A257903 Sequence (a(n)) generated by Algorithm (in Comments) with a(1) = 0 and d(1) = 3.

0, 1, 3, 2, 6, 4, 9, 5, 11, 8, 15, 7, 16, 10, 18, 13, 23, 12, 24, 14, 25, 38, 17, 31, 19, 34, 20, 36, 21, 39, 22, 41, 28, 45, 26, 46, 30, 51, 27, 49, 29, 52, 43, 67, 32, 57, 35, 61, 33, 60, 37, 65, 40, 69, 42, 72, 54, 47, 78, 44, 76, 50, 83, 53, 87, 48, 84

Offset: 1

Author

Clark Kimberling, May 13 2015

Keywords

• nonn
• easy

Comments

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.

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 A257883 for a guide to related sequences.

Examples

			a(1) = 0, d(1) = 3;
a(2) = 1, d(2) = 1;
a(3) = 3, d(3) = 2;
a(4) = 2, d(4) = -1.
		

Links

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

Crossrefs

Cf. A257904, A257883, A257705.

Programs

Mathematica

a[1] = 0; d[1] = 3; 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]]
Table[{h = Min[Intersection[c[k], T[k]]], a[k + 1] = a[k] + h,
   d[k + 1] = h, k = k + 1}, {i, 1, zz}];
u = Table[a[k], {k, 1, zz}]  (* A257903 *)
Table[d[k], {k, 1, zz}]      (* A257904 *)

Formula

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