A171884 -id:A171884 - 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

A048473 a(0)=1, a(n) = 3a(n-1) + 2; a(n) = 23^n - 1.

Original entry on oeis.org

1, 5, 17, 53, 161, 485, 1457, 4373, 13121, 39365, 118097, 354293, 1062881, 3188645, 9565937, 28697813, 86093441, 258280325, 774840977, 2324522933, 6973568801, 20920706405, 62762119217, 188286357653, 564859072961, 1694577218885, 5083731656657, 15251194969973, 45753584909921

Offset: 0

Clark Kimberling

The number of triangles (of all sizes, including holes) in Sierpiński's triangle after n inscriptions. - Lee Reeves, May 10 2004
The sequence is not only related to Sierpiński's triangle, but also to "Floret's cube" and the quaternion factor space Q X Q / {(1,1), (-1,-1)}. It can be written as a_n = ves((A+1)x)^n) as described at the Math Forum Discussions link. - Creighton Dement, Jul 28 2004
Relation to C(n) = Collatz function iteration using only odd steps: If we look for record subsequences where C(n) > n, this subsequence starts at 2^n - 1 and stops at the local maximum of 2*3^n - 1. Examples: [3,5], [7,11,17], [15,23,35,53], ..., [127,191,287,431,647,971,1457]. - Lambert Klasen, Mar 11 2005
Group the natural numbers so that the (2n-1)-th group sum is a multiple of the (2n)-th group containing one term. (1,2),(3),(4,5,6,7,8,9,10,11),(12),(13,14,15,16,17,18,19,...,38),(39),(40,41,...,118,119),(120), (121,122,123,...) ... a(n) = {the sum of the terms of (2n-1)-th group}/{the term of (2n)th group}. The first term of the odd numbered group is given by A003462. The only term of even numbered group is given by A029858. - Amarnath Murthy, Aug 01 2005
a(n)+1 = A008776(n); it appears that this gives the number of terms in the (n+1)-th "gap" of numbers missing in A171884. - M. F. Hasler, May 09 2013
Sum of n-th row of triangle of powers of 3: 1; 1 3 1; 1 3 9 3 1; 1 3 9 27 9 3 1; ... - Philippe Deléham, Feb 23 2014
For n >= 3, also the number of dominating sets in the n-helm graph. - Eric W. Weisstein, May 28 2017
The number of elements of length <= n in the free group on two generators. - Anton Mellit, Aug 10 2017
In general, a first order inhomogeneous recurrence of the form s(0) = a, s(n) = m*s(n-1) + k, n>0, will have a closed form of a*m^n + ((m^n-1)/(m-1))*k. - Gary Detlefs, Jun 07 2024

			a(0) = 1;
a(1) = 1 + 3 + 1 = 5;
a(2) = 1 + 3 + 9 + 3 + 1 = 17;
a(3) = 1 + 3 + 9 + 27 + 9 + 3 + 1 = 53; etc. - _Philippe Deléham_, Feb 23 2014

Theoni Pappas, Math Stuff, Wide World Publ/Tetra, San Carlos CA, page 15, 2002.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Creighton Dement, A paper on floretions and quaternions, some questions, The Math Forum.
Eric Weisstein's World of Mathematics, Dominating Set.
Eric Weisstein's World of Mathematics, Helm Graph.
Index entries for linear recurrences with constant coefficients, signature (4,-3).

a(n)=T(2,n), array T given by A048471.
Cf. A003462, A029858. A column of A119725.
Cf. A008776, A010684, A046055, A112468, A112739, A134347, A171884.

