OFFSET
1,2
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
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 18A (2018), #A4.
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 7.
Benoit Cloitre, The Golden Sieve and its connections to Hiccup sequences and Fraenkel games, arXiv:2602.17735 [math.NT], 2026. See pp. 2, 27-28, 30.
Robbert Fokkink and Gandhar Joshi, On Cloitre's hiccup sequences, Ramanujan J. 69 (2026), 40. See pp. 3, 4 (Table 1), 9 (Lemma 9). See also arXiv:2507.16956 [math.CO], 2025. See pp. 3, 9, 15.
Wolfgang Steiner, Continued fractions and S-adic sequences, in Numeration systems: automata, combinatorics, dynamical systems, number theory, Thesis, Univ. de Paris (France, 2021), see p. 33.
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
a(n) = A080652(n) - 1. [Fokkink-Joshi] - Michael De Vlieger, Feb 02 2026
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 *)
PROG
(Magma) [Floor((1+Sqrt(2))*n-1/Sqrt(2)): n in [1..70]]; // Vincenzo Librandi, Oct 05 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Sep 13 2003
STATUS
approved