A080428 -id:A080428 - cp's OEIS frontend

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

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, 35, 36, 38, 39, 41, 42, 43, 44, 46, 47, 49, 50, 52, 53, 55, 56, 57, 58, 60, 61, 63, 64, 65, 66, 68, 69, 71, 72, 74, 75, 77, 78, 79, 80, 82, 83, 85, 86

Offset: 1

Clark Kimberling

nonn

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
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
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
Positions of 1 in the fixed point of the morphism 0->11, 1->101; see A285430.
Conjecture:  -1 < n*r - a(n) < 2 for n>=1, where r = (1 + sqrt(3))/2. - Clark Kimberling, Apr 29 2017

Carmine Suriano, Table of n, a(n) for n = 1..10000
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See pp. 5, 12.

Cf. A026362, A079255, A080428, A187224, A026364, A004526, A285430.

seqA = Table[Floor[(n+2)/2], {n, 1, 180}] (* A004526 *)
seqB = Table[n, {n, 1, 80}];              (* A000027 *)
jointRank[{seqA_, seqB_}] := {Flatten@Position[#1, {_, 1}],
Flatten@Position[#1, {_, 2}]} &[Sort@Flatten[{{#1, 1} & /@ seqA, {#1, 2} & /@ seqB}, 1]];
limseqU = FixedPoint[jointRank[{seqA, #1[[1]]}] &, jointRank[{seqA, seqB}]][[1]]                                  (* A026363 *)
Complement[Range[Length[seqA]], limseqU] (* A026364 *)
(* Peter J. C. Moses, Mar 10 2011 *)
s = Nest[Flatten[# /. {0 -> {1, 1}, 1 -> {1, 0, 1}}] &, {0}, 13] (* A285430 *)
Flatten[Position[s, 0]]  (* A026364 *)
Flatten[Position[s, 1]]  (* A026363 *)
(* Clark Kimberling, Apr 28 2017 *)

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

A079255 a(n) is taken to be the smallest positive integer greater than a(n-1) such that the condition "n is in the sequence if and only if a(n) is odd and a(n+1) is even" can be satisfied.

Original entry on oeis.org

1, 4, 6, 9, 12, 15, 18, 20, 23, 26, 28, 31, 34, 36, 39, 42, 44, 47, 50, 53, 56, 58, 61, 64, 66, 69, 72, 75, 78, 80, 83, 86, 88, 91, 94, 97, 100, 102, 105, 108, 110, 113, 116, 119, 122, 124, 127, 130, 132, 135, 138, 140, 143, 146, 148, 151, 154, 157, 160, 162, 165, 168

Offset: 1

N. J. A. Sloane and Matthew Vandermast, Feb 04 2003

No two terms in the sequence are consecutive integers (see example for a(3)).

			a(2) cannot be odd; it also cannot be 2, since that would imply that a(2) was odd. 4 is the smallest value for a(2) that creates no contradiction. a(3) cannot be 5, which would imply that a(5) was odd because it is known from 4's being in the sequence that a(4) is odd and a(5) even. 6 is the smallest value for a(3) that creates no contradiction.

B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence (math.NT/0305308)

Cf. A079000, A079259. First differences give A080428.

Cf. A026363, A080428.

With the convention A026363(0)=0 (offset is 1 for this sequence) we have a(n)=A026363(2n)+1; a(n)=(1+sqrt(3))*n+O(1). The sequence satisfies the meta-system for n>=2: a(a(n))=2*a(n)+2*n+2 ; a(a(n)-1)=2*a(n)+2*n-1 ; a(a(n)-2)=2*a(n)+2*n-4 which allows us to have all terms since first differences =2 or 3 only. a(n)=a(n-1)+3 if n is in A026363, a(n)=a(n-1)+2 otherwise (if n is in A026364). - Benoit Cloitre, Apr 23 2008

cp's OEIS Frontend

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

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