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%I #22 Dec 20 2019 08:55:55
%S 0,1,-1,4,-2,2,-4,5,3,6,-10,7,-3,8,-16,15,-5,12,-6,19,-7,-9,-11,22,-8,
%T 9,-15,28,-12,-14,-18,16,14,10,-20,13,11,17,-29,20,18,21,-35,23,-41,
%U 24,-46,57,-13,26,-42,35,-17,25,-55,29,27,30,-54,31,-59,32,-68
%N a(1) = 0, and for any n > 0, a(2*n) = a(n) + k(n) and a(2*n+1) = a(n) - k(n) where k(n) is the least positive integer not leading to a duplicate term in sequence a.
%C The point is that the same k(n) must be used for both a(2*n) and a(2*n+1). - _N. J. A. Sloane_, Dec 17 2019
%C Apparently every signed integer appears in the sequence.
%H Rémy Sigrist, <a href="/A322510/b322510.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = (a(2*n) + a(2*n+1))/2.
%e The first terms, alongside k(n) and associate children, are:
%e n a(n) k(n) a(2*n) a(2*n+1)
%e -- ---- ---- ------ --------
%e 1 0 1 1 -1
%e 2 1 3 4 -2
%e 3 -1 3 2 -4
%e 4 4 1 5 3
%e 5 -2 8 6 -10
%e 6 2 5 7 -3
%e 7 -4 12 8 -16
%e 8 5 10 15 -5
%e 9 3 9 12 -6
%e 10 6 13 19 -7
%o (PARI) lista(nn) = my (a=[0], s=Set(0)); for (n=1, ceil(nn/2), for (k=1, oo, if (!setsearch(s, a[n]+k) && !setsearch(s, a[n]-k), a=concat(a, [a[n]+k, a[n]-k]); s=setunion(s, Set([a[n]+k, a[n]-k])); break))); a[1..nn]
%Y For k(n) see A330337, A330338.
%Y See A305410, A304971 and A322574-A322575 for similar sequences.
%K sign,look
%O 1,4
%A _Rémy Sigrist_, Dec 13 2018
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