A156253 Least k such that A054353(k) >= n.

1, 2, 2, 3, 3, 4, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 15, 15, 16, 17, 18, 18, 19, 19, 20, 21, 21, 22, 23, 24, 24, 25, 26, 26, 27, 27, 28, 29, 30, 30, 31, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 39, 40, 41, 42, 42, 43, 44, 44, 45, 45, 46, 47, 47, 48, 49, 50

Offset: 1

Author

Benoit Cloitre, Feb 07 2009

Keywords

• nonn

Comments

a(n)=1 plus the number of symbol changes in the first n terms of A000002. - Jean-Marc Fedou and Gabriele Fici, Mar 18 2010

From N. J. A. Sloane, Nov 12 2018: (Start)

This seems to be A001462 rewritten so the run lengths are given by A000002. The companion sequence, A000002 rewritten so the run lengths are given by A001462, is A321020.

Note that Kolakoski's sequence A000002 and Golomb's sequence A001462 have very similar definitions, although the asymptotic behavior of A001462 is well-understood, while that of A000002 is a mystery. The asymptotic behavior of the two hybrids A156253 and A321020 might be worth investigating. (End)

To expand upon N. J. A. Sloane's comments, it's worth noting that Golomb's sequence has a formula from Colin Mallows: g(n) = g(n-g(g(n-1))) + 1, which closely resembles a(n) = a(n-gcd(a(a(n-1)),2)) + 1. - Jon Maiga, May 16 2023

Links

• Jon Maiga, Table of n, a(n) for n = 1..10000
• J. M. Fedou and G. Fici, Some remarks on differentiable sequences and recursivity, Journal of Integer Sequences 13(3): Article 10.3.2 (2010).
• Jon Maiga, A Recurrence Related to the Kolakoski Sequence
• N. J. A. Sloane, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, Part I, Part 2, Slides (Mentions this sequence)

Crossrefs

Cf. A000002, A001462, A054353, A156351, A321020.

Programs

Mathematica

a2 = {1, 2, 2}; Do[ a2 = Join[a2, {1 + Mod[n - 1, 2]}], {n, 3, 80}, {i, 1, a2[[n]]}]; a3 = Accumulate[a2]; a[1] = 1; a[n_] := a[n] = For[k = a[n - 1], True, k++, If[a3[[k]] >= n, Return[k]]]; Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Jun 18 2013 *)
a[1] = 1;
a[n_]:=a[n]=a[n-GCD[a[a[n - 1]], 2]]+1
Array[a, 100] (* Jon Maiga, May 16 2023 *)

Formula

Conjecture: a(n) should be asymptotic to 2n/3.

Length of n-th run of the sequence = A000002(n). - Benoit Cloitre, Feb 19 2009

Conjecture: a(n) = (a(a(n-1)) mod 2) + a(n-2) + 1. - Jon Maiga, Dec 09 2021

a(n) = a(n-gcd(a(a(n-1)), 2)) + 1. - Jon Maiga, May 16 2023