

A257908


Sequence (a(n)) generated by Rule 3 (in Comments) with a(1) = 0 and d(1) = 2.


4



0, 1, 4, 2, 6, 3, 8, 7, 15, 5, 11, 23, 9, 19, 10, 21, 13, 27, 12, 25, 14, 29, 16, 33, 17, 35, 18, 37, 30, 24, 20, 41, 22, 45, 40, 28, 57, 26, 53, 31, 63, 34, 69, 32, 65, 38, 77, 36, 73, 39, 79, 43, 87, 42, 85, 46, 93, 44, 89, 47, 95, 48, 97, 49, 99, 55, 111
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OFFSET

1,3


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


MATHEMATICA

{a, f} = {{0}, {2}}; 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, A257909.
Sequence in context: A237056 A285296 A257885 * A097467 A169839 A308712
Adjacent sequences: A257905 A257906 A257907 * A257909 A257910 A257911


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, May 16 2015


STATUS

approved



