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A245219
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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.
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16
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3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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].
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MATHEMATICA
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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 *)
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CROSSREFS
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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
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KEYWORD
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nonn,cofr,easy
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AUTHOR
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STATUS
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approved
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