A257909 Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 2.

2, 1, 3, -2, 4, -3, 5, -1, 8, -10, 6, 12, -14, 10, -9, 11, -8, 14, -15, 13, -11, 15, -13, 17, -16, 18, -17, 19, -7, -6, -4, 21, -19, 23, -5, -12, 29, -31, 27, -22, 32, -29, 35, -37, 33, -27, 39, -41, 37, -34, 40, -36, 44, -45, 43, -39, 47, -49, 45, -42, 48

Offset: 1

Author

Clark Kimberling, May 16 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) = 0, d(1) = 2;
a(2) = 1, d(2) = 1;
a(3) = 4, d(3) = 3;
a(4) = 2, d(4) = -2.
		

Links

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

Crossrefs

Cf. A257905, A257908.

Programs

Mathematica

{a, f} = {{0}, {2}}; 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 *)