40 GOSUB 170
50 PRINT A(N);" ";
60 NEXT N
70 PRINT
80 PRINT "FIRST REPEATED TERM:"
90 GOSUB 170
100 FOR M=0 TO N-1
110 IF A(M)=A(N) THEN 150
120 NEXT M
130 LET N=N+1
140 GOTO 90
150 PRINT "A(";N;") = ";A(N)
160 STOP
170 IF N=0 THEN 280
180 LET X = A(N-1)-N
190 IF X<0 THEN 250
200 FOR M=0 TO N-1
210 IF A(M)=X THEN 250
220 NEXT M
230 LET A(N)=X
240 RETURN
250 LET A(N)=A(N-1)+N
260 RETURN
270 STOP
280 LET A(0)=0
290 RETURN
300 END

MSX Basic

Works with: MSX BASIC version any

The BASIC solution works without any changes.

Quite BASIC

The Minimal BASIC solution works without any changes.

BCPL

get "libhdr"

// Generate the N'th term of the Recaman sequence
// given terms 0 to N-1.
let generate(a, n) be 
    a!n := n=0 -> 0, valof
    $(  let subterm = a!(n-1) - n
        let addterm = a!(n-1) + n
        if subterm <= 0 resultis addterm
        for i=0 to n-1 
            if a!i = subterm resultis addterm
        resultis subterm
    $)      

let start() be
$(  let a = vec 50 and n = 15 and rep = ?

    writef("First %N members:*N", n)
    for i = 0 to n-1
    $(  generate(a, i)
        writef("%N ", a!i)
    $)
    
    writef("*NFirst repeated term:*N")
    rep := valof 
    $(  generate(a, n)
        for i = 0 to n-1
            if a!i = a!n resultis n
        n := n + 1
    $) repeat
    
    writef("a!%N = %N*N", rep, a!rep)
$)
Output:
First 15 members:
0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
First repeated term:
a!24 = 42

C

Library: GLib
Translation of: Go
#include <stdio.h>
#include <stdlib.h>
#include <gmodule.h>

typedef int bool;

int main() {
    int i, n, k = 0, next, *a;
    bool foundDup = FALSE;
    gboolean alreadyUsed;
    GHashTable* used = g_hash_table_new(g_direct_hash, g_direct_equal);
    GHashTable* used1000 = g_hash_table_new(g_direct_hash, g_direct_equal);
    a = malloc(400000 * sizeof(int));
    a[0] = 0;
    g_hash_table_add(used, GINT_TO_POINTER(0));
    g_hash_table_add(used1000, GINT_TO_POINTER(0));

    for (n = 1; n <= 15 || !foundDup || k < 1001; ++n) {
        next = a[n - 1] - n;
        if (next < 1 || g_hash_table_contains(used, GINT_TO_POINTER(next))) {
            next += 2 * n;
        }
        alreadyUsed = g_hash_table_contains(used, GINT_TO_POINTER(next));
        a[n] = next;

        if (!alreadyUsed) {
            g_hash_table_add(used, GINT_TO_POINTER(next));
            if (next >= 0 && next <= 1000) {
                g_hash_table_add(used1000, GINT_TO_POINTER(next));
            }
        }

        if (n == 14) {
            printf("The first 15 terms of the Recaman's sequence are: ");
            printf("[");
            for (i = 0; i < 15; ++i) printf("%d ", a[i]);
            printf("\b]\n");
        }

        if (!foundDup && alreadyUsed) {
            printf("The first duplicated term is a[%d] = %d\n", n, next);
            foundDup = TRUE;
        }
        k = g_hash_table_size(used1000);

        if (k == 1001) {
            printf("Terms up to a[%d] are needed to generate 0 to 1000\n", n);
        }
    }
    g_hash_table_destroy(used);
    g_hash_table_destroy(used1000);
    free(a);
    return 0;
}
Output:
The first 15 terms of the Recaman's sequence are: [0 1 3 6 2 7 13 20 12 21 11 22 10 23 9]
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

C#

Translation of: Kotlin
using System;
using System.Collections.Generic;

namespace RecamanSequence {
    class Program {
        static void Main(string[] args) {
            List<int> a = new List<int>() { 0 };
            HashSet<int> used = new HashSet<int>() { 0 };
            HashSet<int> used1000 = new HashSet<int>() { 0 };
            bool foundDup = false;
            int n = 1;
            while (n <= 15 || !foundDup || used1000.Count < 1001) {
                int next = a[n - 1] - n;
                if (next < 1 || used.Contains(next)) {
                    next += 2 * n;
                }
                bool alreadyUsed = used.Contains(next);
                a.Add(next);
                if (!alreadyUsed) {
                    used.Add(next);
                    if (0 <= next && next <= 1000) {
                        used1000.Add(next);
                    }
                }
                if (n == 14) {
                    Console.WriteLine("The first 15 terms of the Recaman sequence are: [{0}]", string.Join(", ", a));
                }
                if (!foundDup && alreadyUsed) {
                    Console.WriteLine("The first duplicated term is a[{0}] = {1}", n, next);
                    foundDup = true;
                }
                if (used1000.Count == 1001) {
                    Console.WriteLine("Terms up to a[{0}] are needed to generate 0 to 1000", n);
                }
                n++;
            }
        }
    }
}
Output:
The first 15 terms of the Recaman sequence are: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

C++

Translation of: C#
#include <iostream>
#include <ostream>
#include <set>
#include <vector>

template<typename T>
std::ostream& operator<<(std::ostream& os, const std::vector<T>& v) {
    auto i = v.cbegin();
    auto e = v.cend();
    os << '[';
    if (i != e) {
        os << *i;
        i = std::next(i);
    }
    while (i != e) {
        os << ", " << *i;
        i = std::next(i);
    }
    return os << ']';
}

int main() {
    using namespace std;

    vector<int> a{ 0 };
    set<int> used{ 0 };
    set<int> used1000{ 0 };
    bool foundDup = false;
    int n = 1;
    while (n <= 15 || !foundDup || used1000.size() < 1001) {
        int next = a[n - 1] - n;
        if (next < 1 || used.find(next) != used.end()) {
            next += 2 * n;
        }
        bool alreadyUsed = used.find(next) != used.end();
        a.push_back(next);
        if (!alreadyUsed) {
            used.insert(next);
            if (0 <= next && next <= 1000) {
                used1000.insert(next);
            }
        }
        if (n == 14) {
            cout << "The first 15 terms of the Recaman sequence are: " << a << '\n';
        }
        if (!foundDup && alreadyUsed) {
            cout << "The first duplicated term is a[" << n << "] = " << next << '\n';
            foundDup = true;
        }
        if (used1000.size() == 1001) {
            cout << "Terms up to a[" << n << "] are needed to generate 0 to 1000\n";
        }
        n++;
    }

    return 0;
}
Output:
The first 15 terms of the Recaman sequence are: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

CLU

% Recaman sequence
recaman = cluster is new, fetch
    rep = array[int]
    
    new = proc () returns (cvt)
        a: rep := rep$predict(0,1000000)
        rep$addh(a,0)
        return(a)
    end new

    % Find the N'th element of the Recaman sequence
    fetch = proc (a: cvt, n: int) returns (int)
        if n > rep$high(a) then extend(a,n) end
        return(a[n])
    end fetch
    
    % See if N has already been generated
    prev = proc (a: rep, n: int) returns (bool)
        for el: int in rep$elements(a) do
            if el = n then return(true) end
        end
        return(false)
    end prev
    
    % Generate members of the sequence until 'top' is reached
    extend = proc (a: rep, top: int)
        while rep$high(a) < top do
            n: int := rep$high(a) + 1
            sub: int := a[n-1] - n
            add: int := a[n-1] + n
            if sub>0 cand ~prev(a, sub)
                then rep$addh(a, sub)
                else rep$addh(a, add)
            end
        end
    end extend
end recaman


start_up = proc () 
    po: stream := stream$primary_output()
    A: recaman := recaman$new()
    
    % Print the first 15 members
    stream$puts(po, "First 15 items:")
    for i: int in int$from_to(0, 14) do
        stream$puts(po, " " || int$unparse(A[i]))
    end
    
    % Find the first duplicated number
    begin
        i: int := 0
        while true do
            i := i + 1
            for j: int in int$from_to(0, i-1) do
                if A[i]=A[j] then exit found(i, A[i]) end
            end
        end
    end except when found(i, n: int):
        stream$putl(po, "\nFirst duplicated number: A("
                      || int$unparse(i) || ") = " || int$unparse(n))
    end
    
    % Find the amount of terms needed to generate all integers 0..1000
    begin
        seen: array[bool] := array[bool]$fill(0,1001,false)
        left: int := 1001
        n: int := -1
        while left > 0 do
            n := n + 1
            if A[n] <= 1000 cand ~seen[A[n]] then
                left := left - 1
                seen[A[n]] := true
            end
        end
        stream$putl(po, "Terms needed to generate [0..1000]: "
                     || int$unparse(n))
    end        
end start_up
Output:
First 15 items: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
First duplicated number: A(24) = 42
Terms needed to generate [0..1000]: 328002

Comal

0010 DIM a#(0:100)
0020 //
0030 // Print the first 15 items
0040 PRINT "First 15 items: ",
0050 FOR i#:=0 TO 14 DO PRINT reca#(i#);
0060 PRINT
0070 //
0080 // Find and print the first repeated item
0090 i#:=15
0100 WHILE NOT find#(i#,reca#(i#)) DO i#:+1
0110 PRINT "First repeated item: A(",i#,") = ",a#(i#)
0120 //
0130 // Generate the n'th member of the Recaman sequence
0140 FUNC reca#(n#)
0150   IF n#=0 THEN RETURN 0
0160   a#(n#):=a#(n#-1)-n#
0180   IF a#(n#)<=0 OR find#(n#,a#(n#)) THEN a#(n#):=a#(n#-1)+n#
0190   RETURN a#(n#)
0200 ENDFUNC reca#
0210 //
0220 // See if a number occurs before the n'th member of the Recaman sequence
0230 FUNC find#(n#,num#)
0240   FOR x#:=0 TO n#-1 DO IF a#(x#)=num# THEN RETURN x#
0250   RETURN 0
0260 ENDFUNC find#
0270 END
Output:
First 15 items: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
First repeated item: A(24) = 42

COBOL

        IDENTIFICATION DIVISION.
        PROGRAM-ID. RECAMAN.
        
        DATA DIVISION.
        WORKING-STORAGE SECTION.
        01 RECAMAN-SEQUENCE COMP.
           02 A     PIC 999 OCCURS 99 TIMES INDEXED BY I.
           02 N     PIC 999 VALUE 0.
           
