A348080 A variant of Van Eck's sequence: For n >= 1, if there exists an m < n such that a(m) = a(n), take the largest such m and set a(n+1) = n XOR m; otherwise a(n+1) = 0. Start with a(1)=0.

0, 0, 3, 0, 6, 0, 2, 0, 14, 0, 2, 12, 0, 7, 0, 2, 27, 0, 29, 0, 6, 16, 0, 3, 27, 8, 0, 12, 16, 11, 0, 4, 0, 62, 0, 2, 52, 0, 5, 0, 14, 32, 0, 3, 52, 8, 52, 2, 20, 0, 25, 0, 6, 32, 28, 0, 12, 37, 0, 3, 16, 32, 8, 17, 0, 122, 0, 2, 116, 0, 5, 96, 0, 15, 0, 2, 8

Offset

1,3

Comments

XOR denotes the bitwise XOR operator.

This sequence is unbounded, and contains infinitely many 0's.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..8192

Rémy Sigrist, Scatterplot of the first 2^20 terms

Example

The first terms, alongside m, are:
  n   a(n)  m
  --  ----  ---
   1     0  N/A
   2     0    1
   3     1  N/A
   4     0    2
   5     2  N/A
   6     0    4
   7     2    5
   8     2    7
   9     1    3
  10     6  N/A

Prog

(PARI) { p=vector(123); v=0; for (n=1, 77, print1(v", "); [p[1+v], v]=[n, if (p[1+v], bitxor(n, p[1+v]), 0)]) }

Crossrefs

Cf. A181391, A346516.

Sequence in context: A321254 A262605 A092731 * A201567 A161829 A290705

Adjacent sequences: A348077 A348078 A348079 * A348081 A348082 A348083

Keyword

nonn,base,look

Author

Rémy Sigrist, Sep 27 2021

Status

approved