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A257706
Sequence (a(n)) generated by Rule 1 (in Comments) with a(1) = 0 and d(1) = 1.
8
0, 2, 1, 4, 8, 6, 3, 9, 5, 10, 17, 12, 20, 14, 7, 16, 26, 18, 29, 19, 31, 22, 11, 24, 38, 25, 13, 28, 44, 30, 15, 32, 50, 34, 53, 36, 56, 37, 58, 40, 62, 42, 21, 45, 23, 46, 71, 48, 74, 49, 76, 52, 80, 54, 27, 57, 86, 55, 87, 59, 90, 61, 94, 64, 98, 66, 33
OFFSET
1,2
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(n+1) - a(n) = d(n+1) = A131389(n+1) for n >= 1.
EXAMPLE
a(1) = 0, d(1) = 1;
a(2) = 2, d(2) = 2;
a(3) = 1, d(3) = -1;
a(4) = 4, d(4) = 3;
(The sequence d differs from A131389 only in the first 13 terms.)
MATHEMATICA
a[1] = 0; d[1] = 1; 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}] (* A257706 *)
Table[d[k], {k, 1, zz}] (* A131389 shifted *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, May 12 2015
STATUS
approved