A046901 -id:A046901 - 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

A008344 a(1)=0; thereafter a(n+1) = a(n) - n if a(n) >= n otherwise a(n+1) = a(n) + n.

Original entry on oeis.org

0, 1, 3, 0, 4, 9, 3, 10, 2, 11, 1, 12, 0, 13, 27, 12, 28, 11, 29, 10, 30, 9, 31, 8, 32, 7, 33, 6, 34, 5, 35, 4, 36, 3, 37, 2, 38, 1, 39, 0, 40, 81, 39, 82, 38, 83, 37, 84, 36, 85, 35, 86, 34, 87, 33, 88, 32, 89, 31, 90, 30, 91, 29, 92, 28, 93, 27, 94, 26, 95, 25, 96, 24, 97, 23, 98

Offset: 1

N. J. A. Sloane

p^a(n) = A084110(p^(n-1)) for n>1 and p prime. - Reinhard Zumkeller, May 12 2003
For n > 1: a(A029858(n)) = A029858(n) and a(A003462(n)) = 0. - Reinhard Zumkeller, May 09 2012
Absolute first differences of A085059; abs(a(n+1)-a(n)) = n, see also A086283. - Reinhard Zumkeller, Oct 17 2014
For n>3, when a(n) = 3, a(n+1) is in A116970. - Bill McEachen, Oct 03 2023

N. J. A. Sloane, Table of n, a(n) for n = 1..29524 (Up to the 10th 0 term)
Index entries for sequences related to Recamán's sequence

Cf. A046901, A008343, A088230, A085059, A086283.
Equals A085059(n)-1.
Cf. A076042 (based on squares).

a008344 n = a008344_list !! (n-1)
a008344_list = 0 : f 0 [1..] where
   f x (z:zs) = y : f y zs where y = if x < z then x + z else x - z
-- Reinhard Zumkeller, Oct 17 2014, May 08 2012

A008344 := proc(n) option remember; if n = 0 then 0 elif A008344(n-1) >= (n-1) then A008344(n-1)-(n-1) else A008344(n-1)+(n-1); fi; end;

a[1]=0; a[n_] := a[n]=If[a[n-1]>=n-1, a[n-1]-n+1, a[n-1]+n-1]
Transpose[ NestList[ If[First[#]>=Last[#],{First[#]-Last[#],Last[#]+1}, {First[#]+Last[#],Last[#]+1}]&,{0,1},80]][[1]] (* Harvey P. Dale, Jun 20 2011 *)
s = 0; Table[If[s < n, s = s + n, s = s - n], {n, 0, 80}] (* Horst H. Manninger, Dec 03 2018 *)

a(n) = my(expo = logint(2*n+1, 3), res = n - (3^expo-1)/2); if(res==0, 0, if(res%2, (3^expo-res)/2, 3^expo-1+res/2)) \\ Jianing Song, May 25 2021

This is a concatenation S_0, S_1, S_2, ... where S_i = [b_0, b_1, ..., b_{3^(i+1)-1}] with b_0 = 0, b_{2j-1} = k+1-j, b_{2j} = 2k+j (j=1..k), k=(3^(i+1)-1)/2. E.g. S_0 = [0, 1, 3], S_1 = [0, 4, 9, 3, 10, 2, 11, 1, 12].
a((3^n-1)/2) = 0; a((3^n-1)/2 + 2k-1) = (3^n+1)/2 - k for 1 <= k <= (3^n-1)/2; a((3^n-1)/2 + 2k) = 3^n - 1 + k for 1 <= k < (3^n-1)/2. - Benoit Cloitre, Jan 09 2003 [Corrected by Jianing Song, May 25 2021]
a(n) = (n-1+a(n-1)) mod (2*(n-1)). - Jon Maiga, Jul 09 2021

Name edited by Dmitry Kamenetsky, Feb 14 2017

A076042 a(0) = 0; thereafter a(n) = a(n-1) + n^2 if a(n-1) < n^2, otherwise a(n) = a(n-1) - n^2.

