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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. 1

%I

%S 5,15,13,23,21,27,31,37,41,39,43,57,63,61,67,81,76,74,80,70,65,71,81,

%T 95,88,94,87,85,78,92,83,81,95,88,102,97,88,87,101,100,93,103,117,115,

%U 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

%N 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.

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

%C 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).

%C 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?

%H Jean-Marc Falcoz, <a href="/A309620/b309620.txt">Table of n, a(n) for n = 1..65531</a>

%e The sequence S begins with 5,15,13,23,21,27,31,37,...

%e As a(1) = 5 and the 1st digit of S is prime (5), we get a(2) = a(1) + 2*5 = 5 + 10 = 15;

%e 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;

%e as a(3) = 13 and the 3rd digit of S is prime (5), we get a(4) = a(3) + 2*5 = 13 + 10 = 23;

%e 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;

%e as a(5) = 21 and the 5th digit of S is prime (3), we get a(6) = a(5) + 2*3 = 21 + 6 = 27;

%e as a(6) = 27 and the 6th digit of S is prime (2), we get a(7) = a(6) + 2*2 = 27 + 4 = 31;

%e etc.

%Y Cf. A309521 (a variant).

%K base,sign

%O 1,1

%A _Eric Angelini_ and _Jean-Marc Falcoz_, Aug 10 2019

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