A064387 -id:A064387 - cp's OEIS frontend

A005132 Recamán's sequence (or Recaman's sequence): a(0) = 0; for n > 0, a(n) = a(n-1) - n if nonnegative and not already in the sequence, otherwise a(n) = a(n-1) + n.

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, 79, 113, 78, 114, 77, 39, 78, 38, 79, 37, 80, 36, 81, 35, 82, 34, 83, 33, 84, 32, 85, 31, 86, 30, 87, 29, 88, 28, 89, 27, 90, 26, 91, 157, 224, 156, 225, 155

Offset: 0

N. J. A. Sloane and Simon Plouffe, May 16 1991

The name "Recamán's sequence" is due to N. J. A. Sloane, not the author!
I conjecture that every number eventually appears - see A057167, A064227, A064228. - N. J. A. Sloane. That was written in 1991. Today I'm not so sure that every number appears. - N. J. A. Sloane, Feb 26 2017
As of Jan 25 2018, the first 13 missing numbers are 852655, 930058, 930557, 964420, 966052, 966727, 969194, 971330, 971626, 971866, 972275, 972827, 976367, ... For further information see the "Status Report" link. - Benjamin Chaffin, Jan 25 2018
From David W. Wilson, Jul 13 2009: (Start)
The sequence satisfies [1] a(n) >= 0, [2] |a(n)-a(n-1)| = n, and tries to avoid repeats by greedy choice of a(n) = a(n-1) -+ n.
This "wants" to be an injection on N = {0, 1, 2, ...}, as it attempts to avoid repeats by choice of a(n) = a(n-1) + n when a(n) = a(n-1) - n is a repeat.
Clearly, there are injections satisfying [1] and [2], e.g., a(n) = n(n+1)/2.
Is there a lexicographically earliest injection satisfying [1] and [2]? (End)
Answer: Yes, of course: The set of injections satisfying [1] and [2] is not empty, so there's a lexicographically least element. More concretely, it starts with the same 23 terms a(0..22) which are known to be minimal, but after a(22) = 41 it has to go on with a(23) = 41 + 23 = 64, since choosing "-" here necessarily yields a non-injective sequence. See A171884. - M. F. Hasler, Apr 01 2019
It appears that a(n) is also the value of "x" and "y" of the endpoint of the L-toothpick structure mentioned in A210606 after n-th stage. - Omar E. Pol, Mar 24 2012
Of course this is not a permutation of the integers: the first repeated term is 42 = a(24) = a(20). - M. F. Hasler, Nov 03 2014. Also 43 = a(18) = a(26). - Jon Perry, Nov 06 2014
Of all the sequences in the OEIS, this one is my favorite to listen to. Click the "listen" button at the top, set the instrument to "103. FX 7 (Echoes)", click "Save", and open the MIDI file with a program like QuickTime Player 7. - N. J. A. Sloane, Aug 08 2017
This sequence cycles clockwise around the OEIS logo. - Ryan Brooks, May 09 2020

			Consider n=6. We have a(5)=7 and try to subtract 6. The result, 1, is certainly positive, but we cannot use it because 1 is already in the sequence. So we must add 6 instead, getting a(6) = 7 + 6 = 13.

