

A005132


Recamán's sequence (or Recaman's sequence): a(0) = 0; for n > 0, a(n) = a(n1)  n if positive and not already in the sequence, otherwise a(n) = a(n1) + n.
(Formerly M2511)


141



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
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OFFSET

0,3


COMMENTS

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
From David W. Wilson, Jul 13 2009: (Start)
The sequence satisfies [1] a(n) >= 0, [2] a(n)a(n1) = n, and tries to avoid repeats by greedy choice of a(n) = a(n1) + n.
This "wants" to be an injection on N = {0, 1, 2, ...}, as it attempts to avoid repeats by choice of a(n) = a(n1) + n when a(n) = a(n1)  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)
The definition can be written as a(n) = a(n1)  n*(1)^(c(n)+d(n)) where c(n) and d(n) are two flags defined via b(n) = a(n1)n; theta(n) = 1 if n>0, =0 if n<=0; c(n) = theta(1+b(n)); d(n) = theta( product_{j=0..n1}( a(j)b(n))^2 ).  Adriano Caroli, Dec 26 2010
It appears that a(n) is also the value of "x" and "y" of the endpoint of the Ltoothpick structure mentioned in A210606 after nth 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


REFERENCES

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 and Allan Wilks, On sequences of Recaman type, paper in preparation, 2006.


LINKS

N. J. A. Sloane, The first 100000 terms
Benjamin Chaffin, Loglog plot of first 10^230 terms
GBnums, See: A nice OEIS sequence
Gordon Hamilton, Wrecker Ball Sequences, Video, 2013
Nick Hobson, Python program for this sequence
C. L. Mallows, Plot (jpeg) of first 10000 terms
C. L. Mallows, Plot (postscript) of first 10000 terms
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
Omar E. Pol, Illustration of initial terms of A001057, A005132, A000217
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
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


EXAMPLE

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.


MAPLE

h := array(1..100000); maxt := 100000; a := [1]; ad := [1]; su := []; h[1] := 1; for nx from 2 to 500 do t1 := a[nx1]nx; if t1>0 and h[t1] <> 1 then su := [op(su), nx]; else t1 := a[nx1]+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, fou, j ;
if n = 0 then 0;
else
a := procname(n1)n ;
if a <= 0 then return a+2*n ; else fou := false; for j from 0 to n1 do if procname(j) = a then fou :=true; break; end if; end do; if fou then a+2*n; else a ; end if;
end if;
end if;
end proc: # R. J. Mathar, Apr 01 2012


MATHEMATICA

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
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 *)


PROG

(PARI) a(n)=if(n<2, 1, if(abs(sign(a(n1)n)1)+setsearch(Set(vector(n1, i, a(i))), a(n1)n), a(n1)+n, a(n1)n)) \\ Benoit Cloitre
(PARI) A005132(N=1000, show=0)={ my(s, t); for(n=1, N, s=bitor(s, 1<<t += if( t<=n  bittest(s, tn), n, n)); show&&print1(t", ")); t} \\ M. F. Hasler, May 11 2008, updated M. F. Hasler, Nov 03 2014
(MATLAB)
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(n1)(n1);
C=0.^( abs(B+1) + (B+1) );
D=ismember(B, a(1:n1));
a(n)=a(n1)+ (n1) * (1)^(C + D 1);
end
% Adriano Caroli, Dec 26 2010
(Haskell)
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 (xn) : recaman (insert (xn) s) (n+1) (xn)
else (x+n) : recaman (insert (x+n) s) (n+1) (x+n)
 Reinhard Zumkeller, Mar 14 2011


CROSSREFS

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.
Sequence in context: A076543 A274648 A277558 * A064388 A064387 A064389
Adjacent sequences: A005129 A005130 A005131 * A005133 A005134 A005135


KEYWORD

nonn,nice,hear,look


AUTHOR

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


EXTENSIONS

Allan Wilks, Nov 06 2001, computed 10^15 terms of this sequence. At this point the smallest missing number was 852655 = 5*31*5501.
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, Mar 22 2010


STATUS

approved



