cp's OEIS Frontend

A086377 a(1)=1; a(n)=a(n-1)+2 if n is in the sequence; a(n)=a(n-1)+2 if n and (n-1) are not in the sequence; a(n)=a(n-1)+3 if n is not in the sequence but (n-1) is in the sequence.

%I A086377 #50 Jul 29 2025 16:02:58
%S A086377 1,4,6,8,11,13,16,18,21,23,25,28,30,33,35,37,40,42,45,47,49,52,54,57,
%T A086377 59,62,64,66,69,71,74,76,78,81,83,86,88,91,93,95,98,100,103,105,107,
%U A086377 110,112,115,117,120,122,124,127,129,132,134,136,139,141,144,146,148,151
%N A086377 a(1)=1; a(n)=a(n-1)+2 if n is in the sequence; a(n)=a(n-1)+2 if n and (n-1) are not in the sequence; a(n)=a(n-1)+3 if n is not in the sequence but (n-1) is in the sequence.
%C A086377 From Joseph Biberstine (jrbibers(AT)indiana.edu), May 02 2006: (Start)
%C A086377 The continued fraction 4/Pi = 1 + 1/(3 + 4/(5 + 9/(7 + 16/(9 + 25/(11 + ...))))) (see A079037) suggests the recurrence b(n) = 2*n - 1 + n^2/b(n+1) with b(1) = 4/Pi. Solving the above recurrence in the other direction we would have b(n) = (n-1)^2/b(n-1 - 2*n + 3) with b(1) = 4/Pi.
%C A086377 Now consider this last defined sequence {b(n)}. It appears to grow linearly. (1) Does it? (2) What is the limit of b(n)/n as n->oo? (3) How does the limit depend on the initial term b(1)? (End)
%C A086377 From the recurrence relation, it follows that the limit L = lim_{b->oo} b(n)/n satisfies the following quadratic equation: L^2 - 2*L - 1 = 0 implying that L = 1+sqrt(2) or 1-sqrt(2). - _Max Alekseyev_, May 02 2006
%C A086377 Note that b(n)/n decreases, while b(n)/(n+1) increases. I speculate that 4/Pi is the only b(1) value such that b(n)/n converges to 1+sqrt(2) instead of 1-sqrt(2). - _Don Reble_, May 02 2006
%C A086377 It appears that round( b(n) ) = floor((1+sqrt(2))*n - 1/sqrt(2)) = A086377(n) = a(n). This is certainly true for the first 190 terms. Is there a formal proof? - _Paul D. Hanna_, May 02 2006
%C A086377 Is A086377 the sequence of positions of 0 in A189687? - _Clark Kimberling_, Apr 25 2011
%C A086377 The three conjectures by respectively Biberstein, Hanna, and Kimberling have all been proved, see the paper by Bosma et al. in the Links. - _Michel Dekking_, Oct 05 2017
%C A086377 In the Fokkink-Joshi paper, this sequence is the Cloitre (1,1,3,2)-hiccup sequence, - _Michael De Vlieger_, Jul 29 2025
%H A086377 Michael De Vlieger, <a href="/A086377/b086377.txt">Table of n, a(n) for n = 1..10000</a>
%H A086377 Wieb Bosma, Michel Dekking, and Wolfgang Steiner, <a href="https://arxiv.org/abs/1710.01498">A remarkable sequence related to Pi and sqrt(2)</a>, arXiv:1710.01498 [math.NT], 2017.
%H A086377 Wieb Bosma, Michel Dekking, and Wolfgang Steiner, <a href="http://math.colgate.edu/~integers/sjs4/sjs4.Abstract.html">A remarkable sequence related to Pi and sqrt(2)</a>, Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A4.
%H A086377 Benoit Cloitre, <a href="https://arxiv.org/abs/2506.18103">A study of a family of self-referential sequences</a>, arXiv:2506.18103 [math.GM], 2025. See p. 7.
%H A086377 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. 3, 9, 15.
%H A086377 Wolfgang Steiner, <a href="https://www.irif.fr/~steiner/hdr.pdf">Continued fractions and S-adic sequences</a>, Numeration systems: automata, combinatorics, dynamical systems, number theory, Univ. de Paris (France, 2021), see p. 33.
%F A086377 a(n) = floor((1+sqrt(2))*n - 1/sqrt(2)).
%F A086377 n is in the sequence if A004641(n)=1 or A001030(n)=2. a(n) = A080652(n) - 1 = A064437(n+1) - 2 = A081841(n+2) - 3. - _Ralf Stephan_, Feb 23 2004
%t A086377 Block[{c, k}, c[_] := False; k = 1; c[1] = True; {k}~Join~Reap[Do[If[c[n], k += 2, If[c[n - 1], k += 3, k += 2]]; c[k] = True; Sow[k], {n, 2, 120}] ][[-1, 1]] ] (* _Michael De Vlieger_, Jul 02 2025 *)
%o A086377 (Magma) [Floor((1+Sqrt(2))*n-1/Sqrt(2)): n in [1..70]]; // _Vincenzo Librandi_, Oct 05 2017
%Y A086377 Cf. A189687, A081477.
%K A086377 nonn
%O A086377 1,2
%A A086377 _Benoit Cloitre_, Sep 13 2003