A257982 Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 1 and d(1) = 1.

1, 2, -1, 3, 5, -6, 4, -2, 6, -5, 7, -3, 11, -13, 9, 18, -23, 13, -11, 15, -14, 16, -15, 17, -10, -4, 20, -21, 19, -17, 21, -19, 23, -18, 28, -31, 25, -9, -12, 29, -27, 31, -35, 27, -7, -8, 39, -45, 33, -29, 37, -39, 35, -26, 44, -43, 45, -47, 43, -38, 48

Offset: 1

Author

Clark Kimberling, May 19 2015

Keywords

• sign
• easy

Comments

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).

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.

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.

See A257905 for a guide to related sequences and conjectures.

Examples

			a(1) = 1, d(1) = 1;
a(2) = 3, d(2) = 2;
a(3) = 2, d(3) = -1;
a(4) = 5, d(4) = 3.
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A257905, A257981.

Programs

Mathematica

{a, f} = {{1}, {1}}; Do[tmp = {#, # - Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2 - Last[a], -1]]]]];
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 *)