[2*3^n - 1: n in [0..30]]; // Vincenzo Librandi, Sep 23 2011

g:= ((1+x)/(1-3*x)/(1-x)): gser:=series(g, x=0, 43): seq(coeff(gser, x, n), n=0..30); # Zerinvary Lajos, Jan 11 2009; typo fixed by Marko Mihaily, Mar 07 2009

NestList[3 # + 2 &, 1, 30] (* Harvey P. Dale, Mar 06 2012 *)
LinearRecurrence[{4, -3}, {1, 5}, 30] (* Harvey P. Dale, Mar 06 2012 *)
Table[2 3^n - 1, {n, 20}] (* Eric W. Weisstein, May 28 2017 *)
2 3^Range[20] - 1 (* Eric W. Weisstein, May 28 2017 *)

first(m)=vector(m,n,n--;2*3^n - 1) \\ Anders Hellström, Dec 11 2015

n-th difference of a(n), a(n-1), ..., a(0) is 2^(n+1) for n=1, 2, 3, ...
a(0)=1, a(n) = a(n-1) + 3^n + 3^(n-1). - Lee Reeves, May 10 2004
a(n) = (3^n + 3^(n+1) - 2)/2. - Creighton Dement, Jul 31 2004
(1, 5, 17, 53, 161, ...) = Ternary (1, 12, 122, 1222, 12222, ...). - Gary W. Adamson, May 02 2005
Row sums of triangle A134347. Also, binomial transform of A046055: (1, 4, 8, 16, 32, 64, ...); and double binomial transform of A010684: (1, 3, 1, 3, 1, 3, ...). - Gary W. Adamson, Oct 21 2007
G.f.: (1+x)/((1-3*x)*(1-x)). - Zerinvary Lajos, Jan 11 2009; corrected by R. J. Mathar, Jan 21 2009
a(0)=1, a(1)=5, a(n) = 4*a(n-1) - 3*a(n-2). - Harvey P. Dale, Mar 06 2012
a(n) = Sum_{k=0..n} A112468(n,k)*4^k. - Philippe Deléham, Feb 23 2014
E.g.f.: exp(x)*(2*exp(2*x) - 1). - Elmo R. Oliveira, Mar 08 2025

Better description from Amarnath Murthy, May 27 2001

A308021 Start at n=1. Fill in a(n) with a value d > 0 not used earlier such that n-d or n+d is the smallest possible index not visited earlier, then continue with that index.

Original entry on oeis.org

1, 2, 5, 3, 7, 10, 4, 6, 12, 15, 17, 8, 20, 9, 22, 11, 25, 27, 29, 14, 13, 33, 35, 37, 16, 40, 18, 19, 42, 45, 47, 49, 21, 24, 52, 55, 23, 58, 61, 62, 30, 26, 65, 66, 28, 68, 34, 31, 71, 74, 76, 79, 81, 39, 32, 83, 86, 36, 89, 43, 38, 91, 93, 96, 99, 41, 101, 50, 103, 106, 44, 108, 54, 111, 46, 113, 117, 48, 57, 118, 51, 120, 123

Offset: 1

Eric Angelini and Jean-Marc Falcoz, May 09 2019

nonn

A variant of Recamán's sequence: start at n=1, a(1)=1, then iterate: let n' =  n - a(n) if this n' > 0 and was not visited earlier, otherwise n' = n + a(n). Then let n'' <> n' be the smallest index not visited earlier (a(n'') not yet defined) such that the value |n'-n''| was not yet used (meaning not yet a value of any a(i)). Set a(n') = |n'-n''| and continue with n = n'. [Name and Comment suggested by M. F. Hasler at the request of the authors.]
Conjectured to be a permutation of the positive integers. The conjectured inverse permutation is given in A308049. - M. F. Hasler, May 10 2019
From Rémy Sigrist and N. J. A. Sloane, May 13 2019: (Start)
The sequence is given by the following formula. Let R(t) = A081145(t). Then for all t >= 1, a(R(t)) = |R(t+1)-R(t)|.
For example, for t=10, R(10)=20, R(11)=6, and a(R(10)) = a(20) = |6-20| = 14.
Since it is known that {R(t): t>=1} is a permutation of the positive integers (it is the "Slater-Velez permutation of the first kind"), this specifies a(n) for all n.
The connection with the definition as interpreted above by M. F. Hasler is that at step t of the procedure, n' is R(t) = A081145(t), n'' is R(t+1) = A081145(t+1), and we calculate a(R(t)) = |n'-n''| = |R(t)-R(t+1)|.
The conjecture that {a(n)} is a permutation of the positive integers is equivalent to Slater and Velez's conjecture (see references) that the absolute values of the first differences of A081145 are also a permutation of the positive integers. This problem appears to be still unsolved. (End)

