# Generate Recamán's sequence

Recamán's sequence (A005132) is a mathematical sequence, defined as such:

A(0) = 0
A(n) = A(n-1) - n if A(n-1) - n > 0 and is new, else
A(n) = A(n-1) + n


A pretty LaTex version of the above (might be more readable):

$$A(n) = \begin{cases}0 & \textrm{if } n = 0 \\ A(n-1) - n & \textrm{if } A(n-1) - n \textrm{ is positive and not already in the sequence} \\ % Seems more readable than %A(n-1) - n & \textrm{if } A(n-1) > n \wedge \not\exists m < n: A(m) = A(n-1)-n \\ A(n-1) + n & \textrm{otherwise} \end{cases}$$

The first few terms are 0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11

To clarify, is new means whether the number is already in the sequence.

Given an integer n, via function argument or STDIN, return the first n terms of the Recamán sequence.

This is a code-golf challenge, so shortest code wins.

• What does 'is new' mean? – Beta Decay Sep 11 '14 at 16:53
• If a number is new, it means it is not yet in the sequence. Just realized I have typed out the sequence wrong, give me a min to correct it. – James Williams Sep 11 '14 at 16:56
• Corrected the sequence. – James Williams Sep 11 '14 at 17:06
• Can you add the first values of the sequence? – proud haskeller Sep 11 '14 at 17:15
• Added the first few numbers! (And a link to its OEIS page) – James Williams Sep 11 '14 at 17:50

# CJam, 34 33 bytes

0ali{_W=_I-__0<4$@#)|@I+@?+}fI1>  Try it online. ### Example run $ cjam <(echo '0ali{_W=_I-__0<4$@#)|@I+@?+}fI1>') <<< 33 [0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 24 8 25 43 62 42 63 41 18 42 17 43 16 44 15 45 14 46]  ### How it works 0ali " Push S := [ 0 ] and read an integer N from STDIN. "; { }fI " For each I in [ 0 ... (N - 1) ]: "; _W= " X := S[-1]. "; _I- " Y := X - I "; _0< " A := (Y < 0) "; _ 4$@#)               "   B := (Y ∊ S)                                       ";
@I+           "   Z := X + I                                         ";
|   @?         "   C := (A || B) ? Z : Y                              ";
+        "   S += [C]                                           ";
1>  " Push str(S[1:]).                                     ";

• What change did you make? – Soham Chowdhury Sep 11 '14 at 19:55
• My first approach prepended negative numbers to the sequence, so I didn't have to explicitly check if A(i) - i > 0. However, I didn't prepend enough numbers for small values of n. Now, I just do exactly what the spec says. – Dennis Sep 11 '14 at 19:58
• 33 vs. 45. So close and yet so far. :) – Ingo Bürk Sep 12 '14 at 21:07
• Wow, comment without e# in Cjam... tasty cherry. – Chromium Jun 21 '18 at 8:04

l=0:0#1
a§v|a<0||aelemr v=v|1<2=0-v
a#b=a+(a-b)§b:l!!b#(b+1)
r=(takel)


Example usage:

λ> r 20
[0,1,3,6,2,7,13,20,12,21,11,22,10,23,9,24,8,25,43,62]


## Ruby, 71 70 bytes

f=->n{a=[0];(n-1).times{|i|a+=[[b=a[-1]-i-1]-a!=[]&&b>0?b:b+2*i+2]};a}


A very "word-for-word" implementation of the definition.

## Python 2, 787573 69 Bytes

Kudos to xnor and flornquake
Now almost 10 bytes shorter than the initial answer

m=p,=0,
exec"p+=1;k=m[-1]-p;m+=k+2*p*(k*(k>0)in m),;"*input()
print m

• You can shorten [k,k+2*p][bool] to k+2*p*(bool). – xnor Sep 11 '14 at 19:04
• @xnor Thanks, saved 3 Bytes. – Markuz Sep 11 '14 at 19:16
• Also, k in m or k<0 can be k*(k>=0)in m since if k<0, the product is 0, which is in m. – xnor Sep 11 '14 at 19:22
• @xnor Brilliant! Thanks again – Markuz Sep 11 '14 at 19:36
• You can write -1 instead of p-1. Edit: You can also make m a tuple and write m=0, and m+=k+2*p*(k*(k>0)in m),. – flornquake Sep 12 '14 at 13:14

