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%I #33 Jun 08 2021 23:21:15
%S 2,3,4,4,5,5,6,6,6,7,7,7,8,8,8,9,9,9,9,10,10,10,10,11,11,11,11,12,12,
%T 12,12,12,13,13,13,13,13,14,14,14,14,14,15,15,15,15,15,16,16,16,16,16,
%U 16,17,17,17,17,17,17,18,18,18,18,18,18,19,19,19,19,19
%N Take Golomb's sequence A001462 and shorten all the runs by 1.
%C In other words, apply Lenormand's "raboter" transformation (see A318921) to A001462.
%C Each value of n (n >= 2) appears exactly A001462(n)-1 times.
%C There should be a simple formula for a(n), just as there is for Golomb's sequence. - _N. J. A. Sloane_, Nov 15 2018. After 10000 terms, a(n) seems to be growing like constant*n^0.640. - _N. J. A. Sloane_, Jun 04 2021
%H Rémy Sigrist, <a href="/A319434/b319434.txt">Table of n, a(n) for n = 1..10000</a>
%H Brady Haran and N. J. A. Sloane, <a href="https://www.youtube.com/watch?v=R4OvBB9KHMA">Planing Sequences (Le Rabot)</a>, Numberphile video, June 2021.
%H N. J. A. Sloane, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, <a href="https://vimeo.com/314786942">Part I</a>, <a href="https://vimeo.com/314790822">Part 2</a>, <a href="https://oeis.org/A320487/a320487.pdf">Slides.</a> (Mentions this sequence)
%e Golomb's sequence begins 1, 2,2, 3,3, 4,4,4, 5,5,5, ...
%e and we just shorten each run by one term, getting 2, 3, 4,4, 5,5, ...
%Y Cf. A001462, A318921, A319951.
%K nonn
%O 1,1
%A _N. J. A. Sloane_, Oct 02 2018
%E More terms from _Rémy Sigrist_, Oct 04 2018
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