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Conjectured permutation of the positive integers using Sequence (a(n)) generated by Rule 1 (in Comments) with a(1) = 2 and d(1) = 2.

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, every integer occurs exactly once in (d(n)) is a permutation of the integers if d(1) = 0 and every , or of the nonzero integer occurs exactly once in (d(n))) integers if d(1) > 0.

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allocated for Clark KimberlingConjectured permutation of the positive integers using Rule 1 (in Comments) with a(1) = 2 and d(1) = 2

2, 1, 4, 5, 3, 7, 12, 9, 15, 11, 6, 13, 21, 14, 8, 17, 27, 19, 10, 22, 33, 23, 36, 25, 39, 26, 41, 29, 45, 31, 16, 34, 18, 35, 54, 37, 57, 38, 20, 42, 63, 43, 66, 44, 68, 47, 24, 49, 75, 51, 78, 53, 81, 55, 28, 58, 30, 59, 90, 61, 93, 62, 32, 65, 99, 67, 102

1,1

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, every integer occurs exactly once in (d(n)) if d(1) = 0 and every nonzero integer occurs exactly once in (d(n))) if d(1) > 0.

See A257705 for a guide to related sequences.

Clark Kimberling, <a href="/A257882/b257882.txt">Table of n, a(n) for n = 1..1000</a>

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

a(1) = 2, d(1) = 2;

a(2) = 1, d(2) = -1;

a(3) = 4, d(3) = 3;

a(4) = 5, d(4) = 1.

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]];

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}] (* A257882 *)

Table[d[k], {k, 1, zz}] (* A257918 *)

Cf. A257918, A257880, A257705, A081145, A257883, A175498.

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Clark Kimberling, May 13 2015

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allocated for Clark Kimberling

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