# Golfscript (41 45)

Try it online here:

(,1,\{:~1$=~)-:^1<\.^?)!!@|^\{~)2*+}*+}/  ## Explanation This is for the original 45 bytes solution, but it's still pretty much the same: (, # push array [0 .. n-1] [0]\ # push sequence elements as [0] and reverse stack { # foreach element in [0 .. n-1] do: :m; # store current element in m and discard .m= # get the previous sequence element m)-:^ # subtract the current index from it and store in ^ 0> # is that number greater than 0? \.^?)! # is that number new to our sequence? @& # logically and both checks {^} # if true, push ^ {^m)2*+} # otherwise, add the index twice and push if + # add new element to our sequence }/  # make output pretty  Edit #1: Thanks to Dennis for shaving off 4 bytes. # dc, 46 bytes sn[z+z+d0r:a]sF0[pz-d1>Fd;a0<Fddd:azln!<M]dsMx  Try it online! This program takes input from an otherwise empty stack and outputs to stdout (newline delimited). I'm really proud of this one - it's beating everything that isn't a dedicated golfing language, and showcases three of my favorite dc golfing tricks: • Stack size used as an index variable • Refactoring "if A then B else C" into "unconditionally C, and if A then D" where C and D combine to make B • the little-used random access array feature to solve a uniqueness constraint # Explanation sn Stores the input in register n [z+z+0r:a]sF Defines the macro F, which: z+z+ adds twice the stack size/index variable 0r:a resets the "uniqueness" flag to 0 in the array a In context, F is the "D" in my description above, changing A(z-1)-z to A(z-1)+z 0 The main loop starts with the previous sequence member on top of the stack and total stack depth equal to the next index. Pushing a zero accomplishes both of these things. [ Start of the main loop M p Print the previous sequence member, with newline (no pop) z- Calculate A(z-1)-z d1>F If that's nonpositive, (F)ix it to be A(z-1)+z d;a a is my array of flags to see if we've hit this value before 0<F If we have, (F)ix it! (nonzero = flag, since ;a is zero by default, and also zero if we just (F)ixed and therefore don't care about uniqueness right now) ddd Make one copy to keep and two to eat :a Flag this entry as "used" in the uniqueness array a zln!<M If our "index variable" is n or less, repeat! ]dsMx End of main loop - store it and execute  • that's wild, i had no idea dc even existed – don bright Dec 31 '18 at 7:08 ## JavaScript - 81 80 79 70 Kudos to edc65 for helping me save 9 bytes f=n=>{for(a=[x=i=0];++i<n;)a[i]=x+=x>i&a.indexOf(x-i)<0?-i:i;return a}  • -9 :g=n=>{for(a=[x=i=0];++i<n;)a[i]=x+=x>i&a.indexOf(x-i)<0?-i:i;return a} – edc65 Sep 11 '14 at 18:48 • @edc65 Grazie mille :) – William Barbosa Sep 11 '14 at 19:22 ## JavaScript, ES6, 74 69 characters Run the below code in latest Firefox's Web Console. G=n=>(i=>{for(r=[t=0];++i<n;)r[i]=t+=i>t|~r.indexOf(t-i)?i:-i})(0)||r  Will try to golf it more later. Example usage: G(11) -> 0,1,3,6,2,7,13,20,12,21,11  # MATLAB, 83 78 Bytes Save the below as f.m (73 Bytes) A=0;for i=1:n-1 b=A(i)-i;A(i+1)=b+2*i;if b>0&&~any(A==b) A(i+1)=b;end;end  Run from command window (5 bytes) n=9;f  If the above is not legal, then it requires 90 bytes. function A=f(n) A=0;for i=1:n-1 b=A(i)-i;A(i+1)=b+2*i;if b>0&&~any(A==b) A(i+1)=b;end;end  # R: 96 characters Golfed: A=function(s,n,m,i){if(m==n){return(s)}else{t=i-m;if(t%in%s||t<0){t=i+m};s=c(s,t);A(s,n,m+1,t)}}  Ungolfed: A = function(s,n,m,i) { if(m==n){return(s)} else{ t=i-m if(t%in%s||t<0){t=i+m} s=c(s,t) A(s,n,m+1,t) } }  Sample Run: > An(0,34,1) [1] 0 1 3 6 2 7 13 20 12 21 11 22 10 23 9 24 8 [18] 25 43 62 42 63 41 18 42 17 43 16 44 15 45 14 46 79  # JavaScript, 63 bytes n=>(g=y=>n-x?g(a[++x]=a.includes(z=y-x)|z<0?+y+x:z):a)(a=[x=0])  Try it online # Perl 6, 62 57 bytes {(0,{$-@+@*2*($!>@||$-@∈@)given @[*-1]}...*)[^$]} {(0,{($!=@_[*-1])+@_-@_*2*($!>@_&&$!-@_∉@_)}...*)[^$_]}  -5 bytes thanks to Jo King Try it online! • that's amazing... that literally looks like my cat walked across my keyboard. – don bright Dec 31 '18 at 7:10 # 05AB1E, 19 bytes ¾ˆG¯¤N-DŠD0›*åN·*+ˆ  Try it online! Explanation ¾ˆ # Initialize the global list with 0 G # for N in [1, input-1] do: ¯ # push the global list ¤N- # subtract N from the last item in the list D # duplicate Š # move the copy down 2 spots on the stack D # duplicate again 0› # check if it is positive * # multiply, turning negative results to zero å # is the result already present in the list? N·* # multiply by N*2 + # add to the result ˆ # add this to the list  • How does this work? – lirtosiast Jan 26 '19 at 22:39 • @lirtosiast: Been a while since I did this challenge, so this is the best explanation I can do on short notice. Hope it's enough. – Emigna Jan 27 '19 at 15:53 # K (oK), 53 bytes Solution: {$[y>c:#x;o[x,(r;*|x+c)(r in x)|0>r:*|x-c;y];x]}[,0;]


