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Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 3.
3

%I #5 Jun 05 2015 03:42:18

%S 3,1,2,-1,4,-2,5,-4,6,-3,9,-10,8,-5,11,-9,13,-12,14,-13,15,-11,19,-21,

%T 17,-14,20,-19,21,-18,24,-25,23,-17,29,-33,25,-23,27,-22,32,-31,33,

%U -35,31,-27,35,-34,36,-29,-6,37,-30,44,-49,39,-37,41,-26,-7,49

%N Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 3.

%C Rule 3 follows. For k >= 1, let A(k) = {a(1), …, a(k)} and D(k) = {d(1), …, d(k)}. Begin with k = 1 and nonnegative integers a(1) and d(1).

%C Step 1: If there is an integer h such that 1 - a(k) < h < 0 and h is not in D(k) and a(k) + h is not in A(k), let d(k+1) be the least such h, let a(k+1) = a(k) + h, replace k by k + 1, and repeat Step 1; otherwise do Step 2.

%C Step 2: Let h be the least positive integer not in D(k) such that a(k) - h is not in A(k). Let a(k+1) = a(k) + h and d(k+1) = h. Replace k by k+1 and do Step 1.

%C See A257905 for a guide to related sequences and conjectures.

%H Clark Kimberling, <a href="/A257980/b257980.txt">Table of n, a(n) for n = 1..1000</a>

%e a(1) = 0, d(1) = 3;

%e a(2) = 1, d(2) = 1;

%e a(3) = 3, d(3) = 2;

%e a(4) = 2, d(4) = -1.

%t {a, f} = {{0}, {3}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];

%t If[! IntegerQ[tmp[[1]]], tmp = {Last[a] + #, #} &[NestWhile[# + 1 &, 1, ! (! MemberQ[f, #] && ! MemberQ[a, Last[a] - #]) &]]]; AppendTo[a, tmp[[1]]]; AppendTo[f, tmp[[2]]], {120}]; {a, f} (* _Peter J. C. Moses_, May 14 2015 *)

%Y Cf. A257905, A257910.

%K easy,sign

%O 1,1

%A _Clark Kimberling_, May 19 2015