Alex Bellos and Edmund Harriss, Visions of the Universe (2016), Unnumbered pages. Includes Harriss's illustration of the first 65 steps drawn as a spiral.
Benjamin Chaffin, N. J. A. Sloane, and Allan Wilks, On sequences of Recaman type, paper in preparation, 2006.
Bernardo Recamán Santos, letter to N. J. A. Sloane, Jan 29 1991
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, The first 100000 terms
Farid Aliniaeifard and Shu Xiao Li, Peak algebra in noncommuting variables, arXiv:2506.12868 [math.CO], 2025. See p. 45.
Alex Bellos and Brady Haran, The Slightly Spooky Recamán Sequence, Numberphile video, 2018.
Alex Bellos and Brady Haran, Edmund Harriss's illustration of first 62 steps drawn as a spiral, snapshot from Numberphile video "The Slightly Spooky Recamán Sequence" (2018) at 2:37 minutes. [Included with permission of the authors] See also the Harriss link below.
Harlan Brothers, Recamán's Blues (a sonification Recamán's sequence), Animation, Jun 8, 2024.
Benjamin Chaffin, Log-log plot of first 10^230 terms
Benjamin Chaffin, Status Report, Jan 25 2018.
Fabio Deelan Cunden, Marilena Ligabò, and Tommaso Monni, Random matrices associated to Young diagrams, arXiv:2301.13555 [math.PR], 2023. See p. 7.
Michael De Vlieger, video of first 1200 steps of the Recamán sequence, with audio accompaniment generated by aspects of the sequence. Oct 12, 2019.
GBnums, A nice OEIS sequence
Gordon Hamilton, Wrecker Ball Sequences, Video, 2013
Edmund Harriss, The first 65 steps drawn as a spiral
Nick Hobson, Python program for this sequence
Bradley Klee, Code for pdf spiral drawing (golang)
Bradley Klee, The first 65 steps drawn as a spiral (color)
Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625, Dec 08, 2020
C. L. Mallows, Plot (jpeg) of first 10000 terms
C. L. Mallows, Plot (postscript) of first 10000 terms
Joseph Samuel Myers, Richard Schroeppel, Scott R. Shannon, N. J. A. Sloane, and Paul Zimmermann, Three Cousins of Recaman's Sequence, arXiv:2004:14000 [math.NT], April 2020.
Tilman Piesk, First 172 items in a coordinate system [This is a graph of the start of A005132 rotated clockwise by 90 degs. - _N. J. A. Sloane_, Mar 23 2012]
Omar E. Pol, Illustration of initial terms of A001057, A005132, A000217, 2012.
Omar E. Pol, Illustration of initial terms, 2012.
Omar E. Pol, Lateral view of a 3D-model whose front view is formed by spirals, 2022 (using Plot 2 A005132 vs A000004)
Bernardo Recamán Santos and N. J. A. Sloane, Correspondence, 1991.
Scott R. Shannon, Illustration of the first 2000 terms plotted as steps on a 2D square (Ulam) spiral. The colors are graduated across the spectrum from red to violet to show the relative step order.
Muhammad Khubab Siddique, Sequence and Series-I, Unit 8, Mathematics-II, Dept. of Sci. Ed., Allama Iqbal Open Univ. (Islamabad, Pakistan, 2020), 281-313.
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
N. J. A. Sloane, FORTRAN program for A005132, A057167, A064227, A064228
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, pp. 10, 12-13.
Alex van den Brandhof, Een bizarre rij, Pythagoras, 55ste Jaargang, Nummer 2, Nov 2015 (Article in Dutch about this sequence, see page 19 and back cover).
Index entries for sequences related to Recamán's sequence

Cf. A057165 (addition steps), A057166 (subtraction steps), A057167 (steps to hit n), A008336, A046901 (simplified version), A064227 (records for reaching n), A064228 (value of n that take a record number of steps to reach), A064284 (no. of times n appears), A064290 (heights of terms), A064291 (record highs), A119632 (condensed version), A063733, A079053, A064288, A064289, A064387, A064388, A064389, A228474 (bidirectional version).
A065056 gives partial sums, A160356 gives first differences.
A row of A066201.
Cf. A171884 (injective variant).
See A324784, A324785, A324786 for the "low points".
Cf. A330791, A331659, A331670.

import Data.Set (Set, singleton, notMember, insert)
a005132 n = a005132_list !! n
a005132_list = 0 : recaman (singleton 0) 1 0 where
   recaman :: Set Integer -> Integer -> Integer -> [Integer]
   recaman s n x = if x > n && (x - n) `notMember` s
                      then (x-n) : recaman (insert (x-n) s) (n+1) (x-n)
                      else (x+n) : recaman (insert (x+n) s) (n+1) (x+n)
-- Reinhard Zumkeller, Mar 14 2011

function a=A005132(m)
% m=max number of terms in a(n). Offset n:0
a=zeros(1,m);
for n=2:m
    B=a(n-1)-(n-1);
    C=0.^( abs(B+1) + (B+1) );
    D=ismember(B,a(1:n-1));
    a(n)=a(n-1)+ (n-1) * (-1)^(C + D -1);
end
% Adriano Caroli, Dec 26 2010

h := array(1..100000); maxt := 100000; a := [1]; ad := [1]; su := []; h[1] := 1; for nx from 2 to 500 do t1 := a[nx-1]-nx; if t1>0 and h[t1] <> 1 then su := [op(su), nx]; else t1 := a[nx-1]+nx; ad := [op(ad), nx]; fi; a := [op(a),t1]; if t1 <= maxt then h[t1] := 1; fi; od: # a is A005132, ad is A057165, su is A057166
A005132 := proc(n)
    option remember; local a, found, j;
    if n = 0 then return 0 fi;
    a := procname(n-1) - n ;
    if a <= 0 then return a+2*n fi;
    found := false;
    for j from 0 to n-1 while not found do
        found := procname(j) = a;
    od;
    if found then a+2*n else a fi;
end:
seq(A005132(n), n=0..70); # R. J. Mathar, Apr 01 2012 (reformatted by Peter Luschny, Jan 06 2019)

