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A026363 a(n) is the least k such that s(k) = n, where s = A026362.

%I A026363 #38 Jul 30 2025 17:48:39
%S A026363 1,3,4,5,6,8,9,11,12,14,15,17,18,19,20,22,23,25,26,27,28,30,31,33,34,
%T A026363 35,36,38,39,41,42,43,44,46,47,49,50,52,53,55,56,57,58,60,61,63,64,65,
%U A026363 66,68,69,71,72,74,75,77,78,79,80,82,83,85,86
%N A026363 a(n) is the least k such that s(k) = n, where s = A026362.
%C A026363 Or, starting from the natural number, delete successively from the working sequence the term in position 2*a(n). From natural numbers, delete the term in position 2*1, i.e., 2. This leaves 1,3,4,5,6,7,8,9,10,11,... . Delete now the term in position 2*3=6, i.e., 7. This leaves 1,3,4,5,6,8,9,10,11,... . Delete now the term in position 2*4=8, i.e., 10. This leaves 1,3,4,5,6,8,9,11,... and so on. - Philippe Lallouet (philip.lallouet(AT)wanadoo.fr), Aug 20 2007
%C A026363 The term deleted from the n-th working sequence is equal to A026364(n), which means that all integers which are not in the present sequence are in A026364 and no others. - Philippe Lallouet (philip.lallouet(AT)orange.fr), May 05 2008
%C A026363 Complement of A026364; also the rank transform (as at A187224) of A004526 after removal of its first three terms, leaving (1,2,2,3,3,4,4,5,5,6,6,...). - _Clark Kimberling_, Mar 10 2011
%C A026363 Positions of 1 in the fixed point of the morphism 0->11, 1->101; see A285430.
%C A026363 Conjecture:  -1 < n*r - a(n) < 2 for n>=1, where r = (1 + sqrt(3))/2. - _Clark Kimberling_, Apr 29 2017
%H A026363 Carmine Suriano, <a href="/A026363/b026363.txt">Table of n, a(n) for n = 1..10000</a>
%H A026363 Robbert Fokkink and Gandhar Joshi, <a href="https://arxiv.org/abs/2507.16956">On Cloitre's hiccup sequences</a>, arXiv:2507.16956 [math.CO], 2025. See pp. 5, 12.
%F A026363 a(1)=1, then a(n)=a(n-1)+2 if n is even and n/2 is not in the sequence, a(n)=a(n-1)+1 otherwise (in particular a(2k+1)=a(2k)+1). a(n)=(1+sqrt(3))/2*n+O(1). Taking a(0)=0, for n>=1 a(2n)-a(2n-2)=A080428(n). - _Benoit Cloitre_, Apr 23 2008
%t A026363 seqA = Table[Floor[(n+2)/2], {n, 1, 180}] (* A004526 *)
%t A026363 seqB = Table[n, {n, 1, 80}];              (* A000027 *)
%t A026363 jointRank[{seqA_, seqB_}] := {Flatten@Position[#1, {_, 1}],
%t A026363 Flatten@Position[#1, {_, 2}]} &[Sort@Flatten[{{#1, 1} & /@ seqA, {#1, 2} & /@ seqB}, 1]];
%t A026363 limseqU = FixedPoint[jointRank[{seqA, #1[[1]]}] &, jointRank[{seqA, seqB}]][[1]]                                  (* A026363 *)
%t A026363 Complement[Range[Length[seqA]], limseqU] (* A026364 *)
%t A026363 (* _Peter J. C. Moses_, Mar 10 2011 *)
%t A026363 s = Nest[Flatten[# /. {0 -> {1, 1}, 1 -> {1, 0, 1}}] &, {0}, 13] (* A285430 *)
%t A026363 Flatten[Position[s, 0]]  (* A026364 *)
%t A026363 Flatten[Position[s, 1]]  (* A026363 *)
%t A026363 (* _Clark Kimberling_, Apr 28 2017 *)
%Y A026363 Cf. A026362, A079255, A080428, A187224, A026364, A004526, A285430.
%K A026363 nonn
%O A026363 1,2
%A A026363 _Clark Kimberling_