A306807 An irregular fractal sequence: underline a(n) iff the absolute difference |a(n-1) - a(n)| is prime; all underlined terms rebuild the starting sequence.

1, 2, 3, 1, 5, 2, 6, 3, 1, 7, 5, 2, 8, 6, 3, 1, 9, 7, 5, 2, 10, 8, 6, 3, 1, 11, 9, 7, 5, 2, 12, 10, 8, 6, 3, 1, 13, 11, 9, 7, 5, 2, 14, 12, 10, 8, 6, 3, 1, 15, 13, 11, 9, 7, 5, 2, 16, 14, 12, 10, 8, 6, 3, 1, 17, 15, 13, 11, 9, 7, 5, 2, 18, 16, 14, 12, 10, 8, 6, 3, 1, 19, 17, 15, 13, 11, 9, 7, 5, 2, 20, 18, 16, 14, 12, 10, 8, 6, 3, 1

Offset: 1

Author

Alexandre Wajnberg and Eric Angelini, Mar 11 2019

Keywords

• base
• nonn
• look

Comments

The sequence S starts with a(1) = 1 and a(2) = 2. S is extended by duplicating the first term A among the not yet duplicated terms, under the condition that the absolute difference |a(n-1) - a(n)| is prime. If this is not the case, we then extend S with the smallest integer X not yet present in S such that the absolute difference |a(n-1) - a(n)| is not prime. S is the lexicographically earliest sequence with this property.

Examples

			S starts with a(1) = 1 and a(2) = 2
Can we duplicate a(1) to form a(3)? No, as |a(2) - a(3)| would be 1 and 1 is not prime. We thus extend S with the smallest integer X not yet in S such that |a(2) - X| is not prime. We get a(3) = 3.
Can we duplicate a(1) to form a(4)? Yes, as |a(3) - a(4)| = 2, which is prime. We get a(4) = 1.
Can we duplicate a(2) to form a(5)? No, as |a(4) - a(5)| would be 1 and 1 is not prime. We thus extend S with the smallest integer X not yet in S such that |a(4) - X| is not prime; we get a(5) = 5.
Can we duplicate a(2) to form a(6)? Yes, as |a(6) - a(5)| = 3, which is prime; we get a(6) = 2.
Etc.
		

Links

• Jean-Marc Falcoz, Table of n, a(n) for n = 1..10002

Crossrefs

Cf. A306803 (obtained by replacing the absolute difference by the sum in the definition).