a = {1}; Do[ If[ a[ [ -1 ] ] - n > 0 && Position[ a, a[ [ -1 ] ] - n ] == {}, a = Append[ a, a[ [ -1 ] ] - n ], a = Append[ a, a[ [ -1 ] ] + n ] ], {n, 2, 70} ]; a
(* Second program: *)
f[s_List] := Block[{a = s[[ -1]], len = Length@s}, Append[s, If[a > len && !MemberQ[s, a - len], a - len, a + len]]]; Nest[f, {0}, 70] (* Robert G. Wilson v, May 01 2009 *)
RecamanSeq[i_Integer] := Fold[With[{lst=Last@#, len=Length@#}, Append[#, If[lst > len && !MemberQ[#, lst - len], lst - len, lst + len]]] &, {0}, Range@i]; RecamanSeq[10^5] (* Mikk Heidemaa, Nov 02 2024 *)

a(n)=if(n<2,1,if(abs(sign(a(n-1)-n)-1)+setsearch(Set(vector(n-1,i,a(i))),a(n-1)-n),a(n-1)+n,a(n-1)-n))  \\ Benoit Cloitre

A005132(N=1000,show=0)={ my(s,t); for(n=1,N, s=bitor(s,1<M. F. Hasler, May 11 2008, updated M. F. Hasler, Nov 03 2014

l=[0]
for n in range(1, 101):
    x=l[n - 1] - n
    if x>0 and not x in l: l+=[x, ]
    else: l+=[l[n - 1] + n]
print(l) # Indranil Ghosh, Jun 01 2017

def recaman(n):
  seq = []
  for i in range(n):
    if(i == 0): x = 0
    else: x = seq[i-1]-i
    if(x>=0 and x not in seq): seq+=[x]
    else: seq+=[seq[i-1]+i]
  return seq
print(recaman(1000)) # Ely Golden, Jun 14 2018

from itertools import count, islice
def A005132_gen(): # generator of terms
    a, aset = 0, set()
    for n in count(1):
        yield a
        aset.add(a)
        a = b if (b:=a-n)>=0 and b not in aset else a+n
A005132_list = list(islice(A005132_gen(),30)) # Chai Wah Wu, Sep 15 2022

a(k) = A000217(k) - 2*Sum_{i=1..n} A057166(i), for A057166(n) <= k < A057166(n+1). - Christopher Hohl, Jan 27 2019

Allan Wilks, Nov 06 2001, computed 10^15 terms of this sequence. At this point all the numbers below 852655 had appeared, but 852655 = 5*31*5501 was missing.
After 10^25 terms of A005132 the smallest missing number is still 852655. - Benjamin Chaffin, Jun 13 2006
Even after 7.78*10^37 terms, the smallest missing number is still 852655. - Benjamin Chaffin, Mar 28 2008
Even after 4.28*10^73 terms, the smallest missing number is still 852655. - Benjamin Chaffin, Mar 22 2010
Even after 10^230 terms, the smallest missing number is still 852655. - Benjamin Chaffin, 2018
Changed "positive" in definition to "nonnegative". - N. J. A. Sloane, May 04 2020

A064389 Variation (4) on Recamán's sequence (A005132): to get a(n), we first try to subtract n from a(n-1): a(n) = a(n-1) - n if positive and not already in the sequence; if not then we try to add n: a(n) = a(n-1) + n if not already in the sequence; if this fails we try to subtract n+1 from a(n-1), or to add n+1 to a(n-1), or to subtract n+2, or to add n+2, etc., until one of these produces a positive number not already in the sequence - this is a(n).

Original entry on oeis.org

1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 25, 43, 62, 42, 63, 41, 18, 44, 19, 45, 72, 100, 71, 101, 70, 38, 5, 39, 4, 40, 77, 115, 76, 36, 78, 120, 163, 119, 74, 28, 75, 27, 79, 29, 80, 132, 185, 131, 186, 130, 73, 15, 81, 141, 202

Offset: 1

N. J. A. Sloane, Sep 28 2001

This is the nicest of these variations. Is this a permutation of the natural numbers?
The number of steps before n appears is the inverse series, A078758. The height of n is in A126712.
See A078758 for the inverse permutation (in case this is a permutation of the positive integers). - M. F. Hasler, Nov 03 2014
After 10^12 terms, the smallest number which has not appeared is 5191516. - Benjamin Chaffin, Oct 09 2016

Suggested by J. C. Lagarias.