Original entry on oeis.org

0, 1, 5, 14, 30, 5, 41, 90, 26, 107, 7, 128, 272, 103, 299, 74, 330, 41, 365, 4, 404, 845, 361, 890, 314, 939, 263, 992, 208, 1049, 149, 1110, 86, 1175, 19, 1244, 2540, 1171, 2615, 1094, 2694, 1013, 2777, 928, 2864, 839, 2955, 746, 3050, 649, 3149

Offset: 0

Amarnath Murthy, Oct 29 2002

Does not return to zero within first 2^25000 =~ 10^7525 terms.  Define an epoch as an addition followed by a sequence of (addition, subtraction) pairs.  The first epoch has length 1 (+), the second 3 (++-), the third 5 (++-+-), and so forth (cf. A324792).  The epoch lengths increase geometrically by about the square root of 3, and the value at the end of each epoch is the low value in the epoch.  These observations lead to the Python program given. - Tomas Rokicki, Aug 31 2019
Using the Maple program in A324791, I confirmed that a(n) != 0 for 0 < n < 10^2394. See the a- and b-files in A325056 and A324791. - N. J. A. Sloane, Oct 03 2019
'Easy Recamán transform' of the squares. - Daniel Forgues, Oct 25 2019

Tomas Rokicki, Table of n, a(n) for n = 0..20000

Cf. A076039, A003462, A076041, A046901.
See also A325056, A324791, A324792.
Cf. A053461 ('Recamán transform' of the squares).
Cf. A000290, A008344, A008345.

a:= proc(n) option remember; `if`(n<0, 0,
      ((s, t)-> s+`if`(sAlois P. Heinz, Jan 11 2020

a[0] = 0;
a[n_] := a[n] = a[n-1] + If[a[n-1] < n^2, n^2, -n^2];
a /@ Range[0, 50] (* Jean-François Alcover, Apr 11 2020 *)

v=vector(50); v[1]=1; for(n=2,50,if(v[n-1]

More terms from Ralf Stephan, Mar 20 2003
a(0)=0 prepended, at the suggestion of Allan C. Wechsler, by N. J. A. Sloane, Aug 31 2019
Offset set to 0, to cohere with previous action of N. J. A. Sloane, by Allan C. Wechsler, Sep 08 2019

A134931 a(n) = (5*3^n-3)/2.

Original entry on oeis.org

1, 6, 21, 66, 201, 606, 1821, 5466, 16401, 49206, 147621, 442866, 1328601, 3985806, 11957421, 35872266, 107616801, 322850406, 968551221, 2905653666, 8716961001, 26150883006, 78452649021, 235357947066, 706073841201, 2118221523606

Offset: 0

Rolf Pleisch, Jan 29 2008

Numbers n where the recurrence s(0)=1, if s(n-1) >= n then s(n) = s(n-1) - n else s(n) = s(n-1) + n produces s(n)=0. - Hugo Pfoertner, Jan 05 2012
A046901(a(n)) = 1. - Reinhard Zumkeller, Jan 31 2013
Binomial transform of A146523: (1, 5, 10, 20, 40, ...) and double binomial transform of A010685: (1, 4, 1, 4, 1, 4, ...). - Gary W. Adamson, Aug 25 2016
Also the number of maximal cliques in the (n+1)-Hanoi graph. - Eric W. Weisstein, Dec 01 2017
a(n) is the least k such that f(a(n-1)+1) + ... + f(k) > f(a(n-2)+1) + ... + f(a(n-1)) for n > 1, where f(n) = 1/(n+1). Because Sum_{k=1..5*3^(n-1)} 1/(a(n)+3*k-1) + 1/(a(n)+3*k) + 1/(a(n)+3*k+1) - 1/((a(n)+1+5*3^n)*5*3^(n-1)) < Sum_{k=1..5*3^(n-1)} 1/(a(n-1)+k+1) < Sum_{k=1..5*3^(n-1)} 1/(a(n)+3*k-1) + 1/(a(n)+3*k) + 1/(a(n)+3*k+1), we have 1 < 1/3 + 1/4 + ... + 1/7 < 1/8 + 1/9 + ... + 1/22 < ... . - Jinyuan Wang, Jun 15 2020

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Eric Weisstein's World of Mathematics, Hanoi Graph
Eric Weisstein's World of Mathematics, Maximal Clique
Index entries for linear recurrences with constant coefficients, signature (4,-3).

Cf. A003462, A007051, A034472, A024023, A067771, A029858.

Cf. A005030, A010685, A046901, A060816, A116952, A146523, A225918.

