A257876 Sequence (a(n)) generated by Rule 1 (in Comments) with a(1) = 0 and d(1) = 2.

0, 1, 4, 3, 7, 5, 2, 8, 13, 9, 16, 11, 19, 12, 6, 15, 25, 17, 28, 18, 30, 21, 10, 23, 37, 24, 39, 27, 43, 29, 14, 31, 49, 33, 52, 35, 55, 36, 57, 34, 56, 38, 61, 41, 20, 44, 22, 47, 73, 48, 75, 51, 79, 53, 26, 58, 87, 59, 89, 60, 91, 54, 88, 50, 83, 42, 77

Offset: 1

Author

Clark Kimberling, May 12 2015

Keywords

• nonn
• easy

Comments

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).

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.

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.

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.

See A257705 for a guide to related sequences.

Examples

			a(1) = 0, d(1) = 2;
a(2) = 1, d(2) = 1;
a(3) = 3, d(3) = 3;
a(4) = 4, d(4) = -1.
The first terms of (d(n)) are (2,1,3,-1,4,-2,-3,6,5,...), which differs from A131389 only in initial terms.
		

Links

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

Crossrefs

Cf. A131388, A257705, A081145, A257883, A175498.

Programs

Mathematica

a[1] = 0; d[1] = 2; k = 1; z = 10000; zz = 120;
A[k_] := Table[a[i], {i, 1, k}]; diff[k_] := Table[d[i], {i, 1, k}];
c[k_] := Complement[Range[-z, z], diff[k]];
T[k_] := -a[k] + Complement[Range[z], A[k]];
s[k_] := Intersection[Range[-a[k], -1], c[k], T[k]];
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}];
u = Table[a[k], {k, 1, zz}]  (* A257876 *)
Table[d[k], {k, 1, zz}]      (* A131389 essentially *)

Formula

a(k+1) - a(k) = d(k+1) for k >= 1.