A306803 -id:A306803 - cp's OEIS frontend

A306808 An irregular fractal sequence: underline a(n) iff the sum [a(n-1) + a(n)] is a palindrome; all underlined terms rebuild the starting sequence.

1, 9, 3, 1, 11, 2, 9, 4, 3, 1, 12, 5, 7, 6, 8, 10, 13, 14, 15, 16, 18, 17, 19, 20, 21, 22, 11, 23, 24, 25, 26, 27, 29, 28, 30, 31, 2, 9, 32, 33, 34, 35, 36, 37, 38, 40, 4, 3, 1, 39, 41, 42, 43, 12, 44, 45, 46, 47, 48, 49, 51, 52, 50, 5, 53, 54, 55, 57, 56, 58, 59, 7, 60, 6, 61, 62, 63, 64, 65, 67, 66, 68, 69, 8, 70, 72, 71, 73, 74

Offset: 1

Eric Angelini and Jean-Marc Falcoz, Mar 11 2019

The sequence S starts with a(1) = 1 and a(2) = 9. S is extended by duplicating the first term A among the not yet duplicated terms of S, under the condition that the sum [a(n-1) + a(n)] is a palindrome. If this is not the case, we then extend S with the smallest integer X not yet present in S such that the sum [a(n-1) + a(n)] is not a palindrome. S is the lexicographically earliest sequence with this property.

			S starts with a(1) = 1 and a(2) = 9
Can we duplicate a(1) to form a(3)? No, as a(2) + a(3) would be 10 and 10 is not a palindrome. We thus extend S with the smallest integer X not yet in S such that [a(2) + X] is not a palindrome. We get a(3) = 3.
Can we duplicate a(1) to form a(4)? Yes, as a(3) + a(4) = 4, which is a palindrome. We get a(4) = 1.
Can we duplicate a(2) to form a(5)? No, as a(4) + a(5) would be 10 and 10 is not a palindrome. We thus extend S with the smallest integer X not yet in S such that [a(4) + X] is not a palindrome; we get a(5) = 11.
Can we duplicate a(2) to form a(6)? No, as a(5) + a(6) would be 20 and 20 is not a palindrome. We thus extend S with the smallest integer X not yet in S such that [a(5) + X] is not a palindrome; we get a(6) = 2.
Can we duplicate a(2) to form a(7)? Yes, as [a(6) + a(7)] = 11, which is a palindrome. We get a(7) = 9.
Etc.

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

Cf. A306803 (which is obtained by replacing palindrome by prime in the definition).

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.

Original entry on oeis.org

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

Alexandre Wajnberg and Eric Angelini, Mar 11 2019

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.

			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.

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

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

cp's OEIS Frontend

A306808 An irregular fractal sequence: underline a(n) iff the sum [a(n-1) + a(n)] is a palindrome; all underlined terms rebuild the starting sequence.

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