[(5*3^n-3)/2: n in [0..30]]; // Vincenzo Librandi, Jun 05 2011

seq((5*3^n-3)/2, n= 0..25); # Gary Detlefs, Jun 22 2010

a=1; lst={a}; Do[a=a*3+3; AppendTo[lst,a], {n,0,100}]; lst (* Vladimir Joseph Stephan Orlovsky, Dec 25 2008 *)
Table[(5 3^n - 9)/6, {n, 20}] (* Eric W. Weisstein, Dec 01 2017 *)
(5 3^Range[20] - 9)/6 (* Eric W. Weisstein, Dec 01 2017 *)
LinearRecurrence[{4, -3}, {1, 6}, 20] (* Eric W. Weisstein, Dec 01 2017 *)
CoefficientList[Series[(1 + 2 x)/(1 - 4 x + 3 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Dec 01 2017 *)

a(n) = (5*3^n-3)/2; /* Joerg Arndt, Apr 14 2013 */

a(n) = 3*(a(n-1) + 1), with a(0)=1.
From R. J. Mathar, Jan 31 2008: (Start)
O.g.f.: (5/2)/(1-3*x) - (3/2)/(1-x).
a(n) = (A005030(n) - 3)/2. (End)
a(n) = A060816(n+1) - 1. - Philippe Deléham, Apr 14 2013
E.g.f.: exp(x)*(5*exp(2*x) - 3)/2. - Stefano Spezia, Aug 28 2023

More terms from Vladimir Joseph Stephan Orlovsky, Dec 25 2008

A057198 a(n) = (5*3^(n-1)+1)/2.

Original entry on oeis.org

3, 8, 23, 68, 203, 608, 1823, 5468, 16403, 49208, 147623, 442868, 1328603, 3985808, 11957423, 35872268, 107616803, 322850408, 968551223, 2905653668, 8716961003, 26150883008, 78452649023, 235357947068, 706073841203, 2118221523608

Offset: 1

Colin Mallows and N. J. A. Sloane, Sep 16 2000

It appears that if s(n) is a first-order rational sequence of the form s(0)=4, s(n) = (2*s(n-1)+1)/(s(n-1)+2), n > 0, then s(n) = a(n)/(a(n)-1), n > 0.

			G.f. = 3*x + 8*x^2 + 23*x^3 + 68*x^4 + 203*x^5 + 608*x^6 + 1823*x^7 + 5468*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Nathan Fox, A Slow Relative of Hofstadter's Q-Sequence, arXiv:1611.08244 [math.NT], 2016.
Index entries for linear recurrences with constant coefficients, signature (4,-3).

Related to A046901.
Equals A060816 + 1.
Cf. A135423 (bisection), A191450 (2nd row).

[(5*3^(n-1) + 1)/2: n in [1..30]]; // Vincenzo Librandi, Oct 11 2012

Table[(5*3^(n-1) + 1)/2, {n, 30}] (* T. D. Noe, Oct 11 2012 *)

a(n)=(5*3^(n-1)+1)/2 \\ Charles R Greathouse IV, Oct 11 2012

a(n+1) = 3*a(n) - 1 for n > 1. - Reinhard Zumkeller, Jan 22 2011
G.f.: (5/2)*U(0) where U(k) = 1 + 2/(5*3^k + 5*3^k/(1 - 30*x*3^k/(15*x*3^k - 1/U(k+1)))); (continued fraction, 4-step). - Sergei N. Gladkovskii, Nov 01 2012
E.g.f.: (5/2)*U(0) where U(k) = 1 + 2/(5*3^k + 5*3^k/(1 - 30*x*3^k/(15*x*3^k - (k+1)/U(k+1)))); (continued fraction, 4-step). - Sergei N. Gladkovskii, Nov 01 2012
G.f.: x*(3-4*x) / ( (3*x-1)*(x-1) ). - R. J. Mathar, Jan 25 2015
E.g.f.: (5*exp(3*x) + 3*exp(x) - 8)/6. - Stefano Spezia, Aug 28 2023

Incorrect zeroth term removed by Jon Perry, Oct 11 2012

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

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.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

MATLAB

Maple

Mathematica

PARI

PARI

Python

Python

Python

Formula

Extensions

A008344 a(1)=0; thereafter a(n+1) = a(n) - n if a(n) >= n otherwise a(n+1) = a(n) + n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Formula

Extensions

A076042 a(0) = 0; thereafter a(n) = a(n-1) + n^2 if a(n-1) < n^2, otherwise a(n) = a(n-1) - n^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Extensions

A134931 a(n) = (5*3^n-3)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

A057198 a(n) = (5*3^(n-1)+1)/2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

Views

Author

Keywords

Comments