A257912 Sequence (d(n)) generated by Algorithm (in Comments) with a(1) = 2 and d(1) = 2.

2, -1, 3, 1, -2, 4, 5, -6, 7, -5, 6, -4, 8, -9, 10, -8, 9, -3, 11, -13, 12, -11, 13, -7, 14, -15, 16, -14, 15, -12, 17, -19, 18, -17, 19, -10, 20, -24, 21, -20, 22, -21, 23, -22, 24, -18, 25, -29, 26, -16, 27, -32, 28, -27, 29, -23, 30, -31, 32, -30, 31, 33

Offset: 1

Author

Clark Kimberling, Jun 12 2015

Keywords

• easy
• sign

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.

Links

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

Crossrefs

Cf. A257911, A257883.

Programs

Mathematica

a[1] = 2; 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]];
Table[{h = Min[Intersection[c[k], T[k]]], a[k + 1] = a[k] + h, d[k + 1] = h, k = k + 1}, {i, 1, zz}];
Table[a[k], {k, 1, zz}]  (* A257911 *)
Table[d[k], {k, 1, zz}]  (* A257912 *)