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.

1, 4, 6, 8, 11, 13, 16, 18, 21, 23, 25, 28, 30, 33, 35, 37, 40, 42, 45, 47, 49, 52, 54, 57, 59, 62, 64, 66, 69, 71, 74, 76, 78, 81, 83, 86, 88, 91, 93, 95, 98, 100, 103, 105, 107, 110, 112, 115, 117, 120, 122, 124, 127, 129, 132, 134, 136, 139, 141, 144, 146, 148, 151

Offset: 1

Author

Benoit Cloitre, Sep 13 2003

Keywords

• nonn

Comments

From Joseph Biberstine (jrbibers(AT)indiana.edu), May 02 2006: (Start)

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.

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)

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

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

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

Is A086377 the sequence of positions of 0 in A189687? - Clark Kimberling, Apr 25 2011

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

In the Fokkink-Joshi paper, this sequence is the Cloitre (1,1,3,2)-hiccup sequence, - Michael De Vlieger, Jul 29 2025

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Wieb Bosma, Michel Dekking, and Wolfgang Steiner, A remarkable sequence related to Pi and sqrt(2), arXiv:1710.01498 [math.NT], 2017.
• Wieb Bosma, Michel Dekking, and Wolfgang Steiner, A remarkable sequence related to Pi and sqrt(2), Integers, Electronic Journal of Combinatorial Number Theory 18A (2018), #A4.
• Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 7.
• Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, arXiv:2507.16956 [math.CO], 2025. See pp. 3, 9, 15.
• Wolfgang Steiner, Continued fractions and S-adic sequences, Numeration systems: automata, combinatorics, dynamical systems, number theory, Univ. de Paris (France, 2021), see p. 33.

Crossrefs

Cf. A189687, A081477.

Programs

Magma

[Floor((1+Sqrt(2))*n-1/Sqrt(2)): n in [1..70]]; // Vincenzo Librandi, Oct 05 2017

Mathematica

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 *)

Formula

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

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