A308984 If a(n) is not a term of a(0)..a(n-1), then a(n+1) = |a(n) - a(n-1)|, otherwise a(n+1) = a(n) + n - m, where a(m) = a(n), m < n, and m is maximal. a(0)=0, a(1)=1.

0, 1, 1, 2, 1, 3, 2, 5, 3, 6, 3, 5, 9, 4, 5, 8, 3, 9, 14, 5, 10, 5, 7, 2, 19, 17, 2, 5, 11, 6, 26, 20, 6, 9, 25, 16, 9, 12, 3, 25, 30, 5, 19, 37, 18, 19, 22, 3, 12, 23, 11, 33, 22, 28, 6, 28, 30, 46, 16, 39, 23, 34, 11, 23, 26, 60, 34, 39, 47, 8, 62, 54, 8, 11

Offset

0,4

Comments

In other words, if the last term has not appeared previously, take the absolute difference between the previous term and itself to obtain the next one; otherwise, add to the last term the difference between its index and that of the last identical term to obtain the next one.

The sequence is infinite, because loops are impossible.

Links

Table of n, a(n) for n=0..73.

Sean A. Irvine, Java program (github)

Example

a(0)=0 (given).
a(1)=1 (given).
a(2)=1: a(1) is not a term of a(0..0), therefore a(2) = |a(1) - a(0)| = 1 - 0 = 1.
a(3)=2: a(2)=a(1), therefore a(3) = a(2) + 2 - 1 = 1 + 2 - 1 = 2.
a(4)=1: a(3) is not a term of a(0..2), therefore a(4) = |a(3) - a(2)| = 2 - 1 = 1.
a(5)=3: a(4)=a(2), therefore a(5) = a(4) + 4 - 2 = 1 + 4 - 2 = 3.
a(6)=2: a(5) is not a term of a(0..4), therefore a(6) = |a(5) - a(4)| = 3 - 1 = 2.

Mathematica

Nest[Function[{a, n}, Append[a, If[FreeQ[Most@ a, Last@ a], Abs[Subtract @@ a[[-2 ;; -1]] ], Last[a] + n - Position[Most@ a, _?(# == Last@ a &)][[-1, 1]] ]]] @@ {#, Length@ #} &, {0, 1}, 72] (* Michael De Vlieger, Jul 08 2019 *)

Crossrefs

Sequence in context: A363083 A348112 A045747 * A029138 A161051 A161255

Adjacent sequences: A308981 A308982 A308983 * A308985 A308986 A308987

Keyword

nonn

Author

Marc Morgenegg, Jul 04 2019

Status

approved