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 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. 8
 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 (list; graph; refs; listen; history; text; internal format)
 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 = 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 = 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->Infinity? (3) How does the limit depend on the initial term b? (End) From the recurrence relation, it follows that the L=lim 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 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 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. Wolfgang Steiner, Continued fractions and S-adic sequences, Numeration systems: automata, combinatorics, dynamical systems, number theory, Univ. de Paris (France, 2021), see p. 33. FORMULA a(n) = floor((1+sqrt(2))*n-1/sqrt(2)). n is in 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 MATHEMATICA t = Nest[Flatten[# /. {0->{0, 1, 1}, 1->{0, 1}}] &, {0}, 5] (*A189687*) f[n_] := t[[n]] Flatten[Position[t, 0]] (* A086377 conjectured *) Flatten[Position[t, 1]] (* A081477 conjectured *) s[n_] := Sum[f[i], {i, 1, n}]; s = 0; Table[s[n], {n, 1, 120}] (*A189688*) (* Clark Kimberling, Apr 25 2011 *) Table[Floor[(1 + Sqrt) n - 1/Sqrt], {n, 80}] (* Vincenzo Librandi, Oct 05 2017 *) PROG (MAGMA) [Floor((1+Sqrt(2))*n-1/Sqrt(2)): n in [1..70]]; // Vincenzo Librandi, Oct 05 2017 CROSSREFS Cf. A189687, A081477. Sequence in context: A189462 A047290 A225002 * A003662 A132635 A182131 Adjacent sequences:  A086374 A086375 A086376 * A086378 A086379 A086380 KEYWORD nonn AUTHOR Benoit Cloitre, Sep 13 2003 STATUS approved

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Last modified October 23 14:42 EDT 2021. Contains 348214 sequences. (Running on oeis4.)