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A257905 Sequence (a(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 0.

%I A257905 #20 Jun 16 2015 13:53:23
%S A257905 0,1,3,2,5,11,4,9,6,13,7,15,10,8,17,35,12,25,14,29,16,33,18,37,19,39,
%T A257905 20,41,21,43,22,45,23,47,30,26,53,24,49,40,28,57,27,55,31,63,32,65,38,
%U A257905 42,34,69,36,73,48,97,44,89,46,93,51,103,52,105,50,101
%N A257905 Sequence (a(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 0.
%C A257905 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 A257905 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 A257905 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 A257905 Conjecture:  suppose that a(1) is an nonnegative integer and d(1) is an integer.
%C A257905 If a(1) = 0 and d(1) != 1, then (a(n)) is a permutation of the nonnegative integers;
%C A257905 if a(1) = 0 and d(1) = 1, then (a(n)) is a permutation of the nonnegative integers excluding 1;
%C A257905 if a(1) = 1, then (a(n)) is a permutation of the positive integers;
%C A257905 if a(1) > 1, then (a(n)) is a permutation of the integers >1;
%C A257905 if d(1) = 0, then (d(n)) is a permutation of the integers;
%C A257905 if d(1) !=0, then (d(n)) is a permutation of the nonzero integers.
%C A257905 Guide to related sequences:
%C A257905 a(1)  d(1)      (a(n))       (d(n))
%C A257905 0       0      A257905      A258047
%C A257905 0       1      A257906      A257907
%C A257905 0       2      A257908      A257909
%C A257905 0       3      A257910      A257980
%C A257905 1       0      A258046      A258047
%C A257905 1       1      A257981      A257982
%C A257905 1       2      A257983      A257909
%C A257905 2       0      A257985      A257047
%C A257905 2       1      A257986      A257982
%C A257905 2       2      A257987      A257909
%H A257905 Clark Kimberling, <a href="/A257905/b257905.txt">Table of n, a(n) for n = 1..1000</a>
%F A257905 a(n) = A258046(n) - 1 for n >= 1.
%e A257905 a(1) = 0, d(1) = 0;
%e A257905 a(2) = 1, d(2) = 1;
%e A257905 a(3) = 3, d(3) = 2;
%e A257905 a(4) = 2, d(4) = -1.
%t A257905 {a, f} = {{0}, {0}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];
%t A257905 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 *)
%o A257905 (Haskell)
%o A257905 import Data.List ((\\))
%o A257905 a257905 n = a257905_list !! (n-1)
%o A257905 a257905_list = 0 : f [0] [0] where
%o A257905    f xs@(x:_) ds = g [2 - x .. -1] where
%o A257905      g [] = y : f (y:xs) (h:ds) where
%o A257905                   y = x + h
%o A257905                   (h:_) = [z | z <- [1..] \\ ds, x - z `notElem` xs]
%o A257905      g (h:hs) | h `notElem` ds && y `notElem` xs = y : f (y:xs) (h:ds)
%o A257905               | otherwise = g hs
%o A257905               where y = x + h
%o A257905 -- _Reinhard Zumkeller_, Jun 03 2015
%Y A257905 Cf. A258047, A257705, A257883, A175498.
%Y A257905 Cf. A256283 (putative inverse).
%K A257905 nonn,easy
%O A257905 1,3
%A A257905 _Clark Kimberling_, May 16 2015