			a(1) = 1 drives us to the empty cell a(2) since we can't go further to the left. We fill this cell with the number 2 which is the smallest integer not used before and thus allows us to go to the leftmost possible empty cell, 2 + 2 = 4. (There are no empty places to the left and we can't go to 3 = 2 + 1 since a step 1 has already been used.) So we have a(2) = 2.
a(2) = 2 drives us to the empty cell a(4). We see that the leftmost empty cell a(3) cannot be reached from a(4) since a step of 1 has already been used. We thus fill the cell a(4) with the smallest integer not used before, a(4) = 3.
a(4) = 3 drives us to the empty cell a(7). We see that the leftmost empty cell a(3) can now be reached from a(7) if we fill a(7) with 4; we have thus a(7) = 4.
a(7) = 4 drives us to the empty cell a(3), which is the one we wanted to fill. We fill a(3) with 5 which is the smallest integer not leading to a contradiction, whence a(3) = 5.
a(3) = 5 drives us to the empty cell a(8). We would like to fill this cell with 3, as this 3 would allow us to fill the leftmost empty cell of the sequence - but 3 has been used before; thus we'll have a(8) = 6.
a(8) = 6 drives us to the empty cell a(14). We fill a(14) with 9 as this will allow us to reach the leftmost empty cell of the sequence, whence a(14) = 9.
a(14) = 9 drives us to the empty cell a(5). We fill a(5) with 7 as this is the smallest integer not leading to a contradiction, so we have a(5) = 7, etc.

Jean-Marc Falcoz, Table of n, a(n) for n = 1..5564
P. J. Slater and W. Y. Velez, Permutations of the Positive Integers with Restrictions on the Sequence of Differences, Pacific Journal of Mathematics, Vol. 71, No. 1, 1977, 193-196.
P. J. Slater and W. Y. Velez, Permutations of the Positive Integers with Restrictions on the Sequence of Differences, II, Pacific Journal of Mathematics, Vol. 82, No. 2, 1979, 527-531.
William Y. Velez, Research problems 159-160, Discrete Math., 110 (1992), pp. 301-302.

Cf. A005132 (Recamán's sequence), A171884 (injective variant).
Cf. A308049 (conjectured inverse permutation).
See also A081145, A081146, A099004.

{A=vector(N=199); n=1; while (n<=N,S=Set(A); Z=select(t->!t,A,1); for (i=1,#Z,Z[i]!=n||next; setsearch(S, abs(n-z=Z[i]))&& next; A[n]=abs(n-z); n=z; next(2)); break); if(#Z, A[1..Z[1]-!A[Z[1]]], A)} \\ M. F. Hasler, May 09 2019

A336830 The Sydney Opera House sequence: a(0) = 0, a(1) = 1; for n > 0, a(n) = min(a(n-1)/n if n|a(n-1), a(n-1)-n) where a(n) is nonnegative and not already in the sequence. Otherwise a(n) = min(a(n-1)+n, a(n-1)*n) where a(n) is not already in the sequence. Otherwise a(n) = a(n-1) + n.

Original entry on oeis.org

0, 1, 2, 5, 9, 4, 10, 3, 11, 20, 30, 19, 7, 91, 77, 62, 46, 29, 47, 28, 8, 168, 146, 123, 99, 74, 48, 21, 49, 78, 108, 139, 107, 140, 106, 71, 35, 72, 34, 73, 33, 1353, 1311, 1268, 1224, 1179, 1133, 1086, 1038, 989, 939, 888, 836, 783, 729, 674, 618, 561, 503, 444, 384, 323, 261