        01 VARIABLES COMP.
           02 ADDC  PIC S999.
           02 SUBC  PIC S999.
           02 SPTR  PIC 99 VALUE 1.
           
        01 OUTPUT-FORMAT.
           02 OUTI  PIC Z9.
           02 OUTN  PIC BZ9.
           02 OUTS  PIC X(79).
           
        PROCEDURE DIVISION.
        BEGIN.
            PERFORM GENERATE-NEXT-ITEM 15 TIMES.
            PERFORM COLLATE-ITEM VARYING I FROM 1 BY 1
                UNTIL I IS GREATER THAN 15.
            DISPLAY 'First 15 items:' OUTS.
        
        FIND-REPEATING.
            PERFORM GENERATE-NEXT-ITEM.
            SET I TO 1.
            SEARCH A VARYING I
                WHEN I IS NOT LESS THAN N 
                    NEXT SENTENCE
                WHEN A(I) IS EQUAL TO A(N)
                    SUBTRACT 1 FROM N GIVING OUTI
                    MOVE A(N) TO OUTN
                    DISPLAY 'First repeated item: A(' OUTI ') =' OUTN
                    STOP RUN.
            GO TO FIND-REPEATING.
        
        GENERATE-NEXT-ITEM.
            IF N IS EQUAL TO ZERO
                MOVE ZERO TO A(1)
            ELSE
                ADD N, A(N) GIVING ADDC
                SUBTRACT N FROM A(N) GIVING SUBC
                IF SUBC IS NOT GREATER THAN ZERO
                    MOVE ADDC TO A(N + 1)
                ELSE
                    SET I TO 1
                    SEARCH A VARYING I
                        WHEN I IS NOT LESS THAN N
                            MOVE SUBC TO A(N + 1)
                        WHEN A(I) IS EQUAL TO SUBC
                            MOVE ADDC TO A(N + 1).
            ADD 1 TO N.
            
        COLLATE-ITEM.
            MOVE A(I) TO OUTN.
            STRING OUTN DELIMITED BY SIZE INTO OUTS WITH POINTER SPTR.
Output:
First 15 items:  0  1  3  6  2  7 13 20 12 21 11 22 10 23  9
First repeated item: A(24) = 42

D

Translation of: Kotlin
import std.stdio;

void main() {
    int[] a;
    bool[int] used;
    bool[int] used1000;
    bool foundDup;

    a ~= 0;
    used[0] = true;
    used1000[0] = true;

    int n = 1;
    while (n <= 15 || !foundDup || used1000.length < 1001) {
        int next = a[n - 1] - n;
        if (next < 1 || (next in used) !is null) {
            next += 2 * n;
        }
        bool alreadyUsed = (next in used) !is null;
        a ~= next;
        if (!alreadyUsed) {
            used[next] = true;
            if (0 <= next && next <= 1000) {
                used1000[next] = true;
            }
        }
        if (n == 14) {
            writeln("The first 15 terms of the Recaman sequence are: ", a);
        }
        if (!foundDup && alreadyUsed) {
            writefln("The first duplicated term is a[%d] = %d", n, next);
            foundDup = true;
        }
        if (used1000.length == 1001) {
            writefln("Terms up to a[%d] are needed to generate 0 to 1000", n);
        }
        n++;
    }
}
Output:
The first 15 terms of the Recaman sequence are: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

Draco

proc nonrec find([*] int A; word top; int n) bool:
    word i;
    bool found;
    i := 0;
    found := false;
    while i < top and not found do
        found := A[i] = n;
        i := i + 1
    od;
    found
corp

proc nonrec gen_next([*] int A; word n) int:
    int add, sub;
    add := A[n-1] + n;
    sub := A[n-1] - n;
    A[n] := 
        if sub > 0 and not find(A, n, sub) 
            then sub
            else add
        fi;
    A[n]
corp
    
proc nonrec main() void:
    [30] int A;
    word i;
    
    A[0] := 0;
    write("First 15 items:  0");
    for i from 1 upto 14 do write(gen_next(A, i):3) od;
    writeln();
    
    while not find(A, i, gen_next(A, i)) do i := i + 1 od;  
    writeln("First repeated item: A(", i:2, ") = ", A[i]:2)
corp
Output:
First 15 items:  0  1  3  6  2  7 13 20 12 21 11 22 10 23  9
First repeated item: A(24) = 42

EasyLang

arrbase a[] 0
arrbase seen[] 0
len seen[] 100
# 
a[] &= 0
seen[0] = 1
i = 1
repeat
   h = a[i - 1] - i
   if h <= 0 or seen[h] = 1
      h = a[i - 1] + i
   .
   until seen[h] = 1
   seen[h] = 1
   a[] &= h
   if i = 14
      print a[]
   .
   i += 1
.
print h

Forth

Works with: gforth version 0.7.3
: array ( n -- ) ( i -- addr)
  create cells allot
  does> swap cells + ;

100 array sequence

: sequence. ( n -- ) cr 0 ?do i sequence @ . loop ;

: ?unused ( n -- t | n )
  100 0 ?do
    dup i sequence @ = if unloop exit then
  loop drop true ;

: sequence-next ( n -- a[n] )
  dup 0= if 0 0 sequence ! exit then     ( case a[0]=0   )
  dup dup 1- sequence @ swap -           ( a[n]=a[n-1]-n )
  dup dup 0> swap ?unused true = and if
    nip exit then drop
  dup 1- sequence @ swap + ;             ( a[n]=a[n-1]+n )

: sequence-gen ( n -- )
  0 ?do i sequence-next i sequence ! loop ;

: sequence-repeated
  100 0 ?do
    i 0 ?do
      i sequence @ j sequence @ = if
        cr ." first repeated : a[" i . ." ]=a[" j . ." ]=" i sequence @ .  unloop unloop exit then
    loop
  loop ;

100 sequence-gen
15 sequence.
sequence-repeated
Output:
0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 
first repeated : a[20 ]=a[24 ]=42  ok

FreeBASIC

' version 26-01-2019
' compile with: fbc -s console

Dim As UByte used()
Dim As Integer sum, temp
Dim As UInteger n, max, count, i

max = 1000 : ReDim used(max)

Print "The first 15 terms are 0";

For n = 0 To 14
    temp = sum - n
    If temp < 1 OrElse used(temp) = 1 Then
        temp = sum + n
    End If
    If temp <= max Then used(temp) = 1
    sum = temp
    Print sum;
Next


sum = 0 : max = 1000 : ReDim used(max)
Print : Print

For n = 0 To 50
    temp = sum - n
    If temp < 1 OrElse used(temp) = 1 Then
        temp = sum + n
    End If
    If used(temp) = 1 Then
        Print "First duplicated number is a(" + Str(n) + ")"
        Exit For
    End If
    If temp <= max Then used(temp) = 1
    sum = temp
Next


sum = 0 : max = 2000000 : ReDim used(max)
Print : Print

For n = 0 To max
    temp = sum - n
    If temp < 1 OrElse used(temp) = 1 Then
        temp = sum + n
    End If
    If temp <= max Then used(temp) = 1
    If i = temp Then
        While used(i) = 1
            i += 1
            If i > 1000 Then
                Exit For
            End If
        Wend
    End If
    sum = temp
    count += 1
Next

Print "All integers from 0 to 1000 are generated in " & count & " terms"
Print

' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
Output:
The first 15 terms are 0 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9

First duplicated number is a(24)


All integers from 0 to 1000 are generated in 328002 terms

FOCAL

01.10 T "FIRST 15"
01.20 F N=0,14;D 2;T %2,A(N)
01.30 T !"FIRST REPEATED"
01.40 D 2;S Y=1
01.50 F M=0,N-1;S Y=Y*(A(M)-A(N))
01.60 I (Y)1.7,1.8,1.7
01.70 S N=N+1;G 1.4
01.80 T A(N)," AT A(",N,")"!
01.90 Q

02.05 I (N)2.1,2.06,2.1
02.06 A(0)=0;R
02.10 S X=A(N-1)-N
02.20 I (X)2.7
02.30 S Y=1
02.40 F M=0,N-1;S Y=Y*(A(M)-X)
02.50 I (Y)2.6,2.7,2.6
02.60 S A(N)=X;R
02.70 S A(N)=A(N-1)+N
Output:
FIRST 15=  0=  1=  3=  6=  2=  7= 13= 20= 12= 21= 11= 22= 10= 23=  9
FIRST REPEATED= 42 AT A(= 24)


Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website.

In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.

Solution

The following snippet generates the Recaman's sequence of a given number of terms:

File:Fōrmulæ - Recamán's sequence 01.png

Case 1

  • Generate and show here the first 15 members of the sequence.
  • Find and show here, the first duplicated number in the sequence.
  • Optionally. Find and show here, How many terms of the sequence are needed until all the integers 0..1000, inclusive, are generated.

File:Fōrmulæ - Recamán's sequence 02.png

File:Fōrmulæ - Recamán's sequence 03.png

File:Fōrmulæ - Recamán's sequence 04.png

Case 2. Plotting the sequence

File:Fōrmulæ - Recamán's sequence 05.png

File:Fōrmulæ - Recamán's sequence 06.png

Case 3. Drawing the sequence as it was shown in the Numberphile video

File:Fōrmulæ - Recamán's sequence 07.png

File:Fōrmulæ - Recamán's sequence 08.png

File:Fōrmulæ - Recamán's sequence 09.png

Go

package main

import "fmt"

func main() {
    a := []int{0}
    used := make(map[int]bool, 1001)
    used[0] = true
    used1000 := make(map[int]bool, 1001)
    used1000[0] = true
    for n, foundDup := 1, false; n <= 15 || !foundDup || len(used1000) < 1001; n++ {
        next := a[n-1] - n
        if next < 1 || used[next] {
            next += 2 * n
        }
        alreadyUsed := used[next]
        a = append(a, next)

        if !alreadyUsed {
            used[next] = true
            if next >= 0 && next <= 1000 {
                used1000[next] = true
            }
        }

        if n == 14 {
            fmt.Println("The first 15 terms of the Recaman's sequence are:", a)
        }

        if !foundDup && alreadyUsed {
            fmt.Printf("The first duplicated term is a[%d] = %d\n", n, next)
            foundDup = true
        }

        if len(used1000) == 1001 {
            fmt.Printf("Terms up to a[%d] are needed to generate 0 to 1000\n", n)
        }
    }
}
Output:
The first 15 terms of the Recaman's sequence are: [0 1 3 6 2 7 13 20 12 21 11 22 10 23 9]
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

Haskell

Recursion

A basic recursive function for the first N terms,

recaman :: Int -> [Int]
recaman n = fst <$> reverse (go n)
  where
    go 0 = []
    go 1 = [(0, 1)]
    go x =
        let xs@((r, i):_) = go (pred x)
            back = r - i
        in ( if 0 < back && not (any ((back ==) . fst) xs)
               then back
               else r + i
           , succ i) :
           xs

main :: IO ()
main = print $ recaman 15
Output:
[0,1,3,6,2,7,13,20,12,21,11,22,10,23,9]

Conditional iteration

Or, a little more flexibly, a recamanUpto (predicate) function.