Try it online!

Explanation:

Recursive solution.

{$[y>c:#x;o[x,(r;*|x+c)(r in x)|0>r:*|x-c;y];x]}[,0;] / the solution { }[,0;] / lambda with first arg set as list containing 0$[      ;                                  ; ]       / if[condition;true;false]
#x                                             / length of x
c:                                               / save as c
y>                                                 / y greater than? (ie have we produced enough results?)
x        / return x if we are done
o[                             ;y]          / recurse with new x and existing y
x-c             / subtract c from x
*|                / reverse first, aka last
r:                  / save result as r
0>                    / 0 greater than?
|                      / or
(      )                       / do together
r in x                        / r in x?
( ;     )                               / use result to index into this 2-item list
x+c                                / add c to x
*|                                   / reverse first, aka last
r                                      / result
x,                                        / append to x


## Java, 144

int[]f(int n){int[]a=new int[n];a[0]=0;int i,j,k,m;for(i=0;i<n-1;){k=a[i++]-i;m=0;for(j=0;j<i;)if(k==a[j++])m=1;a[i]=m<1&k>0?k:k+2*i;}return a;}


## Lua - 141 135 139 135

function s(n)a,b={1},{[0]=0}for i=1,n do k=b[i-1]-i c=k+i+i if(k>0)and(a[k]==nil)then b[i],a[k]=k,1 else b[i],a[c]=c,1 end end return b end


function s(n)
a,b={1},{[0]=0}
for i=1,n do
k=b[i-1]-i
c=k+i+i
if (k>0) and (a[k]==nil) then
b[i],a[k]=k,1
else
b[i],a[c]=c,1
end
end
return b
end


I use 2 tables, the first one is called a and it is built so that a[i]=1 iff i has already appeared in the sequence, nil otherwise, while the second table actually holds the sequence

• You're right, I didn't look at the question very carefully and assumed it had the same definition at mathworld (starting with 1), I think that won't cost any more character, I'll test and correct it later, I'm writing from my phone now! – user25169 Sep 11 '14 at 19:26

## Python, 73

def f(x,t=0):
if x:t=f(x-1);t+=2*x*(t*(t>0)in map(f,range(x)))
return t


Edit 1: Thanks to @xnor's tips on the other Python answer! (I just realised that both look very similar.)

Edit 2: Thanks again, @xnor.