Cf. A005132, A046901, A064387, A064388. Agrees with A064387 for first 187 terms, then diverges.

h := array(1..100000); maxt := 100000; a := array(1..1000); a[1] := 1; h[1] := 1; for nx from 2 to 1000 do for i from 0 to 100 do t1 := a[nx-1]-nx-i; if t1>0 and h[t1] <> 1 then a[nx] := t1; if t1 < maxt then h[t1] := 1; fi; break; fi; t1 := a[nx-1]+nx+i; if h[t1] <> 1 then a[nx] := t1; if t1 < maxt then h[t1] := 1; fi; break; fi; od; od; evalm(a);

h[1] = 1; h[] = 0; maxt = 100000; a[1] = 1; a[] = 0; For[nx = 2, nx <= 1000, nx++, For[i = 0, i <= 100, i++, t1 = a[nx - 1] - nx - i; If[t1 > 0 && h[t1] != 1, a[nx] = t1; If[t1 < maxt, h[t1] = 1]; Break[]]; t1 = a[nx - 1] + nx + i; If[h[t1] != 1, a[nx] = t1; If[t1 < maxt, h[t1] = 1]; Break[]]]]; Table[a[n], {n, 1, 100}](* Jean-François Alcover, May 09 2012, after Maple *)

A064389(n=1000,show=0)={ my(k,s,t); for(n=1,n, k=n; while( !(t>k && !bittest(s, t-k) && t-=k) && !(!bittest(s, t+k)  && t+=k), k++); s=bitor(s,1<M. F. Hasler, Nov 03 2014

A064388 Variation (3) on Recamán's sequence (A005132): set s (the step size) initially equal to 2; to get a(n), we first try to subtract s from a(n-1): a(n) = a(n-1)-s if positive and not already in the sequence, in which case we change s to s+1; if not, a(n) = a(n-1)+s+i, where i >= 0 is the smallest number such that a(n-1)+s+i has not already appeared and now we change s to s+i+1.

Original entry on oeis.org

1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 25, 43, 62, 42, 63, 41, 18, 44, 17, 45, 16, 46, 15, 47, 14, 48, 83, 119, 82, 120, 81, 121, 80, 38, 84, 37, 85, 36, 86, 35, 87, 34, 88, 33, 89, 32, 90, 31, 91, 30, 92, 29, 93, 28, 94, 27, 95, 26, 96, 167, 239, 166, 240

Offset: 1

N. J. A. Sloane, Sep 28 2001

Variation (4) (A064389) is the nicest of these variations.

I would also like to get the following sequences: number of steps before n appears (or 0 if n never appears), list of numbers that never appear, height of n (cf. A064288, A064289, A064290), etc.

Index entries for sequences related to Recamán's sequence

Cf. A005132, A046901, A064387, A064389.

rem Chipmunk BASIC v3.6.4(b8) http://www.nicholson.com/rhn/basic/
max=1000 : dim a(max)
s=2 : z=1 : a(z)=1 : print str$(z)+",";
for n=1 to 200
x=z-s : if x <= 0 then goto yyy
if a(x)=0 then a(x)=1 : print str$(x)+","; : s=s+1 : z=x : goto xxx
yyy: for i=0 to max
x=z+s+i
if a(x)=0 then a(x)=1 : print str$(x)+","; : s=s+i+1 : z=x : goto xxx
next i
xxx: next n
print : end
rem Jeremy Gardiner, Feb 22 2014

More terms from David Wasserman, Jul 16 2002

A277558 A variation on Recamán's sequence (A005132): to get a(n), we first try to subtract n from a(n-1): a(n) = a(n-1)-n if positive and not already in the sequence; if not then a(n) = a(n-1)+n-i, where i >= 0 is the smallest number such that a(n-1)+n-i has not already appeared.

Original entry on oeis.org

0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 25, 43, 62, 42, 63, 41, 18, 40, 15, 39, 66, 38, 67, 37, 68, 36, 69, 35, 70, 34, 71, 33, 72, 32, 73, 31, 74, 30, 75, 29, 76, 28, 77, 27, 78, 26, 79, 133, 188, 132, 189, 131, 190, 130, 191, 129, 192

Offset: 0

Benjamin Chaffin, Oct 19 2016

nonn

Is it ever impossible to extend the sequence -- meaning there is no number less than a(n-1)+n which has not appeared?

