A309620 Start with a(1) = 5. For n>1, the sequence is always extended with a(n+1) = a(n) + 2*d if d is prime, and with a(n+1) = a(n) - 1 - d if d is not prime, where d is the n-th digit of the sequence.

5, 15, 13, 23, 21, 27, 31, 37, 41, 39, 43, 57, 63, 61, 67, 81, 76, 74, 80, 70, 65, 71, 81, 95, 88, 94, 87, 85, 78, 92, 83, 81, 95, 88, 102, 97, 88, 87, 101, 100, 93, 103, 117, 115, 106, 104, 94, 104, 95, 86, 76, 71, 62, 76, 67, 77, 91, 82, 72, 76, 67, 73, 64, 62, 52, 62, 53, 44, 42, 41, 45, 35, 49, 40, 31, 22, 36, 34, 33, 31

Offset

1,1

Comments

If we start with a positive number less than 5 the sequence enters almost immediately into a 1-term loop.

The authors wanted the sequence to oscillate evenly around zero by giving different weights to the nonprime vs prime digits: 0, 1, 4, 6, 8, 9 have respective weights 1, 2, 5, 7, 9, 10 (total 34) and 2, 3, 5, 7 become 4, 6, 10, 14 (same total 34).

The result is not convincing after 10^5 terms, the sequence staying for almost 95% of the time above zero. But who knows what will happen after 10^10 terms, for instance?

Links

Jean-Marc Falcoz, Table of n, a(n) for n = 1..65531

Example

The sequence S begins with 5,15,13,23,21,27,31,37,...
As a(1) = 5 and the 1st digit of S is prime (5), we get a(2) = a(1) + 2*5 = 5 + 10 = 15;
as a(2) = 15 and the 2nd digit of S is not prime (1), we get a(3) = a(2) - 1 - 1 = 15 - 2 = 13;
as a(3) = 13 and the 3rd digit of S is prime (5), we get a(4) = a(3) + 2*5 = 13 + 10 = 23;
as a(4) = 23 and the 4th digit of S is not prime (1), we get a(5) = a(4) - 1 - 1 = 23 - 2 = 21;
as a(5) = 21 and the 5th digit of S is prime (3), we get a(6) = a(5) + 2*3 = 21 + 6 = 27;
as a(6) = 27 and the 6th digit of S is prime (2), we get a(7) = a(6) + 2*2 = 27 + 4 = 31;
etc.

Crossrefs

Cf. A309521 (a variant).

Sequence in context: A113259 A298265 A366419 * A291794 A321775 A166621

Adjacent sequences: A309617 A309618 A309619 * A309621 A309622 A309623

Keyword

base,sign

Author

Eric Angelini and Jean-Marc Falcoz, Aug 10 2019

Status

approved