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A224994
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Least odd number d such that the Collatz (3x+1) iteration of d has the following property: if the length of the iteration is b and the maximum value occurs at c, the ratio c/b is 1/n.
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1
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1, 3, 5, 21, 13, 53, 67, 141, 93, 61, 37, 149, 101, 65, 261, 173, 693, 461, 305, 209, 813, 541, 2165, 1445, 961, 657, 2315, 1709, 1169, 4557, 3037, 2021, 3659, 5389, 3589, 2413, 5123, 2291, 4253, 2755, 7235, 7557, 5037, 3357, 10123, 8949, 5965, 3973, 15893
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OFFSET
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1,2
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COMMENTS
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This sequence is the first column of the irregular triangle shown in A224537.
For 45 < n <= 300, the length the Collatz sequence of a(n) is 2n. - T. D. Noe, May 01 2013
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LINKS
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EXAMPLE
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The Collatz iteration of 5 is {5, 16, 8, 4, 2, 1}, which has length 6. The maximum occurs at the second position. Note that 2/6 = 1/3. No number less than 5 has this property. Hence a(3) = 5.
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MATHEMATICA
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Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 50; t = Table[0, {nn}]; t[[1]] = 1; n = 1; While[Times @@ t == 0, n = n + 2; c = Collatz[n]; frac = Position[c, Max[c]][[1, 1]]/Length[c]; numer = Numerator[frac]; denom = Denominator[frac]; If[numer == 1 && denom <= nn && t[[denom]] == 0, t[[denom]] = n]]; t
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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