Translation of: JavaScript
import Data.Set (Set, fromList, insert, isSubsetOf, member, size)
import Data.Bool (bool)

firstNRecamans :: Int -> [Int]
firstNRecamans n = reverse $ recamanUpto (\(_, i, _) -> n == i)

firstDuplicateR :: Int
firstDuplicateR = head $ recamanUpto (\(rs, _, set) -> size set /= length rs)

recamanSuperset :: Set Int -> [Int]
recamanSuperset setInts =
  tail $ recamanUpto (\(_, _, setR) -> isSubsetOf setInts setR)

recamanUpto :: (([Int], Int, Set Int) -> Bool) -> [Int]
recamanUpto p = rs
  where
    (rs, _, _) =
      until
        p
        (\(rs@(r:_), i, seen) ->
            let n = nextR seen i r
            in (n : rs, succ i, insert n seen))
        ([0], 1, fromList [0])

nextR :: Set Int -> Int -> Int -> Int
nextR seen i r =
  let back = r - i
  in bool back (r + i) (0 > back || member back seen)

-- TEST ---------------------------------------------------------------
main :: IO ()
main =
  (putStrLn . unlines)
    [ "First 15 Recamans:"
    , show $ firstNRecamans 15
    , []
    , "First duplicated Recaman:"
    , show firstDuplicateR
    , []
    , "Length of Recaman series required to include [0..1000]:"
    , (show . length . recamanSuperset) $ fromList [0 .. 1000]
    ]
Output:
First 15 Recamans:
[0,1,3,6,2,7,13,20,12,21,11,22,10,23,9]

First duplicated Recaman:
42

Length of Recaman series required to include [0..1000]:
328002

Lazy search over infinite lists

For a lazier solution, we could define an infinite series of Recaman sequences of growing length, starting with [0], and simply search through them for the first series of length 15, or the first to include a duplicated integer. For the third task, it would be enough to search through an infinite stream of Recaman-generated integer sets of increasing size, until we find the first that contains [0..1000] as a subset.

import Data.List (find, findIndex, nub)
import Data.Maybe (fromJust)
import Data.Set (Set, fromList, insert, isSubsetOf, member)

--- INFINITE STREAM OF RECAMAN SERIES OF GROWING LENGTH --
rSeries :: [[Int]]
rSeries =
  scanl
    ( \rs@(r : _) i ->
        let back = r - i
            nxt
              | 0 > back || elem back rs = r + i
              | otherwise = back
         in nxt : rs
    )
    [0]
    [1 ..]

----------- INFINITE STREAM OF RECAMAN-GENERATED ---------
--------------- INTEGER SETS OF GROWING SIZE -------------
rSets :: [(Set Int, Int)]
rSets =
  scanl
    ( \(seen, r) i ->
        let back = r - i
            nxt
              | 0 > back || member back seen = r + i
              | otherwise = back
         in (insert nxt seen, nxt)
    )
    (fromList [0], 0)
    [1 ..]

--------------------------- TEST -------------------------
main :: IO ()
main = do
  let setK = fromList [0 .. 1000]
  (putStrLn . unlines)
    [ "First 15 Recamans:",
      show . reverse . fromJust $ find ((15 ==) . length) rSeries,
      [],
      "First duplicated Recaman:",
      show . head . fromJust $ find ((/=) <$> length <*> (length . nub)) rSeries,
      [],
      "Length of Recaman series required to include [0..1000]:",
      show . fromJust $ findIndex (\(setR, _) -> isSubsetOf setK setR) rSets
    ]
Output:
First 15 Recamans:
[0,1,3,6,2,7,13,20,12,21,11,22,10,23,9]

First duplicated Recaman:
42

Length of Recaman series required to include [0..1000]:
328002

J

   positive =: >&0
   unique =: -.@:e.
   condition =: (positive@:] *. unique~) ({: - #)

   NB. with the agenda set by the condition, add or subtract tail with tally
   recaman_term =: ({: + #)`({: - #)@.condition


   NB. generate four hundred thousand terms and display the first 15
   15 {. R=:(, recaman_term)^:400000]0
0 1 3 6 2 7 13 20 12 21 11 22 10 23 9


   NB. plot the sequence to see why numberphile might be interested.
   load'plot'
   plot 470{.R

   NB. binaryish search for first duplicate.
   (-:&# ~.) 100 {. R
0
   (-:&# ~.) 50 {. R
0
   (-:&# ~.) 25 {. R
0
   (-:&# ~.) 12 {. R
1
   (-:&# ~.) 18 {. R
1
   (-:&# ~.) 21 {. R
1
   (-:&# ~.) 23 {. R
1
   (-:&# ~.) 24 {. R
1
   (-:&# ~.) 25 {. R
0

Let's write a binary search adverb.

average =: +/ % #
NB. extra_data u Bsearch bounds
NB. Bsearch returns narrowed bounds depending if u return 0 (left) or 1 (right)
NB. u is called as extra_data u index
NB.   or as        index u index
NB. u is invoked as a dyad
Bsearch =: 1 :'((0 1 + (u <.@:average)) { ({. , <.@:average, {:)@:])^:_'
   NB. f expresses "not all [0, 1000] are in the first y members of list x"
   f =: ([: -. [: *./ (i.1001) e. ~.)@:{.~

   R f Bsearch 0 , #R
328002 328003
   
   (<: 328002 328003) { R
328881 879

The sequence begins 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 . With 0 as the first term, We've learned that 25 terms are required to generate a duplicate, and that 328003 terms are needed to generate 0 through 1000 inclusively.

Java

Translation of: Kotlin
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class RecamanSequence {
    public static void main(String[] args) {
        List<Integer> a = new ArrayList<>();
        a.add(0);

        Set<Integer> used = new HashSet<>();
        used.add(0);

        Set<Integer> used1000 = new HashSet<>();
        used1000.add(0);

        boolean foundDup = false;
        int n = 1;
        while (n <= 15 || !foundDup || used1000.size() < 1001) {
            int next = a.get(n - 1) - n;
            if (next < 1 || used.contains(next)) {
                next += 2 * n;
            }
            boolean alreadyUsed = used.contains(next);
            a.add(next);
            if (!alreadyUsed) {
                used.add(next);
                if (0 <= next && next <= 1000) {
                    used1000.add(next);
                }
            }
            if (n == 14) {
                System.out.printf("The first 15 terms of the Recaman sequence are : %s\n", a);
            }
            if (!foundDup && alreadyUsed) {
                System.out.printf("The first duplicate term is a[%d] = %d\n", n, next);
                foundDup = true;
            }
            if (used1000.size() == 1001) {
                System.out.printf("Terms up to a[%d] are needed to generate 0 to 1000\n", n);
            }
            n++;
        }
    }
}
Output:
The first 15 terms of the Recaman sequence are : [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
The first duplicate term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

JavaScript

Translation of: Haskell
(() => {
    const main = () => {

        console.log(
            'First 15 Recaman:\n' +
            recamanUpto(i => 15 === i)
        );

        console.log(
            '\n\nFirst duplicated Recaman:\n' +
            last(recamanUpto(
                (_, set, rs) => set.size !== rs.length
            ))
        );

        const setK = new Set(enumFromTo(0, 1000));
        console.log(
            '\n\nNumber of Recaman terms needed to generate' +
            '\nall integers from [0..1000]:\n' +
            (recamanUpto(
                (_, setR) => isSubSetOf(setK, setR)
            ).length - 1)
        );
    };

    // RECAMAN --------------------------------------------

    // recamanUpto :: (Int -> Set Int > [Int] -> Bool) -> [Int]
    const recamanUpto = p => {
        let
            i = 1,
            r = 0, // First term of series
            rs = [r];
        const seen = new Set(rs);
        while (!p(i, seen, rs)) {
            r = nextR(seen, i, r);
            seen.add(r);
            rs.push(r);
            i++;
        }
        return rs;
    }

    // Next Recaman number.

    // nextR :: Set Int -> Int -> Int
    const nextR = (seen, i, n) => {
        const back = n - i;
        return (0 > back || seen.has(back)) ? (
            n + i
        ) : back;
    };

    // GENERIC --------------------------------------------

    // enumFromTo :: Int -> Int -> [Int]
    const enumFromTo = (m, n) =>
        m <= n ? iterateUntil(
            x => n <= x,
            x => 1 + x,
            m
        ) : [];

    // isSubsetOf :: Ord a => Set a -> Set a -> Bool
    const isSubSetOf = (a, b) => {
        for (let x of a) {
            if (!b.has(x)) return false;
        }
        return true;
    };

    // iterateUntil :: (a -> Bool) -> (a -> a) -> a -> [a]
    const iterateUntil = (p, f, x) => {
        const vs = [x];
        let h = x;
        while (!p(h))(h = f(h), vs.push(h));
        return vs;
    };

    // last :: [a] -> a
    const last = xs =>
        0 < xs.length ? xs.slice(-1)[0] : undefined;

    // MAIN ------------------------------------------------
    return main();
})();
Output:
First 15 Recaman:
0,1,3,6,2,7,13,20,12,21,11,22,10,23,9

First duplicated Recaman:
42

Number of Recaman terms needed to generate
all integers from [0..1000]:
328002

jq

Works with: jq

Works with gojq, the Go implementation of jq

Two jq solutions are provided here. The first is a monolithic "all-in-one" approach in which a single function or procedure does all the work in one pass. This approach has the disadvantage of neither providing or using reusable components.

The second solution is a stream-oriented one that separates the generation of the sequence from other computations. This approach can be just as efficient as the monolithic approach, but for the sake of illustration, the solution offered here provides separate functions for generation and for each of the three specific tasks.

Monolithic approach

Adapted from Go

The main point of potential interest in this implementation is that the main function only retains the first few elements of the sequence required to print them as an array.

# Let R[n] be the Recaman sequence, n >= 0, so R[0]=0.
# Input: a number, $required, specifying the required range of integers, [1 .. $required]
#        to be covered by R[0] ... R[.n]
# $capture: the number of elements of the sequence to retain.
# Output: an object as described below.
# Note that .a|length will be equal to $capture.
#
def recaman_required($capture):
  . as $required
  | {
      n: 0,
      current: 0,                 # R[.n]
      previous: null,             # R[.n-1]
      a: [0],                     # only maintained up to a[$capture-1]
      used:  { "0": true },       # hash for checking whether a value has already occurred
      found: { "0": true },       # hash for checking how many in [0 .. $required] inclusive have been found
      nfound: 1,                  # .found|length
      foundDup: null,             # the first duplicated entry in the sequence
      foundDupAt: null            # .foundDup == R[.foundDupAt]
     }
  | until ((.n >= $capture) and .foundDup and (.nfound > $required);
      .n += 1
      | .current -= .n
      | if (.current < 1 or .used[.current|tostring]) then .current = .current + 2*.n else . end
      | (.current|tostring) as $s
      | .used[$s] as $alreadyUsed
      | if .n < $capture then .a += [.current] else . end
      | if ($alreadyUsed|not)
        then .used[$s] = true
	| if (.current >= 0 and .current <= $required)
	  then .found[$s] = true | .nfound+=1
	  else . end
        else .
	end
      | if (.foundDup|not) and $alreadyUsed
        then .foundDup = .current
	| .foundDupAt = .n
        else .
	end );
	
1000 as $required
| 15 as $capture
| $required | recaman_required($capture)
| "The first \($capture) terms of Recaman's sequence are: \(.a)",
  "The first duplicated term is a[\(.foundDupAt)] = \(.foundDup)",
  "Terms up to a[\(.n)] are needed to generate 0 to \($required) inclusive."
Output:
The first 15 terms of Recaman's sequence are: [0,1,3,6,2,7,13,20,12,21,11,22,10,23,9]
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000 inclusive.

