A361614 Set a(1)=0 and a(2)=1. For n > 1, if a(n) has already appeared in the sequence, then a(n+1) = number of steps since its first appearance. If a(n) has not appeared before, search instead for a(n)-1, then a(n)-2, etc., until you find a number that has appeared before.

0, 1, 1, 1, 2, 3, 1, 5, 2, 4, 4, 1, 10, 5, 6, 7, 1, 15, 5, 11, 7, 5, 14, 3, 18, 7, 10, 14, 5, 21, 5, 23, 2, 28, 2, 30, 2, 32, 2, 34, 2, 36, 2, 38, 2, 40, 2, 42, 2, 44, 2, 46, 2, 48, 2, 50, 2, 52, 2, 54, 2, 56, 2, 58, 2, 60, 2, 62, 2, 64, 2, 66, 2, 68, 2, 70, 2

Offset

1,5

Comments

The first 32 terms are distributed chaotically, after which the sequence alternates between 2 and n-6 indefinitely.

Links

Table of n, a(n) for n=1..77.

Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Example

We start with a(1) = 0 and a(2) = 1. 1 has not appeared before, so we search for the greatest valid integer less than 1, which in this case is 0. 0 first occurred at a(1), which is 1 term before a(2) so a(3) = 1.
1 first occurred 1 term before, so a(4) = 1.
1 appeared at term a(1), which is 2 terms before a(4), so a(5) = 2.
2 has not appeared before, so we search for 1, which occurred 3 terms before at a(1). a(6) = 3.
And so on.

Crossrefs

Cf. A328294, A181391.

Sequence in context: A075014 A304880 A086686 * A306807 A021816 A055023

Adjacent sequences: A361611 A361612 A361613 * A361615 A361616 A361617

Keyword

nonn,easy

Author

Robin Powell, Mar 17 2023

Status

approved