proposed

approved

editing

proposed

Cf. A257905, A257984A257909.

approved

editing

proposed

approved

editing

proposed

allocated for Clark KimberlingSequence (a(n)) generated by Rule 3 (in Comments) with a(1) = 1 and d(1) = 2.

1, 2, 5, 3, 7, 4, 9, 8, 16, 6, 12, 24, 10, 20, 11, 22, 14, 28, 13, 26, 15, 30, 17, 34, 18, 36, 19, 38, 31, 25, 21, 42, 23, 46, 41, 29, 58, 27, 54, 32, 64, 35, 70, 33, 66, 39, 78, 37, 74, 40, 80, 44, 88, 43, 86, 47, 94, 45, 90, 48, 96, 49, 98, 50, 100, 56

1,2

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.

Clark Kimberling, <a href="/A257983/b257983.txt">Table of n, a(n) for n = 1..1000</a>

a(1) = 1, d(1) = 2;

a(2) = 2, d(2) = 1;

a(3) = 5, d(3) = 3;

a(4) = 3, d(4) = -2.

{a, f} = {{1}, {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 *)

Cf. A257905, A257984.

allocated

nonn,easy

Clark Kimberling, May 19 2015

approved

editing

allocated for Clark Kimberling

allocated

approved