A stream-oriented solution

# Output: the stream of elements in the Recaman sequence, beginning with 0.
def recaman:
  0,
  foreach range(1; infinite) as $i ({used: {"0": true}, current: 0};
      (.current - $i) as $next
      | .current = (if ($next < 1 or .used[$next|tostring]) then $next + 2 * $i else $next end)
      | .used[.current|tostring] = true;
      .current );

# emit [.i, $x] for duplicated terms using IO==0
def duplicated(s):
  foreach s as $x ({used: {}, i: -1};
    .i += 1 
    | ($x|tostring) as $xs
    | if .used[$xs] then .emit = [.i, $x] else .used[$xs] = true end;
    select(.emit) | .emit);

# Input: an integer, $required
# s: a stream of non-negative integers
# Output: the index of the item in the stream s at which the stream up to and including
# that item includes all integers in the closed interval [0 .. $required].
# 
def covers(s):
  . as $required
  | first(foreach s as $x ( { i: -1, found: {}, nfound: 0};
            .i += 1
	    | ($x|tostring) as $xs
            | if .found[$xs] then .
	      elif $x <= $required
	      then .found[$xs] = true | .nfound += 1
	      | if .nfound > $required then .emit=.i else . end
	      else .
	      end;
            select(.emit).emit) );

The three tasks:

"First 15:", limit(15; recaman),

"\First duplicated:", first(duplicated(recaman)),

"\Index of first element to include 0 to 1000 inclusive:",
(1000|covers(recaman))
Output:
First 15:
0
1
3
6
2
7
13
20
12
21
11
22
10
23
9

First duplicated:
[24,42]

Index of first element to include 0 to 1000 inclusive:
328002

Julia

Translation of: Go
function recaman()
    a = Vector{Int}([0])
    used = Dict{Int, Bool}(0 => true)
    used1000 = Set(0)
    founddup = false
    termcount = 1
    while length(used1000) <= 1000
        nextterm = a[termcount] - termcount
        if nextterm < 1 || haskey(used, nextterm)
            nextterm += termcount + termcount
        end
        push!(a, nextterm)
        if !haskey(used, nextterm)
            used[nextterm] = true
            if 1 <= nextterm <= 1000
                push!(used1000, nextterm)
            end
        elseif !founddup
            println("The first duplicated term is a[$(termcount + 1)] = $nextterm.")
            founddup = true
        end
        if termcount == 14
            println("The first 15 terms of the Recaman sequence are $a")
        end
        termcount += 1
    end
    println("Terms up to $(termcount - 1) are needed to generate 0 to 1000.")
end

recaman()
Output:

The first 15 terms of the Recaman sequence are [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
The first duplicated term is a[25] = 42.
Terms up to 328002 are needed to generate 0 to 1000.

Kotlin

Translation of: Go
// Version 1.2.60

fun main(args: Array<String>) {
    val a = mutableListOf(0)
    val used = mutableSetOf(0)
    val used1000 = mutableSetOf(0)
    var foundDup = false
    var n = 1
    while (n <= 15 || !foundDup || used1000.size < 1001) {
        var next = a[n - 1] - n
        if (next < 1 || used.contains(next)) next += 2 * n
        val alreadyUsed = used.contains(next)
        a.add(next)
        if (!alreadyUsed) {
            used.add(next)
            if (next in 0..1000) used1000.add(next)
        }
        if (n == 14) {
            println("The first 15 terms of the Recaman's sequence are: $a")
        }
        if (!foundDup && alreadyUsed) {
            println("The first duplicated term is a[$n] = $next")
            foundDup = true
        }
        if (used1000.size == 1001) {
            println("Terms up to a[$n] are needed to generate 0 to 1000")
        }
        n++
    }
}
Output:
The first 15 terms of the Recaman's sequence are: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

Lua

This runs out of memory determining the final part :(

Translation of: C++
local a = {[0]=0}
local used = {[0]=true}
local used1000 = {[0]=true}
local foundDup = false
local n = 1

while n<=15 or not foundDup or #used1000<1001 do
    local nxt = a[n - 1] - n
    if nxt<1 or used[nxt] ~= nil then
        nxt = nxt + 2 * n
    end
    local alreadyUsed = used[nxt] ~= nil
    table.insert(a, nxt)
    if not alreadyUsed then
        used[nxt] = true
        if 0<=nxt and nxt<=1000 then
            used1000[nxt] = true
        end
    end
    if n==14 then
        io.write("The first 15 terms of the Recaman sequence are:")
        for k=0,#a do
            io.write(" "..a[k])
        end
        print()
    end
    if not foundDup and alreadyUsed then
        print("The first duplicated term is a["..n.."] = "..nxt)
        foundDup = true
    end
    if #used1000 == 1001 then
        print("Terms up to a["..n.."] are needed to generate 0 to 1000")
    end
    n = n + 1
end
Output:
The first 15 terms of the Recaman sequence are: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
The first duplicated term is a[24] = 42
lua: not enough memory

MAD

            NORMAL MODE IS INTEGER
            VECTOR VALUES ELEMF = $2HA(,I2,4H) = ,I2*$
            DIMENSION A(100)
            A(0) = 0
            
            PRINT COMMENT $ FIRST 15 ELEMENTS$
            PRINT FORMAT ELEMF,0,0
            THROUGH EL, FOR I=1, 1, I.GE.15
EL          PRINT FORMAT ELEMF,I,NEXT.(I)
            
            PRINT COMMENT $ $
            PRINT COMMENT $ FIRST REPEATED ELEMENT$
RPT         THROUGH RPT, FOR I=I, 1, FIND.(I,NEXT.(I))
            PRINT FORMAT ELEMF,I,A(I)
            
            INTERNAL FUNCTION(N,TOP)
            ENTRY TO FIND.
            FI=0
SRCH        WHENEVER FI.GE.TOP, FUNCTION RETURN 0B
            WHENEVER A(FI).E.N, FUNCTION RETURN 1B
            FI=FI+1
            TRANSFER TO SRCH
            END OF FUNCTION
            
            INTERNAL FUNCTION(N)
            ENTRY TO NEXT.
            HI=A(N-1)+N
            LO=A(N-1)-N
            WHENEVER LO.L.0
                A(N)=HI
            OR WHENEVER FIND.(LO,N)
                A(N)=HI
            OTHERWISE
                A(N)=LO
            END OF CONDITIONAL
            FUNCTION RETURN A(N)
            END OF FUNCTION
            END OF PROGRAM
Output:
FIRST 15 ELEMENTS
A( 0) =  0
A( 1) =  1
A( 2) =  3
A( 3) =  6
A( 4) =  2
A( 5) =  7
A( 6) = 13
A( 7) = 20
A( 8) = 12
A( 9) = 21
A(10) = 11
A(11) = 22
A(12) = 10
A(13) = 23
A(14) =  9

FIRST REPEATED ELEMENT
A(20) = 42

Mathematica / Wolfram Language

ClearAll[f]
f[s_List] := Block[{a = s[[-1]], len = Length@s}, 
  Append[s, If[a > len && ! MemberQ[s, a - len], a - len, a + len]]]; g = Nest[f, {0}, 70]
g = Nest[f, {0}, 70];
Take[g, 15]
p = Select[Tally[g], Last /* EqualTo[2]][[All, 1]]
p = Flatten[Position[g, #]] & /@ p;
TakeSmallestBy[p, Last, 1][[1]]
Output:
{0,1,3,6,2,7,13,20,12,21,11,22,10,23,9}
{43,42,79,78}
{21,25}

Microsoft Small Basic

Inefficency of associative array allocation in Small Basic ban to provide the optional task.

' Recaman's sequence - smallbasic - 05/08/2015
    nn=15
    TextWindow.WriteLine("Recaman's sequence for the first " + nn + " numbers:")
    recaman()
    TextWindow.WriteLine(Text.GetSubTextToEnd(recaman,2))
    nn="firstdup"
    recaman()
    TextWindow.WriteLine("The first duplicated term is a["+n+"]="+a[n])
 
Sub recaman
    a=""
    b=""
    dup=""
    recaman=""
    firstdup=""
    If nn="firstdup" Then
        nn=1000
        firstdup="True"
    EndIf
    For n=0 To nn-1
        ap=a[n-1]+n 
        If a[n-1]<=n Then 
            a[n]=ap  'a[n]=a[n-1]+n
            b[ap]=1
        Else
            am=a[n-1]-n
            If b[am]=1 Then
                a[n]=ap  'a[n]=a[n-1]+n
                b[ap]=1
            Else
                a[n]=am  'a[n]=a[n-1]-n
                b[am]=1
            EndIf
        EndIf
        If firstdup Then
            If dup[a[n]]=1 Then
                Goto exitsub
            EndIf
            dup[a[n]]=1
        EndIf
        recaman=recaman+","+a[n]
    EndFor
    exitsub:
EndSub
Output:
Recaman's sequence for the first 15 numbers:
0,1,3,6,2,7,13,20,12,21,11,22,10,23,9
The first duplicated term is a[24]=42

Nim

import sequtils, sets, strutils

iterator recaman(num: Positive = Natural.high): tuple[n, a: int; duplicate: bool] =
  var a = 0
  yield (0, a, false)
  var known = [0].toHashSet
  for n in 1..<num:
    var next = a - n
    if next <= 0 or next in known:
      next = a + n
    a = next
    yield (n, a, a in known)
    known.incl a

echo "First 15 numbers in Recaman’s sequence: ", toSeq(recaman(15)).mapIt(it.a).join(" ")

for (n, a, dup) in recaman():
  if dup:
    echo "First duplicate found: a($1) = $2".format(n, a)
    break

var target = toSeq(0..1000).toHashSet
for (n, a, dup) in recaman():
  target.excl a
  if target.card == 0:
    echo "All numbers from 0 to 1000 generated after $1 terms.".format(n)
    break
Output:
First 15 numbers in Recaman’s sequence: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
First duplicate found: a(24) = 42
All numbers from 0 to 1000 generated after 328002 terms.

