

A257879


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


3



2, 1, 3, 4, 7, 5, 9, 6, 11, 17, 13, 8, 15, 23, 16, 10, 19, 29, 21, 12, 24, 14, 25, 38, 27, 41, 28, 43, 31, 47, 33, 18, 35, 53, 37, 20, 39, 59, 40, 22, 44, 65, 45, 68, 46, 70, 49, 26, 51, 77, 52, 79, 55, 83, 57, 30, 60, 32, 61, 92, 63, 95, 64, 34, 67, 101, 69
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OFFSET

1,1


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.


LINKS



FORMULA

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


EXAMPLE

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


MATHEMATICA

a[1] = 2; d[1] = 0; 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}] (* A257879 *)
Table[d[k], {k, 1, zz}] (* A257880 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