After 10^11 terms, the smallest number which has not appeared is 609790506.

			a(23) = 18. To get a(24) we try 18-24, but that is negative; so we try 18+24 = 42, but 42 has already appeared; so we try 18+24-1, but 41 has also already appeared; so we try 18+24-2. 40 is positive and has not yet appeared, and so a(24) = 40.

Benjamin Chaffin, Table of n, a(n) for n = 0..10000

Cf. A005132, A064387 (chooses a(n-1)+n+i instead of a(n-1)+n-i).

A350578 a(0) = 0; for n > 0, if a(n-1) - n >= 0 and a(n-1) - n has appeared the same or fewer times than a(n-1) + n, then a(n) = a(n-1) - n. Otherwise a(n) = a(n-1) + n.

Original entry on oeis.org

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, 79, 113, 78, 114, 77, 39, 0, 40, 81, 123, 80, 36, 81, 35, 82, 34, 83, 33, 84, 32, 85, 31, 86, 30, 87, 29, 88, 28, 89, 27, 90, 26, 91, 157, 224, 156, 225, 155, 226, 154, 227

Offset: 0

Scott R. Shannon, Jan 07 2022

nonn

This sequence is similar to the Recamán sequence A005132 except that here the number chosen to step to, if a(n-1) - n is >= 0, is determined by which of the two possible numbers, a(n-1) - n or a(n-1) + n, has appeared the fewest times. If the number of times these numbers have appeared is the same then a(n) = a(n-1) - n. If a(n-1) - n is less than zero then a(n) = a(n-1) + n.
The first term to differ from A005132 is a(39) = 0. See the examples below. In the first 10^8 terms the largest value is a(99928118) = 842174195, while the smallest number not to have appeared is 366. It is likely all numbers eventually appear although this is unknown.
See the companion sequence A350579 for the first number to appear n times.

			a(3) = 6 as a(2) = 3 giving the two possible numbers for the next step as 0 or 6. But 0 has already appeared once while 6 has not yet appeared, so 6 is chosen.
a(39) = 0 as a(38) = 39 giving the two possible numbers for the next step as 0 or 78. Both 0 and 78 have already appeared once thus the smaller is chosen. This is the first term to differ from A005132.
a(447) = 934 as a(446) = 487 giving the two possible numbers for the next step as 40 or 934. 40 has already appeared twice while 934 has appeared once, so the later is chosen. This is the first time both possible numbers have already appeared and the larger is chosen.
a(2462) = 551 as a(2461) = 3013 giving the two possible numbers for the next step as 551 or 5475. 551 has already appeared once while 5475 has appeared twice, so the former is chosen. This is the first time both possible numbers have already appeared and the smaller is chosen.

Scott R. Shannon, Image of the first 10^8 terms.
Index entries for sequences related to Recamán's sequence

Cf. A005132, A350579, A335299, A064389, A261573, A064387.

a[0]=0;a[n_]:=a[n]=If[a[n-1]-n>=0&&Count[Array[a,n-1,0],a[n-1]-n]<=Count[Array[a,n-1,0],a[n-1]+n],a[n-1]-n,a[n-1]+n];Array[a,74,0] (* Giorgos Kalogeropoulos, Jan 07 2022 *)

from itertools import count, islice
from collections import Counter
def A350578_gen(): # generator of terms
    yield 0
    b, bcounter = 0, Counter({0})
    for n in count(1):
        b += -n if b-n >= 0 and bcounter[b-n] <= bcounter[b+n] else n
        bcounter[b] += 1
        yield b
A350578_list = list(islice(A350578_gen(),30)) # Chai Wah Wu, Jan 08 2022

A350579 The first number in A350578 to appear n times.

Original entry on oeis.org

0, 42, 153, 504, 921, 921, 17729, 17729, 17729, 17729, 60610, 60610, 109617, 109617, 109617, 109617, 109617, 109617, 109599, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 658778, 2184074, 2184074, 2184074

Offset: 1

Scott R. Shannon, Jan 07 2022

nonn

See A350578 for further details.

			a(1) = 0 as A350578(0) = 0, thus 0 is the first number to appear one time.
a(2) = 42 as A350578(20) = A350578(24) = 42, thus 42 is the first number to appear two times. This is also true for A005132.
a(3) = 153 as A350578(74) = A350578(78) = A350578(114) = 153, thus 153 is the first number to appear three times.