• This gives an infinite loop. You need some sort of control flow so that f(x) doesn't always immediately call f(x-1). – xnor Sep 11 '14 at 19:25
• @xnor fixed the code. – Soham Chowdhury Sep 11 '14 at 19:36
• This seems to return the nth term, not the first n terms. – Dennis Sep 11 '14 at 19:55
• Some minor saves: t=0 can go as an optional parameter to f, and t=t+ can be t+=. – xnor Sep 11 '14 at 20:21

## Groovy : 122118 111 chars

Golfed:

m=args[0] as int
a=[0]
(1..m-1).each{n->b=a[n-1];x=b-n;(x>0&!(x in a))?a[n]=x:(a[n]=b+n)}
a.each{print "$it "}  Ungolfed: m = args[0] as int a = [0] (1..m-1).each { n-> b = a[n-1] x = b-n ( x>0 & !(x in a) ) ? a[n] = x : (a[n] = b+n) } a.each{print "$it "}


Sample Run:

bash$groovy Rec.groovy 14 0 1 3 6 2 7 13 20 12 21 11 22 10 23  ## Clojure : 174 chars Golfed: (defn f[m a](let[n(count a)b(last a)x(- b n)y(if(and(> x 0)(not(.contains a x)))x(+ b n))](if(= m n)a(f m(conj a y)))))(println(f(read-string(first *command-line-args*))[0]))  Ungolfed: (defn f[m a] (let [n (count a) b (last a) x (- b n) y (if (and (> x 0) (not (.contains a x))) x (+ b n)) ] (if (= m n) a (f m (conj a y))) ) ) (println (f (read-string (first *command-line-args*)) [0]) )  Sample run: bash$ java -jar clojure-1.6.0.jar rec.clj 14
[0 1 3 6 2 7 13 20 12 21 11 22 10 23]

• I suggest you not to read from STDIN but instead just take an integer argument to the function :) Also you don't get any benefits from defining y on the let form, you can use the expression directly where the value is needed. – NikoNyrh Jul 26 '19 at 11:07

From user perspective, Mathcad is effectively a 2D whiteboard, with expressions evaluated from left-to-right,top-to-bottom. Mathcad does not support a conventional "text" input, but instead makes use of a combination of text and special keys / toolbar / menu items to insert an expression, text, plot or component. For example, type ":" to enter the definition operator (shown on screen as ":=") or "ctl-shft-#" to enter the for loop operator (inclusive of placeholders for the iteration variable, iteration values and one body expression). What you see in the image above is exactly what appears on the user interface and as "typed" in.

For golfing purposes, the "byte" count is the equivalent number of keyboard operations required to enter an expression.

• That's all well and good, but what are the actual keystrokes? – Jo King Jun 18 '18 at 13:14

# C (gcc), 116 111 109 bytes

*f(n){int*a=calloc(4,n),i=0,j,k,m;for(;~i+n;a[i]=k+(m|k<1)*2*i)for(k=a[i++]-i,m=0,j=i;j--;)m=k-a[j]?m:1;n=a;}


Try it online!

# Stax, 19 bytes

É╖C8½ΔL▄░▬L+≡ΩSa⌂¼╧


Run and debug it

Unpacked, ungolfed, and commented, it looks like this. It keeps the sequence so far on the stack, and remembers A(n - 1) in the X register. The iteration index is used for n. The first time through, it's 0, but in that iteration it generates the 0 without any special cases, so there's no need to adjust for the off-by-1 index.

0X      push 0 to main stack and store it in X register, which will store A(n - 1)
z       push an empty array that will be used to store the sequence
,D      pop input from input stack, execute the rest of the program that many times
xi-Y  push (x-register - iteration-index) and store it in the Y register
this is (A(n - 1) - n)
0>    test if (A(n - 1) - n) is greater than 0 (a)
ny#   count number of times (A(n - 1) - n) occurs in the sequence so far (b)
>     test if (a) > (b)
y   (A(n - 1) - n)
xi+ A(n - 1) + n
?     if/else;  choose between the two values based on the condition
X     store the result in the X register
Q     print without popping
+     append to sequence array


Run and debug this one

• interesting. how does this work? – don bright Dec 31 '18 at 7:28
• @donbright: Added some annotations and explanation. – recursive Jan 2 '19 at 21:07