Objeck

Translation of: Java
use Collection.Generic;

class RecamanSequence {
  function : Main(args : String[]) ~ Nil {
    GenerateSequence();
  }
  
  function : native : GenerateSequence() ~ Nil {
    a := Vector->New()<IntHolder>;
    a->AddBack(0);
     
    used := Set->New()<IntHolder>;
    used->Insert(0);

    used1000 := Set->New()<IntHolder>;
    used1000->Insert(0);
 
    foundDup := false;
    n := 1;
    while (n <= 15 | <>foundDup | used1000->Size() < 1001) {
      next := a->Get(n - 1) - n;
      if (next < 1 | used->Has(next)) {
        next += 2 * n;
      };
      alreadyUsed := used->Has(next);
      a->AddBack(next);
      if (<>alreadyUsed) {
        used->Insert(next);
        if (0 <= next & next <= 1000) {
          used1000->Insert(next);
        };
      };
      if (n = 14) {
        str := ToString(a);
        "The first 15 terms of the Recaman sequence are : {$str}"->PrintLine();
      };
      if (<>foundDup & alreadyUsed) {
        "The first duplicate term is a[{$n}] := {$next}"->PrintLine();
        foundDup := true;
      };
      if (used1000->Size() = 1001) {
        "Terms up to a[{$n}] are needed to generate 0 to 1000"->PrintLine();
      };
      n++;
    };
  }
  
  function : ToString(a : Vector<IntHolder>) ~ String {
    out := "[";
    each(i : a) {
      out += a->Get(i)->Get();
      if(i + 1 < a->Size())  {
        out += ',';
      };
    };
    out += ']';

    return out;
  }
}
Output:
The first 15 terms of the Recaman sequence are : [0,1,3,6,2,7,13,20,12,21,11,22,10,23,9]
The first duplicate term is a[24] := 42
Terms up to a[328002] are needed to generate 0 to 1000

Perl

use bignum;

$max = 1000;
$remaining += $_ for 1..$max;

my @recamans = 0;
my $previous = 0;

while ($remaining > 0) {
   $term++;
   my $this = $previous - $term;
   $this = $previous + $term unless $this > 0 and !$seen{$this};
   push @recamans, $this;
   $dup = $term if !$dup and defined $seen{$this};
   $remaining -= $this if $this <= $max and ! defined $seen{$this};
   $seen{$this}++;
   $previous = $this;
}

print "First fifteen terms of Recaman's sequence: " . join(' ', @recamans[0..14]) . "\n";
print "First duplicate at term: a[$dup]\n";
print "Range 0..1000 covered by terms up to a[$term]\n";
Output:
First fifteen terms of Recaman's sequence: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
First duplicate at term: a[24]
Range 0..1000 covered by terms up to a[328002]

Phix

Translation of: D
with javascript_semantics
bool found_duplicate = false
sequence a = {0}, used = {} -- (grows to 1,942,300 entries)
integer all_used = 0, n = 1, next, prev = 0
while n<=15 or not found_duplicate or all_used<1000 do
    next = prev - n
    if next<1 or (next<=length(used) and used[next]) then
        next = prev + n
    end if
    a &= next
    integer padlen = next-length(used)
    bool already_used = padlen<=0 and used[next]
    if not already_used then
        if padlen>0 then used &= repeat(false,padlen) end if
        used[next] = true
        while all_used<length(used) and used[all_used+1] do
            all_used += 1
        end while
    end if
    if length(a)=15 then
        printf(1,"The first 15 terms of the Recaman sequence are: %v\n",{a})
    end if
    if already_used and not found_duplicate then
        printf(1,"The first duplicated term is a[%d] = %d\n", {n, next})
        found_duplicate = true;
    end if
    if all_used>=1000 then
        printf(1,"Terms up to a[%d] are needed to generate 0 to 1000\n", {n});
    end if
    prev = next
    n += 1
end while
Output:
The first 15 terms of the Recaman sequence are: {0,1,3,6,2,7,13,20,12,21,11,22,10,23,9}
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

PHP

Translation of: Java
<?php
$a = array();
array_push($a, 0);

$used = array();
array_push($used, 0);

$used1000 = array();
array_push($used1000, 0);

$foundDup = false;
$n = 1;

while($n <= 15 || !$foundDup || count($used1000) < 1001) {
	$next = $a[$n - 1] - $n;
	if ($next < 1 || in_array($next, $used)) {
		$next += 2 * $n;
	}
	$alreadyUsed = in_array($next, $used);
	array_push($a, $next);
	if (!$alreadyUsed) {
		array_push($used, $next);
		if (0 <= $next && $next <= 1000) {
			array_push($used1000, $next);
		}
	}
	if ($n == 14) {
		echo "The first 15 terms of the Recaman sequence are : [";
		foreach($a as $i => $v) {
			if ( $i == count($a) - 1)
				echo "$v";
			else
				echo "$v, ";
		}
		echo "]\n";
	}
	if (!$foundDup && $alreadyUsed) {
		printf("The first duplicate term is a[%d] = %d\n", $n, $next);
		$foundDup = true;
	}
	if (count($used1000) == 1001) {
		printf("Terms up to a[%d] are needed to generate 0 to 1000\n", $n);
	}
	$n++;
}
Output:
The first 15 terms of the Recaman sequence are : [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
The first duplicate term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

PL/I

recaman: procedure options(main);
    declare A(0:30) fixed;
    
    /* is X in the first N terms of the Recaman sequence? */
    find: procedure(x, n) returns(bit);
        declare (x, n, i) fixed;
        do i=0 to n-1;
            if A(i)=x then return('1'b);
        end;
        return('0'b);
    end find;
    
    /* generate the N'th term of the Recaman sequence */
    generate: procedure(n) returns(fixed);
        declare n fixed;
        if n=0 then
            A(0) = 0;
        else do;
            declare (sub, add) fixed;
            sub = A(n-1) - n;
            add = A(n-1) + n;
            /* A(n-1) - n not positive? */
            if sub <= 0 then 
                A(n) = add;
            /* A(n-1) - n already generated? */
            else if find(sub, n) then
                A(n) = add;
            else 
                A(n) = sub;
        end;
        return(A(n));
    end generate;
    
    declare i fixed;
    put skip list('First 15 members:');
    do i=0 to 14;
        put edit(generate(i)) (F(3));
    end;
    
    put skip list('First repeated term: ');
    do i=15 repeat(i+1) while(^find(generate(i), i)); end;
    put edit('A(',i,') = ',A(i)) (A,F(2),A,F(2));
end recaman;
Output:
First 15 members:  0  1  3  6  2  7 13 20 12 21 11 22 10 23  9
First repeated term: A(24) = 42

PL/M

100H:
BDOS: PROCEDURE(F,A); DECLARE F BYTE, A ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
PRINT$CH: PROCEDURE(C); DECLARE C BYTE; CALL BDOS(2,C); END PRINT$CH;
PRINT$STR: PROCEDURE(S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT$STR;

/* PRINT NUMBER */
PRINT$NUM: PROCEDURE(N);
    DECLARE (N, P) ADDRESS, C BASED P BYTE;
    DECLARE S(6) BYTE INITIAL('.....$');
    P = .S(5);
DIGIT:
    P = P-1;
    C = '0' + N MOD 10;
    N = N/10;
    IF N>0 THEN GO TO DIGIT;
    CALL PRINT$STR(P);
END PRINT$NUM;

/* IS X IN THE FIRST N TERMS OF THE SEQUENCE */
FIND: PROCEDURE(SEQ,X,N) BYTE;
    DECLARE SEQ ADDRESS, (I, X, N, A BASED SEQ) BYTE;
    DO I=0 TO N-1;
        IF A(I)=X THEN RETURN 0FFH;
    END;
    RETURN 0;
END FIND;

/* GENERATE THE N'TH TERM OF THE SEQUENCE */
GENERATE: PROCEDURE(SEQ,N) BYTE;
    DECLARE SEQ ADDRESS, (N, A BASED SEQ) BYTE;
    IF N=0 THEN
        A(N)=0;
    ELSE DO;
        DECLARE (SUB, ADD) BYTE;
        SUB = A(N-1) - N;
        ADD = A(N-1) + N;
        /* A(N-1) - N NEGATIVE? */
        IF A(N-1) <= N THEN
            A(N) = ADD;
        /* A(N-1) - N ALREADY GENERATED? */
        ELSE IF FIND(SEQ,SUB,N) THEN
            A(N) = ADD;
        ELSE
            A(N) = SUB;
    END;
    RETURN A(N);
END GENERATE;

DECLARE I BYTE, A(30) BYTE;
CALL PRINT$STR(.'FIRST 15 MEMBERS: $');
DO I=0 TO 14;
    CALL PRINT$NUM(GENERATE(.A, I));
    CALL PRINT$CH(' ');
END;
CALL PRINT$STR(.(13,10,'FIRST REPEATED TERM: A($'));

I=15;
DO WHILE NOT FIND(.A, GENERATE(.A, I), I);
    I = I+1;
END;

CALL PRINT$NUM(I);
CALL PRINT$STR(.') = $');
CALL PRINT$NUM(A(I));
CALL PRINT$STR(.(13,10,'$'));
CALL EXIT;
EOF
Output:
FIRST 15 MEMBERS: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
FIRST REPEATED TERM: A(24) = 42

PureBasic

#MAX=500000
Dim a.i(#MAX)
Dim b.b(1)
Dim c.b(1000)
FillMemory(@c(),1000,1,#PB_Byte)

If OpenConsole() : Else : End 1 : EndIf
For n_Count=0 To #MAX  
  If n_Count=0     
    a(n_Count)=0
  ElseIf a(n_Count-1)-n_Count>0 And b(a(n_Count-1)-n_Count)=0
    a(n_Count)=a(n_Count-1)-n_Count
  Else
    a(n_Count)=a(n_Count-1)+n_Count
  EndIf  
  If ArraySize(b())<a(n_Count) : ReDim b(a(n_Count)) : EndIf
  If b(a(n_Count))=1 And fitD=0 : fitD=n_Count : EndIf
  b(a(n_Count))=1  
  If CompareMemory(@b(),@c(),1000) And fit1000=0 : fit1000=n_Count : Break : EndIf  
Next

Print("First 15 terms: ") : For i=0 To 14 : Print(RSet(Str(a(i)),4)) : Next : PrintN("")
PrintN("First duplicate term : a("+Str(fitD)+") = "+Str(a(fitD)))
PrintN("Number of Recaman terms needed to generate all integers from [0..1000]: "+Str(fit1000))
Input()
End
Output:
First 15 terms:    0   1   3   6   2   7  13  20  12  21  11  22  10  23   9
First duplicate term : a(24) = 42
Number of Recaman terms needed to generate all integers from [0..1000]: 328002