Offset: 0

Scott R. Shannon, Aug 05 2020

nonn

This sequence is similar to the Recamán sequence A005132 except that division and multiplication by n are also permitted. This leads to larger variations in the values of the terms while minimizing the repetition of previously visited terms.
To determine a(n), initially a(n-1)-n is calculated if a(n-1)-n is nonnegative, along with a(n-1)/n if n|a(n-1). If one or both of these have not already appeared in the sequence then a(n) is set to the minimum of these candidates. If neither are candidates then both a(n-1)+n and a(n-1)*n are calculated. If one or both of these have not already appeared in the sequence then a(n) is set to the minimum of these candidates. If neither are candidates, i.e., all of a(n-1)-n, a(n-1)/n, a(n-1)+n, a(n-1)*n are either invalid or have already been visited, then a(n) = a(n-1)+n. However for the first 100 million terms no instance is found where all four options are unavailable, although it is unknown if this eventually occurs for very large n.
For the first 100 million terms the smallest value not appearing is 6. As with the Recamán sequence it is unknown if this and other small unseen terms eventually appear. The largest term is a(50757703) = 6725080695952885. In the same range, division, subtraction, addition, and multiplication are chosen for the next term 38, 99965692, 34188, and 81 times, respectively.

			a(2) = 2. As a(1) = 1, which is not divisible by 2 nor greater than 2, a(2) must be the minimum of 1*2=2 and 1+2=3, so multiplication is chosen.
a(5) = 4. As a(4) = 9, which is not divisible by 5, and 4 has not appeared previously in the sequence, a(5) = a(4)-5 = 9-5 = 4.
a(82) = 52. As a(81) = 4264 one candidate is 4264-82 = 4182. However 82|4264 and 4264/82 = 52. Neither of these candidates has previously appeared in the sequence, but 52 is the minimum of the two. This is the first time a division operation is used for a(n).

Michael S. Branicky, Table of n, a(n) for n = 0..20000
Scott R. Shannon, Line graph of the first 10 million terms.
Index entries for sequences related to Recamán's sequence

Cf. A005132, A113880, A171884, A330791, A362398, A362399.

global arr
arr = []
def a(n):
    # Case 1
    if n == 0:
        return 0
    a_prev = arr[-1]
    cand = []
    # Case 2
    x = a_prev - n
    y = a_prev / n
    if x > 0 and not x in arr:
        cand.append(x)
    if y == int(y) and not y in arr:
        cand.append(y)
    if cand != []:
        return min(cand)
    # Case 3
    cand = []
    x = a_prev + n
    y = a_prev * n
    if not x in arr:
        cand.append(x)
    if not y in arr:
        cand.append(y)
    if cand != []:
        return min(cand)
    # Case 4
    return a_prev + n
def seq(n):
    for i in range(n):
        print("{}, ".format(a(i)), end="")
        arr.append(a(i))
seq(60)
# Christoph B. Kassir, Apr 08 2022

from itertools import count, islice
def A336830(): # generator of terms
    aset, an, oo = {0, 1}, 1, float('inf')
    yield from [0, 1]
    for n in count(2):
        v1, v2 = an - n if an >= n else oo, an//n if an%n == 0 else oo
        v = min((vi for vi in [v1, v2] if vi not in aset), default=oo)
        if v != oo: an = v
        else:
            v3, v4 = an+n, an*n
            v = min((vi for vi in [v3, v4] if vi not in aset), default=oo)
            if v != oo: an = v
            else: an = an+n
        yield an
        aset.add(an)
print(list(islice(A336830(), 60))) # Michael S. Branicky, Apr 15 2023

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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A048473 a(0)=1, a(n) = 3*a(n-1) + 2; a(n) = 2*3^n - 1.

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A308021 Start at n=1. Fill in a(n) with a value d > 0 not used earlier such that n-d or n+d is the smallest possible index not visited earlier, then continue with that index.

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A336830 The Sydney Opera House sequence: a(0) = 0, a(1) = 1; for n > 0, a(n) = min(a(n-1)/n if n|a(n-1), a(n-1)-n) where a(n) is nonnegative and not already in the sequence. Otherwise a(n) = min(a(n-1)+n, a(n-1)*n) where a(n) is not already in the sequence. Otherwise a(n) = a(n-1) + n.

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A355492 a(n) = 7*3^n - 2.

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A048473 a(0)=1, a(n) = 3a(n-1) + 2; a(n) = 23^n - 1.