

A257982


Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 1 and d(1) = 1.


4



1, 2, 1, 3, 5, 6, 4, 2, 6, 5, 7, 3, 11, 13, 9, 18, 23, 13, 11, 15, 14, 16, 15, 17, 10, 4, 20, 21, 19, 17, 21, 19, 23, 18, 28, 31, 25, 9, 12, 29, 27, 31, 35, 27, 7, 8, 39, 45, 33, 29, 37, 39, 35, 26, 44, 43, 45, 47, 43, 38, 48
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OFFSET

1,2


COMMENTS

Rule 3 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 least 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.
See A257905 for a guide to related sequences and conjectures.


LINKS

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


EXAMPLE

a(1) = 1, d(1) = 1;
a(2) = 3, d(2) = 2;
a(3) = 2, d(3) = 1;
a(4) = 5, d(4) = 3.


MATHEMATICA

{a, f} = {{1}, {1}}; Do[tmp = {#, #  Last[a]} &[Min[Complement[#, Intersection[a, #]]&[Last[a] + Complement[#, Intersection[f, #]] &[Range[2  Last[a], 1]]]]];
If[! IntegerQ[tmp[[1]]], tmp = {Last[a] + #, #} &[NestWhile[# + 1 &, 1, ! (! MemberQ[f, #] && ! MemberQ[a, Last[a]  #]) &]]]; AppendTo[a, tmp[[1]]]; AppendTo[f, tmp[[2]]], {120}]; {a, f} (* Peter J. C. Moses, May 14 2015 *)


CROSSREFS

Cf. A257905, A257981.
Sequence in context: A239738 A058202 A327452 * A275705 A217036 A127201
Adjacent sequences: A257979 A257980 A257981 * A257983 A257984 A257985


KEYWORD

sign,easy


AUTHOR

Clark Kimberling, May 19 2015


STATUS

approved