Python

Conditional iteration over a generator

from itertools import islice

class Recamans():
    "Recamán's sequence generator callable class"
    def __init__(self):
        self.a = None   # Set of results so far
        self.n = None   # n'th term (counting from zero)
    
    def __call__(self):
        "Recamán's sequence  generator"
        nxt = 0
        a, n = {nxt}, 0
        self.a = a
        self.n = n
        yield nxt
        while True:
            an1, n = nxt, n + 1
            nxt = an1 - n
            if nxt < 0 or nxt in a:
                nxt = an1 + n
            a.add(nxt)
            self.n = n
            yield nxt

if __name__ == '__main__':
    recamans = Recamans()
    print("First fifteen members of Recamans sequence:", 
          list(islice(recamans(), 15)))

    so_far = set()
    for term in recamans():
        if term in so_far:
            print(f"First duplicate number in series is: a({recamans.n}) = {term}")
            break
        so_far.add(term)
    
    n = 1_000
    setn = set(range(n + 1))    # The target set of numbers to be covered
    for _ in recamans():
        if setn.issubset(recamans.a):
            print(f"Range 0 ..{n} is covered by terms up to a({recamans.n})")
            break
Output:
First fifteen members of Recamans sequence: [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
First duplicate number in series is: a(24) = 42
Range 0 ..1000 is covered by terms up to a(328002)

Parameterised query predicates

Passing different query predicates to a single more general function:

( This turns out to be c. 8X faster than than the iteration over generator approach above, on a simple start to end measure using time.time())

'''Recaman sequence'''


# recamanUntil :: (Int -> Set Int > [Int] -> Bool) -> [Int]
def recamanUntil(p):
    '''All terms of the Recaman series before the
       first term for which the predicate p holds.'''
    n = 1
    r = 0  # First term of series
    rs = [r]
    seen = set(rs)
    blnNew = True
    while not p(seen, n, r, blnNew):
        r = recamanSucc(seen, n, r)
        blnNew = r not in seen
        seen.add(r)
        rs.append(r)
        n = 1 + n
    return rs


# recamanSucc :: Set Int -> Int -> Int
def recamanSucc(seen, n, r):
    '''The successor for a given Recaman term,
       given the set of Recaman terms seen so far.'''
    back = r - n
    return n + r if 0 > back or (back in seen) else back


# ------------------------- TEST -------------------------
# main :: IO ()
def main():
    '''Test'''
    print(
        'First 15 Recaman:\r',
        recamanUntil(
            lambda seen, n, r, _: 15 == n
        )
    )
    print(
        'First duplicated Recaman:\r',
        recamanUntil(
            lambda seen, n, r, blnNew: not blnNew
        )[-1]
    )
    setK = set(enumFromTo(0)(1000))
    print(
        'Number of Recaman terms needed to generate',
        'all integers from [0..1000]:\r',
        len(recamanUntil(
            lambda seen, n, r, blnNew: (
                blnNew and 1001 > r and setK.issubset(seen)
            )
        )) - 1
    )


# ----------------------- GENERIC ------------------------

# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
    '''Integer enumeration from m to n.'''
    return lambda n: range(m, 1 + n)


if __name__ == '__main__':
    main()
Output:
First 15 Recaman:
 [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
First duplicated Recaman:
 42
Number of Recaman terms needed to generate all integers from [0..1000]:
 328002

Alternatively, we can use query predicates in combination with the iteration of a function over a tuple,
and encapsulate the querying in the generic abstractions iterate and until.

This additional level of abstraction reduces the amount of new code that we have to write, facilitates refactoring, and turns out to have insignificant cost.

( This version is still c. 8X faster than the conditional iteration over generator version, as measured by a simple start and end test using time.time() ).

'''Recaman by iteration of a function over a tuple.'''

from itertools import (islice)


# recamanTupleSucc :: Set Int -> (Int, Int, Bool) -> (Int, Int, Bool)
def recamanTupleSucc(seen):
    '''The Nth in a series of Recaman tuples,
       (N, previous term, boolPreviouslySeen?)
       given the set of all terms seen so far.'''
    def go(n, r, _):
        back = r - n
        nxt = n + r if 0 > back or (back in seen) else back
        bln = nxt in seen
        seen.add(nxt)
        return (1 + n, nxt, bln)
    return lambda tpl: go(*tpl)


# ------------------------- TEST -------------------------
# main :: IO()
def main():
    '''First 15, and first duplicated Recaman.'''
    f = recamanTupleSucc(set([0]))
    print(
        'First 15 Recaman:\n',
        list(map(
            snd,
            take(15)(iterate(f)((1, 0, False)))
        ))
    )
    f = recamanTupleSucc(set([0]))
    print(
        'First duplicated Recaman:\n',
        until(lambda x: x[2])(f)(
            (1, 0, False)
        )[1]
    )

    sk = set(enumFromTo(0)(1000))
    sr = set([0])
    f = recamanTupleSucc(sr)
    print(
        'Number of Recaman terms needed to generate',
        'all integers from [0..1000]:\n',
        until(
            lambda x: not x[2] and 1001 > x[1] and sk.issubset(sr)
        )(f)(
            (1, 0, False)
        )[0] - 1
    )


# ----------------- GENERIC ABSTRACTIONS -----------------

# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
    '''Integer enumeration from m to n.'''
    return lambda n: range(m, 1 + n)


# iterate :: (a -> a) -> a -> Gen [a]
def iterate(f):
    '''An infinite list of repeated
       applications of f to x.
    '''
    def go(x):
        v = x
        while True:
            yield v
            v = f(v)
    return go


# snd :: (a, b) -> b
def snd(tpl):
    '''Second component of a tuple.'''
    return tpl[1]


# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
    '''The prefix of xs of length n,
       or xs itself if n > length xs.'''
    return lambda xs: (
        xs[0:n]
        if isinstance(xs, list)
        else islice(xs, n)
    )


# until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
    '''The result of repeatedly applying f until p holds.
       The initial seed value is x.
    '''
    def go(f):
        def g(x):
            v = x
            while not p(v):
                v = f(v)
            return v
        return g
    return go


# MAIN ---
if __name__ == '__main__':
    main()
First 15 Recaman:
 [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
First duplicated Recaman:
 42
Number of Recaman terms needed to generate all integers from [0..1000]:
 328002

Quackery

  [ stack 0 ]              is seennumbers (     --> s   )

  [ bit seennumbers take 
    | seennumbers put ]    is seen        (   n -->     )

  [ dup 0 < iff
      [ drop true ] done
    bit seennumbers share
    & 0 != ]                is seen?       (   n --> b   )

  [ 1+ bit 1 - 
    seennumbers share
    over & = ]              is allseen?    (   n --> b   )

  [ stack [ ] ]             is repeats     (     --> s   )

  [ 1 seennumbers replace
    [] repeats replace
    ' [ 0 ] 1 ]             is startseq    (     --> [ n )

  [ over -1 peek 
    over - dup seen? if
      [ over 2 * +
        dup seen? if
          [ repeats take
            over join 
            repeats put ] ]
    dup seen
    swap dip join 
    1+ ]                    is nextterm    ( [ n --> [ n )
 
  say "first 15 terms: "
  startseq
  14 times nextterm
  drop echo cr
 
  say "first duplicated term: "
  startseq
  [ repeats share [] = while
    nextterm
    again ]
  drop -1 peek echo cr
 
  say "terms needed to generate 0 to 1000: "
  startseq
  [ nextterm
    1000 allseen? until ] 
  nip 1 - echo cr
Output:
first 15 terms: [ 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 ]
first duplicated term: 42
terms needed to generate 0 to 1000: 328002


R

A bit slow because the append() function is expensive.

visited <- vector('logical', 1e8)

terms <- vector('numeric')

in_a_interval <- function(v) {
	visited[[v+1]] 
}

add_value <- function(v) {
	visited[[v+1]] <<- TRUE
	terms <<- append(terms, v)
}

add_value(0)
step <- 1
value <- 0

founddup <- FALSE

repeat {
	if ((value-step>0) && (!in_a_interval(value-step))) {
		value <- value - step
	} else {
		value <- value + step
	}
	if (in_a_interval(value) && !founddup) {
		cat("The first duplicated term is a[",step,"] = ",value,"\n", sep = "")
		founddup <- TRUE
	}
	add_value(value)
	if (all(visited[1:1000])) {
		cat("Terms up to a[",step,"] are needed to generate 0 to 1000\n",sep = "")
		break
	}
	step <- step + 1
	if (step == 15) { 
		cat("The first 15 terms are :")
		for (aterm in terms) { cat(aterm," ", sep = "") }
		cat("\n")
	}
}
Output:
The first 15 terms are :0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

Raku

(formerly Perl 6)

Works with: Rakudo version 2018.06
my @recamans = 0, {
   state %seen;
   state $term;
   $term++;
   my $this = $^previous - $term;
   $this = $previous + $term unless ($this > 0) && !%seen{$this};
   %seen{$this} = True;
   $this
} … *;

put "First fifteen terms of Recaman's sequence: ", @recamans[^15];

say "First duplicate at term: a[{ @recamans.first({@recamans[^$_].Bag.values.max == 2})-1 }]";

my @seen;
my int $i = 0;
loop {
    next if (my int $this = @recamans[$i++]) > 1000 or @seen[$this];
    @seen[$this] = 1;
    say "Range 0..1000 covered by terms up to a[{$i - 1}]" and last if ++$ == 1001;
}
Output:
First fifteen terms of Recaman's sequence: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
First duplicate at term: a[24]
Range 0..1000 covered by terms up to a[328002]

REXX

version 1

Instead of using a subroutine to perform the three tasks with one invocation,   the subroutine was used three times
(once for each of the task's three requirements).

Programmer's note:   the short-circuited   if   REXX statement   (lines 14 & 15):

  if z<0  then              z= _ + #
          else if !.z  then z= _ + #

could've been replaced with:

  if !.z | z<0         then z= _ + #
/*REXX pgm computes a Recamán sequence up to N; the 1st dup; # terms for a range of #'s.*/
parse arg N h .                                  /*obtain optional arguments from the CL*/
if N=='' | N==","  then N=   15                  /*Not specified?  Then use the default.*/
if h=='' | h==","  then h= 1000                  /* "      "         "   "   "     "    */
      say "Recamán's sequence for the first "        N         " numbers: "    recaman(N)
say;  say "The first duplicate number in the Recamán's sequence is: "          recaman(0)
say;  say "The number of terms to complete the range  0───►"h    ' is: '       recaman(-h)
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
recaman: procedure; parse arg y,,d.; $=0;  !.=0;   _=0;   !.0=1  /*init. array and vars.*/
                    r= y<0;          Reca= 0;    hi= abs(y)      /*for the 2nd invoke.  */
                    o= y==0;         if y<1  then y= 1e8         /* "   "  3rd    "     */
           do #=1  for y-1;          z= _ - #                    /*next # might be < 0. */
           if z<0  then              z= _ + #                    /*this is faster than: */
                   else if !.z  then z= _ + #                    /*if !.z | z<0 then ···*/
           !.z= 1;                      _= z                     /*mark it;  add to seq.*/
           if r  then do;  if z>hi      then iterate             /*ignore #'s too large.*/
                           if d.z==''   then Reca= Reca + 1      /*Unique? Bump counter.*/
                           d.z= .                                /*mark # as a new low. */
                           if Reca>=hi  then return #            /*list is complete ≥ HI*/
                           iterate
                      end                                        /* [↑]  a range of #s. */
           if o  then do;  if d.z==.  then return z;  d.z=.;  iterate  /*check if dup #.*/
                      end
           $= $ z                                                /*add number to $ list?*/
           end   /*#*/;                    return $              /*return the  $  list. */
output   when using the default input:

Run time was about   1/6   of a second,   which is over   230%   times faster than version 2.

