OFFSET
1,3
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.
Guide to related sequences:
a(1) d(1) (a(n)) (d(n))
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
FORMULA
a(k+1) - a(k) = d(k+1) for k >= 1.
Also, a(k) = A131388(n)-1.
EXAMPLE
a(2) = a(1) + d(2) = 0 + 1 = 1;
a(3) = a(2) + d(3) = 1 + 2 = 3;
a(4) = a(3) + d(4) = 3 + (-1) = 2.
MATHEMATICA
a[1] = 0; 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}] (* A257705 *)
Table[d[k], {k, 1, zz}] (* A131389 *)
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 12 2015
STATUS
approved