

A257980


Sequence (d(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 3.


3



3, 1, 2, 1, 4, 2, 5, 4, 6, 3, 9, 10, 8, 5, 11, 9, 13, 12, 14, 13, 15, 11, 19, 21, 17, 14, 20, 19, 21, 18, 24, 25, 23, 17, 29, 33, 25, 23, 27, 22, 32, 31, 33, 35, 31, 27, 35, 34, 36, 29, 6, 37, 30, 44, 49, 39, 37, 41, 26, 7, 49
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


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


MATHEMATICA

{a, f} = {{0}, {3}}; 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, A257910.
Sequence in context: A066743 A257915 A257904 * A203531 A324885 A046645
Adjacent sequences: A257977 A257978 A257979 * A257981 A257982 A257983


KEYWORD

easy,sign


AUTHOR

Clark Kimberling, May 19 2015


STATUS

approved