Recamán's sequence for the first  15  numbers:  0 1 3 6 2 7 13 20 12 21 11 22 10 23 9

The first duplicate number in the Recamán's sequence is:  42

The number of terms to complete the range  0───►1000  is:  328002

version 2

/*REXX program computes & displays the Recaman sequence           */
/*improved using version 1's method for task 3                    */
Call time 'R'                  /* Start timer                     */
Parse Arg n
If n='' Then n=15
Say 'the first' n 'elements:' recaman(n)
Say ans.2
Say ans.3
Say time('E') 'seconds elapsed'
Exit

recaman:
Parse Arg n                    /* Wanted number of elements       */
have.=0                        /* Number not yet in sequence      */
e.0=0                          /* First element                   */
have.0=1                       /*   is in the sequence            */
s=0                            /* Sequence to be shodn            */
done=0                         /* turn on first duplicate switch  */
d.=0
d.0=1
dn=1                           /* number of elements <=1000       */
 Do i=1 until dn==1001         /* Loop until all found            */
  ip=i-1                       /* previous index                  */
  temp=e.ip-i                  /* potential next element          */
  If temp>0 & have.temp=0 Then /*   to be used                    */
    Nop
  Else                         /* compute the alternative         */
    temp=e.ip+i
  e.i=temp                     /* Set next element                */
  If words(s)<n Then           /* not enough in output            */
    s=s temp                   /* add the element to the output   */
  If temp<=1000 Then Do        /* eligible for task 3             */
    If d.temp=0 Then Do        /* not yet encountered             */
      d.temp=1                 /* Remember it's there             */
      dn=dn+1                  /* count of integers<=1000 found   */
      End
    End
  If done=0 & have.temp=1 Then Do
    ans.2='First duplicate ('temp') added in iteration' i,
          'elapsed:' time('E') 'seconds'
    done=1
    End
  ans.3='Element number' i 'is the last to satisfy task 3. It is' temp 
  Have.temp=1
  End
Return s
Output:
the first 15 elements: 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
First duplicate (42) added in iteration 24 elapsed: 0 seconds
Element number 328002 is the last to satisfy task 3. It is 879
7.126000 seconds elapsed

Ring

load "zerolib.ring"

recaman = Z(0:50)
duplicate = 0
dup = []
recDuplicate = []
recnum = 0

see "working..." + nl
see "the first 15 Recaman's numbers are:" + nl

for n = 1 to len(recaman) - 1
    if n = 1
       recaman[0] = 0
       add(dup,0)
       see "" + recaman[0] + " "
    ok
    recaman[n] = recaman[n-1] - n
    if recaman[n] <= 0 
       recaman[n] = recaman[n-1] + n
    ok
    fnrec = find(dup,recaman[n])
    if fnrec > 0
       del(dup,fnrec)
       recaman[n] = recaman[n-1] + n
       add(dup,recaman[n])
    else
       add(dup,recaman[n])
    ok
    recnum = recnum + 1
    if recnum < 15
       see "" + recaman[n] + " "
    ok
    add(recDuplicate,recaman[n])
next
see nl

see "the first duplicated term is a[" 
for n = len(recDuplicate) to 2 step -1
    for m = n-1 to 1 step -1
        if recDuplicate[n] = recDuplicate[m]
           duplicate = recDuplicate[n]
           dupnr = n
        ok
    next
next

see "" + dupnr + "] = " + duplicate + nl
see "done..." + nl
Output:
working...
the first 15 Recaman's numbers are:
0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 
the first duplicated term is a[24] = 42
done...

RPL

Works with: Halcyon Calc version 4.2.8
RPL code Comment
 ≪ 
    1 + RKMSQ SWAP OVER SIZE
    IF DUP2 ≤ 
    THEN DROP GET
    ELSE SWAP 1 - FOR j 
        DUP DUP SIZE GET j - 
        IF DUP2 ABS POS OVER 0 < OR 
        THEN j 2 * + END
        + NEXT 
        DUP ‘RKMSQ‘ STO DUP SIZE GET END
≫ ‘RECAM’ STO

≪ 0 { 0 } ‘RKMSQ’ STO
   DO 
      1 + RKMSQ OVER RECAM 
   UNTIL POS END RECAM 
≫ ‘TASK2’ STO
 RECAM ( n --  a(n) )
get m = size of sequence in memory
if n+1 ≤ m 
then recall sequence(n+1)=a(n)
else for j=m to n
    get a(j-1)-n
    if already in sequence or <0
       then a(j) = (a(j-1)-n)+2*n
    add to sequence
    store updated sequence and recall a(n)
.
.
Initialize variables
Loop
  get a(n)
until a(n) already in sequence
.
Input:
{ 0 } 'RKMSQ’ STO
15 RECAM 
RKMSQ 
{ 0 } ‘RKMSQ’ STO
TASK2
Output:
3: 24
2: { 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 24 }
1: 42

Ruby

Translation of: Kotlin
require 'set'

a = [0]
used = Set[0]
used1000 = Set[0]
foundDup = false
n = 1
while n <= 15 or not foundDup or used1000.size < 1001
    nxt = a[n - 1] - n
    if nxt < 1 or used === nxt then
        nxt = nxt + 2 * n
    end
    alreadyUsed = used === nxt
    a << nxt
    if not alreadyUsed then
        used << nxt
        if nxt >= 0 and nxt <= 1000 then
            used1000 << nxt
        end
    end
    if n == 14 then
        print "The first 15 terms of the Recaman's sequence are ", a, "\n"
    end
    if not foundDup and alreadyUsed then
        print "The first duplicated term is a[", n, "] = ", nxt, "\n"
        foundDup = true
    end
    if used1000.size == 1001 then
        print "Terms up to a[", n, "] are needed to generate 0 to 1000\n"
    end
    n = n + 1
end
Output:
The first 15 terms of the Recaman's sequence are [0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

Rust

use std::collections::HashSet ;

fn main() {
   let mut recamans : Vec<i32> = Vec::new( ) ;
   let mut reca_set : HashSet<i32> = HashSet::new() ;
   let mut first_nums : HashSet<i32> = HashSet::new( ) ;
   for i in 0i32..=1000 {
      first_nums.insert( i ) ;
   }
   recamans.push( 0 ) ;
   reca_set.insert( 0 ) ;
   let mut current : i32 = 0 ;
   while ! first_nums.is_subset( &reca_set ) {
      current += 1 ;
      let mut nextnum : i32 = recamans[( current as usize ) - 1] - current ;
      if nextnum < 0 || reca_set.contains( &nextnum ) {
         nextnum = recamans[(current as usize ) - 1 ] + current ;
      }
      recamans.push( nextnum ) ;
      reca_set.insert( nextnum ) ;
      if current == 15 {
         println!("The first 15 numbers of the Recaman sequence are:" ) ;
         println!("{:?}" , recamans ) ;
      }
   }
   println!("To generate all numbers from 0 to 1000 , one has to go to element {}" , current) ;
}
Output:
The first 15 numbers of the Recaman sequence are:
[0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24]
To generate all numbers from 0 to 1000 , one has to go to element 328002

Scala

Output:
Best seen in running your browser either by ScalaFiddle (ES aka JavaScript, non JVM) or Scastie (remote JVM).
import scala.collection.mutable

object RecamansSequence extends App {
  val (a, used) = (mutable.ArrayBuffer[Int](0), mutable.BitSet())
  var (foundDup, hop, nUsed1000) = (false, 1, 0)

  while (nUsed1000 < 1000) {
    val _next = a(hop - 1) - hop
    val next = if (_next < 1 || used.contains(_next)) _next + 2 * hop else _next
    val alreadyUsed = used.contains(next)

    a += next
    if (!alreadyUsed) {
      used.add(next)
      if (next <= 1000) nUsed1000 += 1
    }
    if (!foundDup && alreadyUsed) {
      println(s"The first duplicate term is a($hop) = $next")
      foundDup = true
    }
    if (nUsed1000 == 1000)
      println(s"Terms up to $hop are needed to generate 0 to 1000")

    hop += 1
  }

  println(s"The first 15 terms of the Recaman sequence are : ${a.take(15)}")

}

Scheme

Works with: Chez Scheme
; Create a dynamically resizing vector (a "dynvec").
; Returns a procedure that takes a variable number of arguments:
; 0 : () --> Returns the vector from index 0 through the maximum index set.
; 1 : (inx) --> Returns the value at the given index.
; 2 : (inx val) --> Sets the given value into the given index, and returns the value.

(define make-dynvec
  (lambda (init-size extra-fact init-val)
    (let ((vec (make-vector init-size init-val)) (maxinx -1))
      (lambda args
        (if (null? args)
          (let ((retvec (make-vector (1+ maxinx))))
            (do ((index 0 (1+ index)))
                ((> index maxinx) retvec)
              (vector-set! retvec index (vector-ref vec index))))
          (let ((inx (car args)))
            (when (>= inx (vector-length vec))
              (let ((newvec (make-vector
                              (inexact->exact (ceiling (* extra-fact inx)))
                              init-val)))
                (do ((index 0 (1+ index)))
                    ((>= index (vector-length vec)))
                  (vector-set! newvec index (vector-ref vec index)))
                (set! vec newvec)))
            (when (pair? (cdr args))
              (when (> inx maxinx) (set! maxinx inx))
              (vector-set! vec inx (cadr args)))
            (vector-ref vec inx)))))))

; Generate the Recaman's sequence.
; Generate the terms of Recaman's sequence until the given "stop" procedure
; returns a true value; that returned value becomes the value of this procedure.
; The arguments to the "stop" procedure are:  n, the value of the n'th term,
; #t if that term was seen before, #t if the term was arrived at by addition,
; the Recaman's sequence so far (as a dynvec), and a dynvec of the n's at which
; a value was first seen or #f if not previously seen ("seen1st").

