%I #14 May 09 2019 17:39:19
%S 0,1,3,2,5,9,4,10,6,11,8,15,7,16,14,22,12,23,17,27,13,25,18,31,19,33,
%T 20,35,24,40,21,38,29,47,26,45,28,48,30,51,36,58,32,55,39,63,34,59,37,
%U 64,41,67,42,70,43,72,44,74,50,81,46,78,111,49,83,52,87
%N Sequence (a(n)) generated by Algorithm (in Comments) with a(1) = 0 and d(1) = 0.
%C Algorithm: 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). Let h be the least integer > -a(k) such that h is not in D(k) and 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 repeat inductively.
%C Conjecture: if a(1) is a 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 Guide to related sequences:
%C a(1) d(1) (a(n)) (d(n))
%C 0 0 A257883 A175499 except for initial terms
%C 0 1 A257884 A175499
%C 0 2 A257885 A257902
%C 0 3 A257903 A257904
%C 1 0 A175498 A175499 except for first term
%C 1 1 A257905 A175499
%C 2 0 A257908 A257909
%C 2 1 A257910 A257909 except for initial terms
%C 2 2 A257911 A257912
%H Clark Kimberling, <a href="/A257883/b257883.txt">Table of n, a(n) for n = 1..1000</a>
%F a(k+1) - a(k) = d(k+1) for k >= 1.
%F Also, A257883(n) = -1 + A175498(n) for n >= 1.
%e a(1) = 0, d(1) = 0;
%e a(2) = 1, d(2) = 1;
%e a(3) = 3, d(3) = 2;
%e a(4) = 2, d(4) = -1.
%t a[1] = 0; d[1] = 0; 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 Table[{h = Min[Intersection[c[k], T[k]]], a[k + 1] = a[k] + h,
%t d[k + 1] = h, k = k + 1}, {i, 1, zz}];
%t u = Table[a[k], {k, 1, zz}] (* A257883, = -1 + A175498 *)
%t Table[d[k], {k, 1, zz}] (* A175499 except that here first term is 0 *)
%Y Cf. A081145, A175498, A257705.
%K nonn,easy
%O 1,3
%A _Clark Kimberling_, May 13 2015