cp's OEIS Frontend

A081835 a(1)=1, a(n) = a(n-1) + 5 if n is already in the sequence, a(n) = a(n-1) + 4 otherwise.

%I A081835 #21 Jul 29 2025 15:57:13
%S A081835 1,5,9,13,18,22,26,30,35,39,43,47,52,56,60,64,68,73,77,81,85,90,94,98,
%T A081835 102,107,111,115,119,124,128,132,136,140,145,149,153,157,162,166,170,
%U A081835 174,179,183,187,191,196,200,204,208,212,217,221,225,229,234,238,242
%N A081835 a(1)=1, a(n) = a(n-1) + 5 if n is already in the sequence, a(n) = a(n-1) + 4 otherwise.
%C A081835 In the Fokkink-Joshi paper, this sequence is the Cloitre (0,1,5,4)-hiccup sequence. - _Michael De Vlieger_, Jul 29 2025
%H A081835 Muniru A Asiru, <a href="/A081835/b081835.txt">Table of n, a(n) for n = 1..2000</a>
%H A081835 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 p. 3.
%F A081835 a(n) = floor(rn-(4r-1)/(r+1)) where r=2+sqrt(5).
%e A081835 a(2) = a(1)+4 = 5 because 2 is not already in the sequence;
%e A081835 a(3) = a(2)+4 = 9 because 3 is not already in the sequence;
%e A081835 a(4) = a(3)+4 = 13 because 4 is not already in the sequence;
%e A081835 a(5) = a(4)+5 = 18 because 5 is already in the sequence.
%p A081835 r:=2+sqrt(5): seq(floor(r*n-(4*r-1)/(r+1)),n=1..60); # _Muniru A Asiru_, Jun 06 2018
%t A081835 Module[{r=2+Sqrt[5],c},c=(4r-1)/(r+1);Table[Floor[r*n-c],{n,60}]] (* _Harvey P. Dale_, Feb 19 2013 *)
%Y A081835 Cf. A064437, A081843.
%K A081835 nonn
%O A081835 1,2
%A A081835 _Benoit Cloitre_, Apr 11 2003