A080652 a(1)=2; for n>1, a(n)=a(n-1)+3 if n is already in the sequence, a(n)=a(n-1)+2 otherwise.

2, 5, 7, 9, 12, 14, 17, 19, 22, 24, 26, 29, 31, 34, 36, 38, 41, 43, 46, 48, 50, 53, 55, 58, 60, 63, 65, 67, 70, 72, 75, 77, 79, 82, 84, 87, 89, 92, 94, 96, 99, 101, 104, 106, 108, 111, 113, 116, 118, 121, 123, 125, 128, 130, 133, 135, 137, 140, 142, 145

Offset

1,1

Comments

In the Fokkink-Joshi paper, this sequence is the Cloitre (0,2,3,2)-hiccup sequence. - Michael De Vlieger, Jul 29 2025

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..5000

Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 7.

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

Benoit Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, arXiv:math/0305308 [math.NT], 2003.

Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, Ramanujan J. 69 (2026), 40. See pp. 4 (Table 1), 9. See also arXiv:2507.16956 [math.CO], 2025. See pp. 3, 9.

Formula

a(n) = floor(n*r + 1/(1+r)) where r = 1+sqrt(2).

a(n) = A086377(n) + 1. [Fokkink-Joshi] - Michael De Vlieger, Feb 02 2026

Mathematica

a[1] = 2;
a[n_] := a[n] = If[MemberQ[Array[a, n-1], n], a[n-1] + 3, a[n-1] + 2];
Array[a, 60] (* Jean-François Alcover, Oct 01 2018 *)
Table[Floor[n (1 + Sqrt[2]) + 1 / (1 + (1 + Sqrt[2]))], {n, 60}] (* Vincenzo Librandi, Oct 02 2018 *)

Prog

(PARI) a(n) = my(r=sqrt(2)+1); (r*(r+1)*n+1)\(r+1); \\ Altug Alkan, Oct 01 2018
(Magma) [Floor(n*(1+Sqrt(2)) + 1/(1+(1+Sqrt(2)))): n in [1..60]]; // Vincenzo Librandi, Oct 02 2018

Crossrefs

Cf. A080455-A080458, A080036, A080037, A086377. Apart from start, equals A064437 - 1.

Sequence in context: A184007 A003256 A029901 * A187344 A342741 A260480

Adjacent sequences: A080649 A080650 A080651 * A080653 A080654 A080655

Keyword

nonn

Author

N. J. A. Sloane, Mar 23 2003

Status

approved