Index entries for sequences related to Recamán's sequence

Cf. A350578, A005132, A335299, A064389, A261573, A064387.

a[0]=0;a[n_]:=a[n]=If[a[n-1]-n>=0&&Count[Array[a,n-1,0],a[n-1]-n]<=Count[Array[a,n-1,0],a[n-1]+n],a[n-1]-n,a[n-1]+n];
Table[k=0;While[Max[Last/@(c=Tally@Array[a,++k,0])]!=i];a[k-1],{i,6}] (* Giorgos Kalogeropoulos, Jan 07 2022 *)

from itertools import count
from collections import Counter
def A350579(n):
    b, bcounter = 0, Counter({0})
    for m in count(1):
        if bcounter[b] == n: return b
        b += -m if b-m >= 0 and bcounter[b-m] <= bcounter[b+m] else m
        bcounter[b] += 1 # Chai Wah Wu, Jan 08 2022

A383730 a(0) = 0, a(n) = a(n-1) + A002260(n) * (-1)^(n-1) if not already in the sequence, otherwise a(n) = a(n-1) - A002260(n) * (-1)^(n-1).

Original entry on oeis.org

0, 1, 2, 4, 3, 5, 8, 9, 7, 10, 6, 5, 7, 4, 8, 13, 12, 14, 11, 15, 20, 26, 25, 27, 24, 28, 23, 29, 22, 21, 19, 16, 20, 15, 21, 14, 22, 21, 23, 20, 24, 19, 25, 32, 40, 49, 48, 50, 47, 51, 46, 52, 45, 53, 44, 54, 55, 57, 60, 64, 59, 65, 58, 66, 75, 85, 74, 73, 71

Offset: 0

Markel Zubia, May 06 2025

This sequence is a bidirectional form of Recamán's sequence.
Another way to define the sequence: starting at 0, take steps of size 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, ... alternating left and right while avoiding repeated values (negative values are allowed).
The sequence is unbounded either above or below.
Conjecture: the sequence is unbounded both above and below.
Conjecture: each integer appears finitely often.
It exhibits a mix of chaotic and periodic behavior, including long plateaus and sudden large jumps.
Around term 25000, the sequence settles near -3000 in a visually fractal structure. After ~1.85 million terms, it appears to settle again near -500000. Astonishingly, after ~66.7 million steps, it jumps sharply from around -500000 to +4.51 million.

			a(1) = 0 + 1.
a(2) = 0 + 1 + 1 = 2, since 0 + 1 - 1 = 0 already appears in the sequence, as a(0) = 0.
a(6) = 0 + 1 + 1 + 2 - 1 + 2 + 3 = 8.

Markel Zubia, Table of n, a(n) for n = 0..9999
Markel Zubia, Plot of the first 100k terms
Markel Zubia, Close-up of the fractal-like pattern
Markel Zubia, Plot of the first 10M terms
Markel Zubia, Plot of the first 1B terms

Cf. A005132.
Other bidirectional extensions of Recamán's sequence: A063733, A079053, A064288, A064289, A064387, A064388, A064389, A228474.
Cf. A002260.

def a(n):
    t, k, curr = 1, 1, 0
    seen = set()
    for i in range(n):
        seen.add(curr)
        step = (-1)**i * t
        if curr + step not in seen:
            curr = curr + step
        else:
            curr = curr - step
        t += 1
        if t > k:
            t, k = 1, k + 1
    return curr

cp's OEIS Frontend

A005132 Recamán's sequence (or Recaman's sequence): a(0) = 0; for n > 0, a(n) = a(n-1) - n if nonnegative and not already in the sequence, otherwise a(n) = a(n-1) + n.

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A277558 A variation on Recamán's sequence (A005132): to get a(n), we first try to subtract n from a(n-1): a(n) = a(n-1)-n if positive and not already in the sequence; if not then a(n) = a(n-1)+n-i, where i >= 0 is the smallest number such that a(n-1)+n-i has not already appeared.

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A350578 a(0) = 0; for n > 0, if a(n-1) - n >= 0 and a(n-1) - n has appeared the same or fewer times than a(n-1) + n, then a(n) = a(n-1) - n. Otherwise a(n) = a(n-1) + n.

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A350579 The first number in A350578 to appear n times.

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A383730 a(0) = 0, a(n) = a(n-1) + A002260(n) * (-1)^(n-1) if not already in the sequence, otherwise a(n) = a(n-1) - A002260(n) * (-1)^(n-1).

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