## Pyth, 31 bytes

VQ=+Y?Y?|>NeYhxY-eYN+eYN-eYNZ)Y


# Pyth, 24 bytes

tu+G-eG_W|g0J-eGH}JGHQ]0


Try it online!

tu+G-eG_W|g0J-eGH}JGHQ]0   Implicit: Q=eval(input())
u                   Q     Reduce [0-Q)...
]0   ... with initial value G=[0], next value as H:
eG             Last value of G (sequence so far)
-  H            Take H from the above
J                Store in J
g0J                0 >= J
}JG         Is J in G?
|                   Logical OR of two previous results
_W           H        If the above is true, negate H, otherwise leave as positive
-eG                      Subtract the above from last value in G
+G                         Append the above to G
The result of the reduction is the sequence with an extra leading 0
t                          Remove a leading 0, implicit print


### Powershell (103)

$n=Read-Host;$a=@(0);$n-=1;1..$n|%{$x=$a[-1]-$_;if($x-gt0-and!($a-like$x)){$a+=$x}else{$a+=$x+2*$_}};$a


Another 'word-for-word' implementation down here as well. Surprisingly readable for PowerShell, too.

Sequence is stored in the array $a, and printed out one term per line. For$n=20 if we run the statement $a-join"," we get 0,1,3,6,2,7,13,20,12,21,11,22,10,23,9,24,8,25,43,62  # C#: 140 characters int i,w,t,y;int[]F(int n){var r=new int[n--];for(;i<n;y=0){w=r[i++]-i;for(t=0;y<i&&t<1;)t=w==r[y++]?1:0;r[i]=w>0&&t<1?w:r[i-1]+i;}return r;}  # C++: 180 characters (158 without cin and cout statements) int a[5000000][2]={0},i,k,l;a[0][0]=0;a[0][1]=1;cin>>k;for(i=1;i<=k;i++){l=a[i-1][0];if(l-i>0&&a[l-i][1]!=1){ a[i][0]=l-i;a[l-i][1]=1;}else{ a[i][0]=l+i;a[l+i][1]=1;}cout<<a[i][0]<<endl;  • Welcome to Programming Puzzles & Code Golf Stack Exchange! Please edit the character/byte count of your solution into your header, as shown in the other answers here. Also, please golf your code (ex. remove whitespace to reduce the character count) as much as possible. Thanks! – Doorknob Aug 17 '15 at 16:00 • Sure thing, I'll do that. – Abhay Jain Aug 18 '15 at 17:41 # Mathematica - 81 bytes Fold[#~Append~(#[[-1]]+If[#[[-1]]>#2&&FreeQ[#,#[[-1]]-#2],-#2,#2])&,{0},Range@#]&  ## Usage Fold[#~Append~(#[[-1]]+If[#[[-1]]>#2&&FreeQ[#,#[[-1]]-#2],-#2,#2])&,{0},Range@#]&[30] {0,1,3,6,2,7,13,20,12,21,11,22,10,23,9,24,8,25,43,62,42,63,41,18,42,17,43,16,44,15,45}  # PHP, 89 bytes $f=function($n){for(;$i<$n;$s[$r[$i++]=$p=$m]=1)if($s[$m=$p-$i]|0>$m)$m=$p+$i;return$r;};  Try it online! Ungolfed: $f = function ($n) { for (;$i < $n;$s[$r[$i++] = $p =$m] = 1) {
if ($s[$m = $p -$i] | 0 > $m) {$m = $p +$i;
}
}

return $r; };  • $r for my result
• $s for tracking seens • $p previous value
• \$m mext value

## Common LISP (139 bytes)

(defun r(n)(do*(s(i 0(1+ i))(a 0(car s))(b 0(- a i)))((> i n)(nreverse s))(push(cond((= 0 i)0)((and(> b 0)(not(find b s)))b)(t(+ a i)))s)))


Ungolfed:

(defun recaman (n)
(do*
(series               ; starts as empty list
(i 0 (1+ i))         ; index variable
(last 0 (car s))     ; last number in the series
(low 0 (- last i)))

((> i n)              ; exit condition
(nreverse series))   ; return value

(push                ; loop body
(cond
((= 0 i) 0)       ; first pass
((and
(> low 0) (not (find low s)))
low)
(t (+ last i)))
series)))