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%I #23 Aug 20 2019 06:25:19
%S 2,4,8,16,17,11,10,17,16,17,16,16,17,10,11,5,4,11,10,16,17,11,10,17,
%T 16,16,17,16,21,17,16,17,16,16,17,11,10,17,16,17,16,16,17,10,11,5,4,
%U 10,11,4,5,-1,-3,-4,-3,-10,-9,-15,-16,-9,-10,-4,-3,-9,-10
%N Start with a(1)=2; thereafter the sequence is always extended by adding the n-th digit of the sequence to a(n) if a(n) is even, else subtracting it.
%C Among the first 10^8 terms, the last positive value occurs at n=28823742. - _Lars Blomberg_, Aug 10 2019
%H Jean-Marc Falcoz, <a href="/A309529/b309529.txt">Table of n, a(n) for n = 1..42917</a>
%H Lars Blomberg, <a href="/A309529/a309529.png">Graph of 10^8 terms</a>
%H Lars Blomberg, <a href="/A309529/a309529_1.png">Graph of accumulated sums of 10^8 terms</a>
%e The sequence begins with 2,4,8,16,17,11,10,17,...
%e As a(1) = 2 (even), we have a(2) = a(1) + [the 1st digit of the seq] = 2 + 2 = 4;
%e as a(2) = 4 (even), we have a(3) = a(2) + [the 2nd digit of the seq] = 4 + 4 = 8;
%e as a(3) = 8 (even), we have a(4) = a(3) + [the 3rd digit of the seq] = 8 + 8 = 16;
%e as a(4) = 16 (even), we have a(5) = a(4) + [the 4th digit of the seq] = 16 + 1 = 17;
%e as a(5) = 17 (odd), we have a(6) = a(5) - [the 5th digit of the seq] = 17 - 6 = 11;
%e as a(6) = 11 (odd), we have a(7) = a(6) - [the 6th digit of the seq] = 11 - 1 = 10;
%e etc.
%Y Cf. A309521 (same idea, but dealing with primes instead of even numbers).
%K sign,base
%O 1,1
%A _Eric Angelini_ and _Jean-Marc Falcoz_, Aug 06 2019
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