(define recaman-sequence
  (lambda (stop-proc)
    (let ((recaman (make-dynvec 10 2 0))
          (seen1st (make-dynvec 10 2 #f)))
      (do ((n 0 (1+ n)) (done-retval #f))
          (done-retval done-retval)
        (if (= n 0)
          (begin
            (recaman n 0)
            (seen1st 0 n)
            (set! done-retval (stop-proc n 0 #f #f recaman seen1st)))
          (let ((try-sub (- (recaman (1- n)) n)))
            (if (and (> try-sub 0) (not (seen1st try-sub)))
              (begin
                (recaman n try-sub)
                (seen1st try-sub n)
                (set! done-retval (stop-proc n try-sub #f #f recaman seen1st)))
              (let* ((val-add (+ (recaman (1- n)) n)) (seen-prev (seen1st val-add)))
                (recaman n val-add)
                (unless (seen1st val-add) (seen1st val-add n))
                (set! done-retval
                      (stop-proc n val-add seen-prev #t recaman seen1st))))))))))

; Generate and display the first 15 Recaman's numbers.

(printf "First 15 Recaman's numbers: ~a~%"
        (recaman-sequence (lambda (n val seen-prev by-add recaman seen1st)
                            (and (>= n (1- 15)) (recaman)))))

; Find and display the first duplicated Recaman's number.
; The only way to be a duplicate is if the number was arrived
; at by adding 'n' and the number has been seen before.

(let ((dup-n-val-1st
        (recaman-sequence (lambda (n val seen-prev by-add recaman seen1st)
                            (and by-add seen-prev (list n val (seen1st val)))))))
  (printf "First duplicate Recaman's number: a[~a] = a[~a] = ~a~%"
          (caddr dup-n-val-1st) (car dup-n-val-1st) (cadr dup-n-val-1st)))

; Find and display how many terms of the sequence are needed
; for all the integers 0..1000, inclusive, to be generated.

(let* ((all-first 1001)
       (terms-to-gen-all (recaman-sequence
                           (lambda (n val seen-prev by-add recaman seen1st)
                             (do ((inx 0 (1+ inx)))
                                 ((or (>= inx all-first) (not (seen1st inx)))
                                  (and (>= inx all-first) (1+ n))))))))
  (printf
    "Terms of Recaman's sequence to generate all integers 0..~a, inclusive: ~a~%"
    (1- all-first) terms-to-gen-all))
Output:
First 15 Recaman's numbers: #(0 1 3 6 2 7 13 20 12 21 11 22 10 23 9)
First duplicate Recaman's number: a[20] = a[24] = 42
Terms of Recaman's sequence to generate all integers 0..1000, inclusive: 328003

SETL

program recaman;
    a := {[0,0]};

    loop for i in [1..14] do
        extend(a);
    end loop;

    print("First 15:", [a(n) : n in [0..14]]);

    loop
        doing n := extend(a);
        until #(rept:=[[r,i] : r = a(i) | r=n]) > 1
        do pass;
    end loop;

    print("First repetition:", n, "at", {x:x in rept}{n});

    proc extend(rw a);
        n := max/ domain a;
        t := a(n) - n-1;
        if t<0 or t in range a then
            t := a(n) + n+1;
        end if;
        return a(n+1) := t;
    end proc;
end program;
Output:
First 15: [0 1 3 6 2 7 13 20 12 21 11 22 10 23 9]
First repetition: 42 at {20 24}

Sidef

func recamans_generator() {

    var term = 0
    var prev = 0
    var seen = Hash()

    {
        var this = (prev - term)

        if ((this <= 0) || seen{this}) {
            this = (prev + term)
        }

        prev = this
        seen{this} = true
        term++
        this
    }
}

with (recamans_generator()) { |r|
    say ("First 15 terms of the Recaman's sequence: ", 15.of { r.run }.join(', '))
}

with (recamans_generator()) {|r|
    var seen = Hash()
    Inf.times {|i|
        var n = r.run
        if (seen{n}) {
            say "First duplicate term in the series is a(#{i}) = #{n}"
            break
        }
        seen{n} = true
    }
}

with (recamans_generator()) {|r|
    var seen = Hash()
    Inf.times {|i|
        var n = r.run
        if ((n <= 1000) && (seen{n} := true) && (seen.len == 1001)) {
            say "Terms up to a(#{i}) are needed to generate 0 to 1000"
            break
        }
    }
}
Output:
First 15 terms of the Recaman's sequence: 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9
First duplicate term in the series is a(24) = 42
Terms up to a(328002) are needed to generate 0 to 1000

uBasic/4tH

a = 0                                  ' the first one is free ;-)
Print "First 15 numbers:"

For i = 1 Step 1                       ' start loop
  If i<16 Then Print a,                ' print first 15 numbers
  b = Iif ((a-i<1) + (Func(_Peek(Max(0, a-i)))), a+i, a-i)
  If Func(_Peek(b)) Then Print "\nFirst repetition: ";b : Break
  Proc _Set(Set(a, b))                 ' set bit in bitmap
Next
End                                    ' terminate program
                                       ' bitmap functions
_Set  Param(1) : Let @(a@/32) = Func(_Poke(a@/32, a@%32)) : Return
_Poke Param(2) : Return (Or(@(a@), Shl(1, b@)))
_Peek Param(1) : Return (And(@(a@/32), Shl(1, a@%32))>0)
Output:
First 15 numbers:
0       1       3       6       2       7       13      20      12      21      11      22      10      23      9       
First repetition: 42

0 OK, 0:459 

VBScript

Translation of: Rexx

To run in console mode with cscript.

' Recaman's sequence - vbscript - 04/08/2015
	nx=15
	h=1000
	Wscript.StdOut.WriteLine "Recaman's sequence for the first " & nx & " numbers:"
	Wscript.StdOut.WriteLine recaman("seq",nx)
	Wscript.StdOut.WriteLine "The first duplicate number is: " & recaman("firstdup",0)
	Wscript.StdOut.WriteLine "The number of terms to complete the range 0--->"& h &" is: "& recaman("numterm",h)
	Wscript.StdOut.Write vbCrlf&".../...": zz=Wscript.StdIn.ReadLine()
	
function recaman(op,nn)
	Dim b,d,h
	Set b = CreateObject("Scripting.Dictionary")
	Set d = CreateObject("Scripting.Dictionary")
    list="0" : firstdup=0
	if op="firstdup" then
		nn=1000 : firstdup=1
	end if
	if op="numterm" then
		h=nn : nn=10000000 : numterm=1
	end if
	ax=0  'a(0)=0
	b.Add 0,1  'b(0)=1
	s=0
	for n=1 to nn-1
        an=ax-n
		if an<=0 then 
			an=ax+n
		elseif b.Exists(an) then 
			an=ax+n
		end if
		ax=an  'a(n)=an
		if not b.Exists(an) then b.Add an,1  'b(an)=1
		if op="seq" then
			list=list&" "&an
		end if
		if firstdup then
			if d.Exists(an) then
				recaman="a("&n&")="&an
				exit function
			else
				d.Add an,1  'd(an)=1
			end if
		end if
		if numterm then
			if an<=h then
				if not d.Exists(an) then
					s=s+1
					d.Add an,1  'd(an)=1
				end if
				if s>=h then
					recaman=n
					exit function
				end if
			end if
		end if
	next 'n
	recaman=list
end function 'recaman
Output:
Recaman's sequence for the first 15 numbers:
0 1 3 6 2 7 13 20 12 21 11 22 10 23 9
The first duplicate number is: a(24)=42
The number of terms to complete the range 0--->1000 is: 328002

Visual Basic .NET

Translation of: C#
Imports System
Imports System.Collections.Generic

Module Module1
    Sub Main(ByVal args As String())
        Dim a As List(Of Integer) = New List(Of Integer)() From { 0 },
            used As HashSet(Of Integer) = New HashSet(Of Integer)() From { 0 },
            used1000 As HashSet(Of Integer) = used.ToHashSet(),
            foundDup As Boolean = False
        For n As Integer = 1 to Integer.MaxValue
            Dim nv As Integer = a(n - 1) - n
            If nv < 1 OrElse used.Contains(nv) Then nv += 2 * n
            Dim alreadyUsed As Boolean = used.Contains(nv) : a.Add(nv)
            If Not alreadyUsed Then used.Add(nv) : If nv > 0 AndAlso nv <= 1000 Then used1000.Add(nv)
            If Not foundDup Then
                If a.Count = 15 Then _
                    Console.WriteLine("The first 15 terms of the Recamán sequence are: ({0})", String.Join(", ", a))
                If alreadyUsed Then _
                    Console.WriteLine("The first duplicated term is a({0}) = {1}", n, nv) : foundDup = True
            End If
            If used1000.Count = 1001 Then _
                Console.WriteLine("Terms up to a({0}) are needed to generate 0 to 1000", n) : Exit For
        Next
    End Sub
End Module
Output:
The first 15 terms of the Recamán sequence are: (0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9)
The first duplicated term is a(24) = 42
Terms up to a(328002) are needed to generate 0 to 1000

Wren

Translation of: Kotlin
var a = [0]
var used = { 0: true }
var used1000 = { 0: true }
var foundDup = false
var n = 1
while (n <= 15 || !foundDup || used1000.count < 1001) {
    var next = a[n-1] - n
    if (next < 1 || used[next]) next = next + 2*n
    var alreadyUsed = used[next]
    a.add(next)
    if (!alreadyUsed) {
        used[next] = true
        if (next >= 0 && next <= 1000) used1000[next] = true
    }
    if (n == 14) System.print("The first 15 terms of the Recaman's sequence are:\n%(a)")
    if (!foundDup && alreadyUsed) {
        System.print("The first duplicated term is a[%(n)] = %(next)")
        foundDup = true
    }
    if (used1000.count == 1001) {
        System.print("Terms up to a[%(n)] are needed to generate 0 to 1000")
    }
    n = n + 1
}
Output:
The first 15 terms of the Recaman's sequence are:
[0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9]
The first duplicated term is a[24] = 42
Terms up to a[328002] are needed to generate 0 to 1000

zkl

fcn recamanW{  // -->iterator -->(n,a,True if a is a dup)
   Walker.tweak(fcn(rn,rp,d){
      n,p,a := rn.value, rp.value, p - n;
      if(a<=0 or d.find(a)) a+=2*n;
      d.incV(a); rp.set(a);
      return(rn.inc(),a,d[a]>1);
   }.fp(Ref(0),Ref(0),Dictionary()) )
}
print("First 15 members of Recaman's sequence: ");
recamanW().walk(15).apply("get",1).println();

n,a := recamanW().filter1("get",2);  // ie filter(a[n].dup)
println("First duplicate number in series is: a(%d) = %d".fmt(n,a));

rw,ns,n,a,dup := recamanW(),1000,0,0,0;
do{ n,a,dup=rw.next(); if(not dup and a<1000) ns-=1; }while(ns);
println("Range 0..1000 is covered by terms up to a(%,d)".fmt(n));
Output:
First 15 members of Recamans sequence: L(0,1,3,6,2,7,13,20,12,21,11,22,10,23,9)
First duplicate number in series is: a(24) = 42
Range 0..1000 is covered by terms up to a(328,002)