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

Author

Robin Powell, Mar 17 2023

Keywords

• nonn
• easy

Comments

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A328294, A181391.