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%I #6 May 14 2015 11:59:30
%S 2,1,3,6,4,8,5,10,16,12,7,14,22,15,9,18,28,20,11,23,13,24,37,26,40,27,
%T 42,30,46,32,17,34,52,36,19,38,58,39,21,43,64,44,67,45,69,48,25,50,76,
%U 51,78,54,82,56,29,59,31,60,91,62,94,63,33,66,100,68,35
%N Sequence (a(n)) generated by Rule 1 (in Comments) with a(1) = 2 and d(1) = 1.
%C Rule 1 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 greatest 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 Conjecture: if a(1) is an nonnegative integer and d(1) is an integer, then (a(n)) is a permutation of the nonnegative integers (if a(1) = 0) or a permutation of the positive integers (if a(1) > 0). Moreover, (d(n)) is a permutation of the integers if d(1) = 0, or of the nonzero integers if d(1) > 0.
%C See A257705 for a guide to related sequences.
%H Clark Kimberling, <a href="/A257881/b257881.txt">Table of n, a(n) for n = 1..1000</a>
%F a(k+1) - a(k) = d(k+1) for k >= 1.
%e a(1) = 2, d(1) = 0;
%e a(2) = 1, d(2) = -1;
%e a(3) = 3, d(3) = 2;
%e a(4) = 6, d(4) = 3.
%t a[1] = 2; d[1] = 1; k = 1; z = 10000; zz = 120;
%t A[k_] := Table[a[i], {i, 1, k}]; diff[k_] := Table[d[i], {i, 1, k}];
%t c[k_] := Complement[Range[-z, z], diff[k]];
%t T[k_] := -a[k] + Complement[Range[z], A[k]];
%t s[k_] := Intersection[Range[-a[k], -1], c[k], T[k]];
%t Table[If[Length[s[k]] == 0, {h = Min[Intersection[c[k], T[k]]], a[k + 1] = a[k] + h, d[k + 1] = h, k = k + 1}, {h = Max[s[k]], a[k + 1] = a[k] + h, d[k + 1] = h, k = k + 1}], {i, 1, zz}];
%t u = Table[a[k], {k, 1, zz}] (* A257881 *)
%t Table[d[k], {k, 1, zz}] (* essentially A257880 *)
%Y Cf. A257880, A257705, A081145, A257883, A175498.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, May 13 2015
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