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 A245219 Continued fraction expansion of the constant c in A245218; c = sup{f(n,1)}, where f(1,x) = x + 1 and thereafter f(n,x) = x + 1 if n is in A001951, else f(n,x) = 1/x. 16
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 OFFSET 1,1 COMMENTS See Comments at A245215. Likely a duplicate of A097509. - R. J. Mathar, Jul 21 2014 Theorem: Referring to Problem B6 in the 81st William Lowell Putnam Mathematical Competition (see link), in the notation of the first solution, the sequence {c_i} equals A245219. This proves the conjecture in the previous comment. - Manjul Bhargava, Kiran Kedlaya, and Lenny Ng, Sep 09 2021. LINKS Manjul Bhargava, Kiran Kedlaya, and Lenny Ng, Solutions to the 81st William Lowell Putnam Mathematical Competition N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence) EXAMPLE c = 3.43648484... ; the first 12 numbers f(n,1) comprise S(12) = {1, 2, 3, 1/3, 4/3, 7/3, 3/7, 10/7, 17/7, 24/7, 7/24, 31/24}; max(S(12)) = 24/7, with continued fraction [3,2,3]. MATHEMATICA tmpRec = \$RecursionLimit; \$RecursionLimit = Infinity; u[x_] := u[x] = x + 1; d[x_] := d[x] = 1/x; r = Sqrt[2]; w = Table[Floor[k*r], {k, 2000}]; s[1] = 1; s[n_] := s[n] = If[MemberQ[w, n - 1], u[s[n - 1]], d[s[n - 1]]]; max = Max[N[Table[s[n], {n, 1, 3000}], 200]] (* A245217 *) ContinuedFraction[max, 120] (* A245219 *) CROSSREFS Cf. A226080 (infinite Fibonacci tree), A245217, A245218, A245222, A245225. The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A003151 as the parent: A003151, A001951, A001952, A003152, A006337, A080763, A082844 (conjectured), A097509, A159684, A188037, A245219 (conjectured), A276862. - N. J. A. Sloane, Mar 09 2021 Sequence in context: A082844 A279124 A101406 * A097509 A095206 A344129 Adjacent sequences:  A245216 A245217 A245218 * A245220 A245221 A245222 KEYWORD nonn,cofr,easy AUTHOR Clark Kimberling, Jul 13 2014 STATUS approved

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Last modified August 8 23:57 EDT 2022. Contains 356016 sequences. (Running on oeis4.)